Analysis

W. Ted Mahavier, Fall 2011 - Spring 2012, Lamar University

Video Diary

There are certain combinations of operating systems and browsers where the mathematics does not display properly.  It does appear to behave properly under Chrome v24, IE9 and Firefox v16.

You may wish to visit the additional resources before viewing videos in order to peruse the course notes, syllabus, blog and the other resources available there. Specifically, the introduction to the course notes informs the students how the course will be taught and the book mentioned there gives details on all aspects of the Moore Method -- developing notes, grading, class culture, frequently-asked-questions and more. If you want to skip straight to the videos, you can search this page for "commentary," as each video clip is incorporated into the commentary that follows the diary entry for the days for which I chose to include the video.

A Caveat


A colleague of mine, Charles Coppin, once told me that he would not allow faculty to visit his classes, unless they were willing to sit in on the semester. His reasoning is pertinent. There is some danger in putting individual clips on the web, because the student growth that occurs in a Moore Method class is not visible on any one day. Students are not jumping up and down or collaborating at the board excitedly. Rather, they have been working hard at home and are struggling with deep mathematical concepts that they are forced to independently discover and deliver. Such progress is painfully slow as any research mathematician knows. Would viewing my work in my office on any given day demonstrate the excitement that is felt when a nice result comes to fruition? Hardly. Thus, what you will see are students (and me) struggling to make progress forward through the notes. What I have seen over the many years that I have done this and what students have reported back to me, is that they grow significantly in such classes. They learn to struggle with difficult problems without help. They learn to write mathematics. They learn that they are capable of doing mathematics. They develop a deep intuition that, according to those who have gone on to PhD programs, have enabled them to excel in those programs. It is not clear to me that this will come out in any one video, or even if it will be evident to the devoted student who carefully follows the clips from the first to the last. Still, I have done my best.

The Students

Here are most of the students in the class, labeled from left to right by row.

Row 1. Me (invisible), Katie, Jessica (well, half of Jessica)

Row 2. Weston, Shaymal, Barron

Row 3. Michael, Anthony

Row 4. Rabbi (behind Michael), Chris, Kimberly, Clint G (behind Barron).

Row 5. Milagro, Aaron, Clint W., John (behind Kimberly)

Window. Brandy, the devoted videographer!


Missing from this picture are Amber (you'll see her in the first video clip as the first presenter), Lucas (now pursuing his Ph.D. at N.C. State, you'll see him in the clips sitting beside me), Jacob (seated beside Michael and out of the shot), Han and Dexter.

Videos


My goal is to illustrate for you the Moore Method by choosing a handful of select days from the two-semester course. I have chosen classes that are respresentative of what occurred on a daily basis in the class. What follows is the diary that I kept during the two semesters, supplemented by commentary that I added, typically wherever I included a video clip. The commentary expands on the diary entry or tells what you might observe by viewing that particular clip. Placing your mouse over any problem should show you the statement of that problem.

Diary 8.22.11, the first day of class

I probably spent too much time at the board because of permission forms and me talking too much, but in thirty minutes I covered the definitions of open interval OPEN INTERVAL
The statement that the point set \(\begin{equation}O\end{equation}\) is an open interval means that there are two points \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\) such that \(\begin{equation}O\end{equation}\) is the set consisting of all points between \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\).
, closed interval CLOSED INTERVAL
The statement that the point set \(\begin{equation}I\end{equation}\) is a closed interval means that there are two points \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\) such that \(\begin{equation}I\end{equation}\) is the set consisting of the points \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\) and all points between \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\).
, Axiom 1 AXIOM 1
\(\begin{equation}\mathbb{R}\end{equation}\) is linearly ordered.
, Axiom 2 AXIOM 2
If \(\begin{equation}p\end{equation}\) is a point, then there is a point less than \(\begin{equation}p\end{equation}\) and a point greater than \(\begin{equation}p\end{equation}\).
, Axiom 3 AXIOM 3
If \(\begin{equation}p\end{equation}\) and \(\begin{equation}q\end{equation}\) are two points then there is a point between them, for example, \(\begin{equation}(p+q)/2\end{equation}\).
, and the definition of a limit point LIMIT POINT
If \(\begin{equation}M\end{equation}\) is a point set and \(\begin{equation}p\end{equation}\) is a point, the statement that \(\begin{equation}p\end{equation}\) is a limit point of the point set \(\begin{equation}M\end{equation}\) means that every open interval containing \(\begin{equation}p\end{equation}\) contains a point of \(\begin{equation}M\end{equation}\) different from \(\begin{equation}p\end{equation}\).
. Barron is energetic and it will take work to channel that energy into productive statements. Amber showed \(\begin{equation}p=4\end{equation}\) is not lp of \(\begin{equation}M=\{3\}\end{equation}\). Chris showed \(\begin{equation}p=3\end{equation}\) is not a lp of \(\begin{equation}M=\{3\}\end{equation}\). Clint W. showed that if \(\begin{equation}p > 3\end{equation}\) then \(\begin{equation}p\end{equation}\) not lp of \(\begin{equation}M=\{3\}\end{equation}\) with help from the class. I left them with three questions, not in the notes, for Wednesday. Q1: Is \(\begin{equation}p = 2\end{equation}\) lp of \(\begin{equation}M=[0,1)\end{equation}\)? Q2: Is \(\begin{equation}p = 1\end{equation}\) lp of \(\begin{equation}M=[0,1)\end{equation}\)? Q3: What are the limit points of \(\begin{equation}M=[0,1)\end{equation}\)?

Commentary 8.22.11, the first day of class

In my experience, the first day of an IBL course is crucial to its success. Often I am asked how I get students to the board and this clip is the answer -- I put them at ease on the first day. While students appear to be coming in late, the class officially starts at 9:05 and I started talking at 9:00, as is my habit. I learn a lot about students in the five minutes before and after each class. You’ll see that I attempt to put them at ease, learn the few names I do not know, and pass out permission slips and a survey. Perhaps nothing is more important than learning all the names on the first day. We begin with very low level mathematics because I want them all to have the opportunity to speak and contribute. We discuss some properties of the real numbers and some notation, including Axiom 1 AXIOM 1
\(\begin{equation}\mathbb{R}\end{equation}\) is linearly ordered.
, Axiom 2 AXIOM 2
If \(\begin{equation}p\end{equation}\) is a point, then there is a point less than \(\begin{equation}p\end{equation}\) and a point greater than \(\begin{equation}p\end{equation}\).
, and Axiom 3 AXIOM 3
If \(\begin{equation}p\end{equation}\) and \(\begin{equation}q\end{equation}\) are two points then there is a point between them, for example, \(\begin{equation}(p+q)/2\end{equation}\).
. Because a sudent suggests that \(\begin{equation}(p+q)/3\end{equation}\) is also between \(\begin{equation}p\end{equation}\) and \(\begin{equation}q\end{equation}\), we investigate that and introduce the idea of a counterexample. This is not something I planned to discuss, or needed to, but once a student contributes I always show respect by addressing the mathematics that they have suggested. Then I define limit point LIMIT POINT
If \(\begin{equation}M\end{equation}\) is a point set and \(\begin{equation}p\end{equation}\) is a point, the statement that \(\begin{equation}p\end{equation}\) is a limit point of the point set \(\begin{equation}M\end{equation}\) means that every open interval containing \(\begin{equation}p\end{equation}\) contains a point of \(\begin{equation}M\end{equation}\) different from \(\begin{equation}p\end{equation}\).
and ask if \(\begin{equation}p=4\end{equation}\) is a limit point of \(\begin{equation}M=\{3\}\end{equation}\). Because I already suspect that Lucas will be able to do many of the early problems, having had him in previous classes, I subtly ignore his correct statements and follow other students' statements. The key moment in the entire clip occurs around thirty seven minutes into the clip when after some discussion Amber demonstrates understanding, I toss her the marker, and I know that this class is off to a good start. My pedagogical goals for the first day are:

1. to get multiple students to the board,
2. to introduce the axioms and the definition of limit point,
3. to emphasize the importance of defending statements using axioms and definitions,
4. to start training them in the parsing of definitions (English!), and
5. to leave them with problems that a majority of students have a high probability of solving to bring back and present
    during the next class period.

Even though the mathematics is very low level, I would call this a successful day. My mathematical goal for the first day, albeit one I don't always reach, is to investigate whether \(\begin{equation}0\end{equation}\) is a limit point of the set of reciprocals of the natural numbers. I conclude the first day by telling them a bit more about the class -- that a problem is due every Friday. My syllabus is on the web site as are the other mundane details, allowing the entire class period to be devoted to mathematics. One aspect of my teaching that has changed since this clip was filmed, was that I now have conceded the value of the smart phones and they are welcome to use them in class to take their own pictures or look at the notes.

Diary 8.24.11

I collected permission forms, collected student interview questions, reiterated working individually, and mandated coming to see me to talk about personal goals. This is a hook to get them to see me about the mathematics. I offered to answer any questions or go over anything we did in class yesterday, and then asked for solutions. Kimberly resolved Q1 Q1
Is \(\begin{equation}p = 2\end{equation}\) lp of \(\begin{equation}M=[0,1)\end{equation}\)?
. Rabbi attempted Q2 Q2
Is \(\begin{equation}p = 1\end{equation}\) lp of \(\begin{equation}M=[0,1)\end{equation}\)?
, but claimed that \(\begin{equation}p=1\end{equation}\) was not a lp of \(\begin{equation}M\end{equation}\). Clint G. resolved Q2 Q2
Is \(\begin{equation}p = 1\end{equation}\) lp of \(\begin{equation}M=[0,1)\end{equation}\)?
. Amber solved Q3 Q3
What are the limit points of \(\begin{equation}M=[0,1)\end{equation}\)?
and showed that \(\begin{equation}x>1\end{equation}\) was not a lp of \(\begin{equation}[0,1)\end{equation}\) and \(\begin{equation}x < 0\end{equation}\) was not lp of \(\begin{equation}[0,1)\end{equation}\) and was ready to start that \(\begin{equation}[0,1]\end{equation}\) are all lps of \(\begin{equation}[0,1)\end{equation}\), but I let Barron start P1 P1
Show that if \(\begin{equation}M\end{equation}\) is the open interval \(\begin{equation}(a,b)\end{equation}\), and \(\begin{equation}p\end{equation}\) is in \(\begin{equation}M\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\).
in order to get one more student to the board. After stating the hypothesis well, he did not select an open interval containing \(\begin{equation}p\end{equation}\). Here is where the art of the Method is important. Amber is very likely to have success, having started off well. Barron had a rocky start. Many have not been to the board yet. Thus, even though I knew Amber could have completed Q3 Q3
What are the limit points of \(\begin{equation}M=[0,1)\end{equation}\)?
with input from the class at the board, I let her do two parts and then stopped and said we'd look at the third part on Friday, without assuring that she would get it. When Barron got stuck, I told him I expected that he would be able to complete it on Friday. Even though he had a rough start, this expectation caused him to work hard on it and come see me on Thursday. Thus I expect that on Friday I'll get one new person to the board on Q3 Q3
What are the limit points of \(\begin{equation}M=[0,1)\end{equation}\)?
and will see Barron have a success. Since Q3 Q3
What are the limit points of \(\begin{equation}M=[0,1)\end{equation}\)?
and P1 P1
Show that if \(\begin{equation}M\end{equation}\) is the open interval \(\begin{equation}(a,b)\end{equation}\), and \(\begin{equation}p\end{equation}\) is in \(\begin{equation}M\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\).
hinge around the same concept of choosing an arbitrary interval around the limit point, if we see Q3 Q3
What are the limit points of \(\begin{equation}M=[0,1)\end{equation}\)?
first, it will increase the odds that Barron will get P1 P1
Show that if \(\begin{equation}M\end{equation}\) is the open interval \(\begin{equation}(a,b)\end{equation}\), and \(\begin{equation}p\end{equation}\) is in \(\begin{equation}M\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\).
right or that the class will be able to help.

Diary 8.26.11

First I picked up problems. Several students were late and since homework was due at the beginning of class, they'll lose a letter grade. I'll point it out on Monday when I return them. Amber was the only student ready to complete Q2 Q2
Is \(\begin{equation}p = 1\end{equation}\) lp of \(\begin{equation}M=[0,1)\end{equation}\)?
and she did not consider an arbitrary interval, but did consider two possible intervals. I am going to consider it resolved. Barron attempted P1 P1
Show that if \(\begin{equation}M\end{equation}\) is the open interval \(\begin{equation}(a,b)\end{equation}\), and \(\begin{equation}p\end{equation}\) is in \(\begin{equation}M\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\).
again and after twenty minutes of writing on the board, he had filled three boards and there was a lot of redundancy. We ran out of time before he had handled the first case, so I went over it with him at the board. Not sure if it helped him or the class.

Diary 8.28.11

Handed back papers. Pointed out that the late ones lost a letter grade because they were due at the beginning of class, not five or ten minutes later. Then I suggested that we did not need to write everything on the board, if we could just get an argument out that the class agreed upon. With some hesitation, I asked Barron to return to the board for a third try at P1 P1
Show that if \(\begin{equation}M\end{equation}\) is the open interval \(\begin{equation}(a,b)\end{equation}\), and \(\begin{equation}p\end{equation}\) is in \(\begin{equation}M\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\).
. Perhaps because of my statement, he simply drew pictures and made valid argument, although there were still missing cases even after the class concluded that there weren't. The seed of the proof was fine and we'll worry about details on the weekly written work. Barron’s off to a rough start, but he’s strong and proud and I'm optimistic that his hard work will pay off. I borrowed from Dad’s toolbox of immediately questioning false statements that were made in class, even as I made a few myself! Then Lucas showed P2 P2
Show that if \(\begin{equation}M\end{equation}\) is the closed interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}p\end{equation}\) is not in \(\begin{equation}M\end{equation}\), then \(\begin{equation}p\end{equation}\) is not a limit point of \(\begin{equation}M\end{equation}\).
. A good example since he started with a specific example, but corrected himself at the board when challenged. Milagro resolved P3 P3
Show that if \(\begin{equation}M\end{equation}\) is a point set having a limit point, then \(\begin{equation}M\end{equation}\) contains 2 points. Must \(\begin{equation}M\end{equation}\) contain 3 points? 4 points?
with help from me and class. The class helped him and I believe that many in the class understood it, but I wasn't sure Milagro did. Sure enough, he came by my office Friday before class, still struggling.

Diary 8.30.11

Jessica did P4 P4
Show that if \(\begin{equation}M\end{equation}\) is the set of all positive integers, then no point is a limit point of \(\begin{equation}M\end{equation}\).
efficiently and Lucas answered P5 P5
Assume \(\begin{equation}M\end{equation}\) is a point set and \(\begin{equation}p\end{equation}\) is a point of \(\begin{equation}M\end{equation}\). Create a definition for ``\(\begin{equation}q\end{equation}\) is the first point to the left of \(\begin{equation}p\end{equation}\) in \(\begin{equation}M\end{equation}\)" by completing the following. ``If \(\begin{equation}M\end{equation}\) is a point set and \(\begin{equation}p\end{equation}\) is a point in \(\begin{equation}M\end{equation}\)..."
efficiently. They seem to have caught on more quickly than previous classes to the notions of first point to the right FIRST POINT
The statement that \(\begin{equation}p\end{equation}\) is the first point to the right of the point set \(\begin{equation}M\end{equation}\) means that \(\begin{equation}p\end{equation}\) is greater than every point of \(\begin{equation}M\end{equation}\) and if \(\begin{equation}q\end{equation}\) is a point less than \(\begin{equation}p\end{equation}\), then \(\begin{equation}q\end{equation}\) is not greater than every point of \(\begin{equation}M\end{equation}\).
and right most point. RIGHT MOST POINT
The statement that \(\begin{equation}p\end{equation}\) is the right-most point of M means that \(\begin{equation}p\end{equation}\) is in \(\begin{equation}M\end{equation}\) and no point of \(\begin{equation}M\end{equation}\) is greater than \(\begin{equation}p\end{equation}\).
There were no solutions to P6 P6
Assume \(\begin{equation}M\end{equation}\) is a point set such that if \(\begin{equation}p\end{equation}\) is a point of \(\begin{equation}M\end{equation}\), there is a first point to the left of \(\begin{equation}p\end{equation}\) in \(\begin{equation}M\end{equation}\) and a first point to the right of \(\begin{equation}p\end{equation}\) in \(\begin{equation}M\end{equation}\). Is it true that \(\begin{equation}M\end{equation}\) cannot have a limit point?
and then Weston attempted P7 P7
Show that if \(\begin{equation}H\end{equation}\) is a point set and \(\begin{equation}K\end{equation}\) is a point set and \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H \cap K\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H\end{equation}\) and \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\).
, but tried to give a counterexample. In my discrete mathematics course, which he took, I occasionally placed false statements in the notes intentionally as learning tools, but I don't do that in Analysis and I told them so after he understood why his example was not a counterexample to the statement.

Commentary 8.30.11

In developing Moore Method notes, it is very important to begin with very elementary problems so that students develop an understanding of the concepts while having success at the board. It is equally important to have some challenging problems, such as P6 P6
Assume \(\begin{equation}M\end{equation}\) is a point set such that if \(\begin{equation}p\end{equation}\) is a point of \(\begin{equation}M\end{equation}\), there is a first point to the left of \(\begin{equation}p\end{equation}\) in \(\begin{equation}M\end{equation}\) and a first point to the right of \(\begin{equation}p\end{equation}\) in \(\begin{equation}M\end{equation}\). Is it true that \(\begin{equation}M\end{equation}\) cannot have a limit point?
, early in the notes so that the student who is capable of doing the easy ones is challenged. This keeps that student from tuning out and it is often the case that a bright student will go to the board with an attempt that is incorrect. Many of my brightest students have never been challenged and are not used to making a mistake. This is a valuable lesson and gets them even more interested in the problems once they know that some problems are worthy of their intellect.

Diary 9.2.11

Weston attempted P7 P7
Show that if \(\begin{equation}H\end{equation}\) is a point set and \(\begin{equation}K\end{equation}\) is a point set and \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H \cap K\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H\end{equation}\) and \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\).
again, and was only handling the possible cases where \(\begin{equation}H\end{equation}\) and \(\begin{equation}K\end{equation}\) were open intervals, closed intervals or half-open intervals. Interestingly, this was the exact same mistake made at Emory University by DeMitchell in the Foundations Seminar that my father taught and filmed there. Amber attempted a counter example, but her choices for \(\begin{equation}H\end{equation}\) and \(\begin{equation}K\end{equation}\) were such that \(\begin{equation}H \cap K\end{equation}\) had no limit points. Shaymal attempted P10 P10
Show that if \(\begin{equation}M\end{equation}\) is the set of all reciprocals of positive integers, then \(\begin{equation}0\end{equation}\) (zero) is a limit point of \(\begin{equation}M\end{equation}\).
, but appeared to believe that the half open interval \(\begin{equation}(0,1]\end{equation}\) was equal to the set of all reciprocals of natural numbers?! Still, he had a good start and I helped him state a lemma that will resolve the problem if he can prove the lemma.

Diary 9.5.11

Labor Day, no class.

Diary 9.7.11

Aaron attempted P6 P6
Assume \(\begin{equation}M\end{equation}\) is a point set such that if \(\begin{equation}p\end{equation}\) is a point of \(\begin{equation}M\end{equation}\), there is a first point to the left of \(\begin{equation}p\end{equation}\) in \(\begin{equation}M\end{equation}\) and a first point to the right of \(\begin{equation}p\end{equation}\) in \(\begin{equation}M\end{equation}\). Is it true that \(\begin{equation}M\end{equation}\) cannot have a limit point?
, showing only the special case that no point in \(\begin{equation}M\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\). He must still consider the possibility that there is no limit point of \(\begin{equation}M\end{equation}\) outside of \(\begin{equation}M\end{equation}\). Barron attempted P15 P15
Show that if \(\begin{equation}p \ne 0\end{equation}\), then \(\begin{equation}p\end{equation}\) is not a limit point of the set \(\begin{equation}\{ 1 , {1 \over 2} , {1 \over 3} , \dots \}\end{equation}\).
, to show that the set of reciprocals has no limit point other than 0. Because of the write-ups that were turned in, I showed another proof of P1 P1
Show that if \(\begin{equation}M\end{equation}\) is the open interval \(\begin{equation}(a,b)\end{equation}\), and \(\begin{equation}p\end{equation}\) is in \(\begin{equation}M\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\).
to show how it can be done in two cases.

Diary 9.9.11

Today was a good day with two people at the board at one time, and another contributing from his desk. Aaron absent and no one had progress on P6 P6
Assume \(\begin{equation}M\end{equation}\) is a point set such that if \(\begin{equation}p\end{equation}\) is a point of \(\begin{equation}M\end{equation}\), there is a first point to the left of \(\begin{equation}p\end{equation}\) in \(\begin{equation}M\end{equation}\) and a first point to the right of \(\begin{equation}p\end{equation}\) in \(\begin{equation}M\end{equation}\). Is it true that \(\begin{equation}M\end{equation}\) cannot have a limit point?
. Weston worked on P7 P7
Show that if \(\begin{equation}H\end{equation}\) is a point set and \(\begin{equation}K\end{equation}\) is a point set and \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H \cap K\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H\end{equation}\) and \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\).
and with help from Amber, I feel we had a proof on the board. Shaymal attempted P10 P10
Show that if \(\begin{equation}M\end{equation}\) is the set of all reciprocals of positive integers, then \(\begin{equation}0\end{equation}\) (zero) is a limit point of \(\begin{equation}M\end{equation}\).
again but worked two examples and did not do the general case. Anthony, Amber and Lucas took turns attempting to resolve P10 P10
Show that if \(\begin{equation}M\end{equation}\) is the set of all reciprocals of positive integers, then \(\begin{equation}0\end{equation}\) (zero) is a limit point of \(\begin{equation}M\end{equation}\).
and by the end, I felt that the team had resolved it. This is an excellent example of collaboration during class, after people have thought about a problem at home. I concluded the class by using the last few minutes to talk about functions and sequences.

Commentary 9.9.11

This  clip demonstrates a productive class with multiple students engaged in the discussion. There are successes at the board and there are failures that I believe offered good learning opportunities for both those at the board and those at their seats. In particular, I would say there was optimal opportunity for learning when Shaymal demonstrates by example the validity of P10 P10
Show that if \(\begin{equation}M\end{equation}\) is the set of all reciprocals of positive integers, then \(\begin{equation}0\end{equation}\) (zero) is a limit point of \(\begin{equation}M\end{equation}\).
and then Amber, Anthony and Lucas further discuss the problem.

As an aside, many IBL instructors like to have students working on new material during classes, but my perspective is different. Allowing students to work on new material in class benefits primarily the quick and the bright. Allowing students to collaborate in class after the material has been digested and the problems attempted, as was done on this day, allows everyone to benefit maximally. I also don't believe the classroom is the place to do that which should have been done at home. If students do not work on the upcoming problems at home before class and can't participate, then that is their choice and it’s a poor one. Students need to know that they are responsible for coming to class prepared. Allowing in class collaboration on new material is often fun for many of the students, but I think it sacrifices too much in terms of putting the responsbility for learning on their shoulders and in terms of the depth of learning that can occur when the material is processed outside of class first.

Diary 9.12.11

Aaron attempts P6 P6
Assume \(\begin{equation}M\end{equation}\) is a point set such that if \(\begin{equation}p\end{equation}\) is a point of \(\begin{equation}M\end{equation}\), there is a first point to the left of \(\begin{equation}p\end{equation}\) in \(\begin{equation}M\end{equation}\) and a first point to the right of \(\begin{equation}p\end{equation}\) in \(\begin{equation}M\end{equation}\). Is it true that \(\begin{equation}M\end{equation}\) cannot have a limit point?
but attempts to show that \(\begin{equation}M\end{equation}\) can't have a lp. He handles two cases nicely -- \(\begin{equation}p \in M\end{equation}\) and \(\begin{equation}p\end{equation}\) between two consecutive elements of \(\begin{equation}M\end{equation}\). I point out how difficult research is and tell them that \(\begin{equation}M\end{equation}\) can have a limit point and I know from a comment Lucas made in class, which I tactfully ignored, that he has resolved this by recognizing that the reciprocals of the naturals (almost) have this property and have a limit point. Lucas then attempts P8 P8
Show that if \(\begin{equation}H\end{equation}\) is a point set and \(\begin{equation}K\end{equation}\) is a point set and every point of \(\begin{equation}H\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\) and \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\).
, but recognizes his flaw, that he does not know that his point of \(\begin{equation}K\end{equation}\) is distinct from his limit point. Finally, even though he has turned it in and knows he has a mistake, I ask Han to present P9 P9
If \(\begin{equation}H\end{equation}\) is a point set and \(\begin{equation}K\end{equation}\) is a point set and \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H \cup K\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H\end{equation}\) or \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\).
and he writes very well, making the traditional mistake which I then clarify for the class.

Diary 9.14.11

Lucas resolved P6 P6
Assume \(\begin{equation}M\end{equation}\) is a point set such that if \(\begin{equation}p\end{equation}\) is a point of \(\begin{equation}M\end{equation}\), there is a first point to the left of \(\begin{equation}p\end{equation}\) in \(\begin{equation}M\end{equation}\) and a first point to the right of \(\begin{equation}p\end{equation}\) in \(\begin{equation}M\end{equation}\). Is it true that \(\begin{equation}M\end{equation}\) cannot have a limit point?
by demonstrating the set \(\begin{equation}\{ 1-\frac{1}{n} : n =1,2,3,\dots \} \cup \{ -1+\frac{1}{n} : n =1,2,3,\dots \}\end{equation}\) which has the property that every point has a first point to the left in the set and a first point to the right in the set and has two limit points. Then he put back up what he had worked on for P8 P8
Show that if \(\begin{equation}H\end{equation}\) is a point set and \(\begin{equation}K\end{equation}\) is a point set and every point of \(\begin{equation}H\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\) and \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\).
and Amber helped him resolve it by creating the open interval he needed. Kim asked if I would discuss what was wrong with Han’s proof of P9 P9
If \(\begin{equation}H\end{equation}\) is a point set and \(\begin{equation}K\end{equation}\) is a point set and \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H \cup K\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H\end{equation}\) or \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\).
and I wanted to walk the class through his proof, so I asked, "What is a natural way to start this problem?" I expected, "let p be a limit point of \(\begin{equation}H \cup K\end{equation}\) and \(\begin{equation}(a,b)\end{equation}\) be an open interval containing \(\begin{equation}p\end{equation}\)." What happened was that Barron made the astute observation that the contra-positive was a natural way to solve it and so we worked as a class on it and I believe that it should be easily resolved tomorrow.

Diary 9.16.11

I started out before class started with a motivational speech in which I passed out some of my own research which showed where I wrote, "Need" and "Need To Show" and "So I don't need" and had doodles of sailboats when I was completely stuck. I also told them that next week they need to start turning in problems that they have not seen presented in class. They are slowly improving on the written work. Almost all grades were "C" last week, which means they are getting the hang of things. Today Barron attempted P9 P9
If \(\begin{equation}H\end{equation}\) is a point set and \(\begin{equation}K\end{equation}\) is a point set and \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H \cup K\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H\end{equation}\) or \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\).
, but tried to show a counter-example to show that it was not true. This led to good discussion of negation and of contra-positive and to a good discussion of how one might prove it either via contra-positive, as Barron suggested, or by assuming that \(\begin{equation}p\end{equation}\) was a limit point of \(\begin{equation}H \cup K\end{equation}\) and not a limit point of \(\begin{equation}H\end{equation}\) and showing that it must be a limit point of \(\begin{equation}K\end{equation}\), as Jessica suggested.

Commentary 9.19.11

One of my students later told me that they often came to class with pictures of sailboats and held them up to one another as a signal for "I don't have anything, do you?"

Diary 9.19.11

I discussed the definition of convergence CONVERGENCE
The statement that the point sequence \(\begin{equation}p_1,p_2,\dots\end{equation}\) converges to the point \(\begin{equation}x\end{equation}\) means that if \(\begin{equation}S\end{equation}\) is an open interval containing \(\begin{equation}x\end{equation}\) then there is a positive integer \(\begin{equation}N\end{equation}\) such that if \(\begin{equation}n\end{equation}\) is a positive integer and \(\begin{equation}n \ge N\end{equation}\) then \(\begin{equation}p_n \in S\end{equation}\).
in response to a question from Barron about P15 P15
Show that if \(\begin{equation}p \ne 0\end{equation}\), then \(\begin{equation}p\end{equation}\) is not a limit point of the set \(\begin{equation}\{ 1 , {1 \over 2} , {1 \over 3} , \dots \}\end{equation}\).
. No one tried P9 P9
If \(\begin{equation}H\end{equation}\) is a point set and \(\begin{equation}K\end{equation}\) is a point set and \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H \cup K\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H\end{equation}\) or \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\).
, but Shaymal has a solution to P11 P11
For each positive integer \(\begin{equation}n\end{equation}\), let \(\begin{equation}p_n= 1 - 1/n\end{equation}\). Show that the sequence \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) converges to \(\begin{equation}1\end{equation}\).
which was very nice for our first proof about convergence. Because the class seemed unable to answer my question as to what Shaymal needed to show, I interrupted the presentation and gave a specific interval and showed how to find the specific value of \(\begin{equation}N\end{equation}\) for which \(\begin{equation}p_N\end{equation}\) was in the interval. Shaymal had come to my office to ask a question about it and once he understood the idea of an arbitrary interval and the changing of \(\begin{equation}N\end{equation}\) based on that interval, he was off! Success!

Diary 9.21.11

No one had anything and I explained that their choices were either to come to class prepared to present or to have me assign problems to individuals to be presented on specific dates. Kim then said that she had something on P14 P14
Show that if the sequence \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) converges to the point \(\begin{equation}x\end{equation}\), and, for each positive integer \(\begin{equation}n\end{equation}\), \(\begin{equation}p_n \ne p_{n+1}\end{equation}\), then \(\begin{equation}x\end{equation}\) is a limit point of the set which is the range of the sequence.
but was not confident that it was correct. As a class, we worked successfully through a proof of it. She had a very good start and with help from Amber, Barron, Weston, Jessica, Lucas, and me was able to get a proof on the board.

Commentary 9.21.11

Some days are diamonds, some days are stones... It seems worthwhile to show how slowly we may progress on occasion, and this clip is best viewed in conjunction with the one for the next class period as the two show together how one slow day may precede one very productive day. This clip shows a day that a visitor to my class might consider a "bad" day. We spent almost the entire class on one problem, with much of the time simply working on an undersatnding of precise notation associated with sequences. Considerable time is spent on understanding how to write proofs, making sure that the objects we use in a line have been defined either in that line or in previous lines. While this period appears wasted, with only one solution, it creates a flurry of successes in the next few class meetings.

This also demonstrates that even when no member of the class volunteers, it is often the case that someone has a good start on something, but is afraid to go to the board. Often I simply ask incredulously, "you mean nobody has anything? Not even a good start?" Then I go name by name looking at each and asking which one they are working on and where they are stuck. This serves two purposes: it tells me which ones are thinking hard enough about a problem to have it well in mind, and it identifies those who look down at their paper because they are not working. Typically, at some point in this process I can say, "that sounds like a good start, will you please write it on the board so that we can look at it as a class?" Then we have something to work on and it is usually something that a majority of the students are struggling with. Other times, if I fear they are not working, I'll issue the statement I made in my diary above and discuss how much harder it is to be assigned a random problem than to be able to work on several until you get one. Another strategy that I have used on rare occassion, is to tell the class that I would come in each day and call on individual students by name to show what they had. This forced every student to attempt to have something on every day. But I prefer the volunteer method, where they know they must volunteer to get a good grade, but the pressure is not daily.

Diary 9.23.11

I can finally say we had a great day! John made a good attempt at P19 P19
Show that if \(\begin{equation}M\end{equation}\) is a point set and there is a point \(\begin{equation}p\end{equation}\) which is the first point to the right of \(\begin{equation}M\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\).
. Jessica P18 P18
Show that if \(\begin{equation}M\end{equation}\) is a point set, then there cannot be both a right-most point of \(\begin{equation}M\end{equation}\) and a first point to the right of \(\begin{equation}M.\end{equation}\)
. Han P13 P13
For each positive integer \(\begin{equation}n\end{equation}\),
let \(\begin{equation}p_{2n} = 1/(2n-1)\end{equation}\), and
let \(\begin{equation}p_{2n-1}= 1/2n\end{equation}\). Show that the
sequence \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) converges to \(\begin{equation}0\end{equation}\).
. Perhaps it is the response to my wake-up call on Wednesday, threatening to assign problems. Perhaps it is the fact that write-ups now must cover material that has not been presented, which tends to get them working ahead. Either way, there was good class discussion because people had read and thought ahead. It makes me wonder if I should change my policy of allowing write-ups to be material that has been presented during the first few weeks and make them turn in something that has not been presented each week from the first week. Sadly, today I lost one student in my Calculus class, who I was counting on pulling her grade up and doing well. I've not managed that class as well to date. Will continue to try today.

Commentary 9.23.11

Although the camera work in this clip starts out disastrously, we can see good progress and excellent class discussion. I continue my motivational talks to keep them working, explaining that these problems are not going to all be solved the first time they sit down and attempt them. We are now one month into the course and we see significant progress on multiple problems. John's difficulty with P19 P19
Show that if \(\begin{equation}M\end{equation}\) is a point set and there is a point \(\begin{equation}p\end{equation}\) which is the first point to the right of \(\begin{equation}M\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\).
shows just how much difficulty students have in reading and processing precise mathematical statements. This is a necessary skill in upper-division lecture courses and courses based on books. Thus even though he does not have it right, we are developing this skill and I can see on the final exams each year, just how much this skill has developed in my students. Both this and the previous clip demonstrate that my students are comfortable at the board, regardless of whether they have a full solution or not, as mistakes are viewed as precursors to success, not as failures. This also shows why we define right most point RIGHT MOST POINT
The statement that \(\begin{equation}p\end{equation}\) is the right-most point of M means that \(\begin{equation}p\end{equation}\) is in \(\begin{equation}M\end{equation}\) and no point of \(\begin{equation}M\end{equation}\) is greater than \(\begin{equation}p\end{equation}\).
and first point to the right FIRST POINT
The statement that \(\begin{equation}p\end{equation}\) is the first point to the right of the point set \(\begin{equation}M\end{equation}\) means that \(\begin{equation}p\end{equation}\) is greater than every point of \(\begin{equation}M\end{equation}\) and if \(\begin{equation}q\end{equation}\) is a point less than \(\begin{equation}p\end{equation}\), then \(\begin{equation}q\end{equation}\) is not greater than every point of \(\begin{equation}M\end{equation}\).
rather than least upper bound.

My analysis classes almost always feel painfully slow to me at the beginning, but this is a concession I am willing to make to assure that everyone has ample chance to participate early on. And it allows them to get a handle on how I conduct class. One thing I do at the beginning is to allow the weekly write-ups, which constitute one-third of their grades, to be any problem that has been presented. A few weeks into the course, as they start to produce well-written arguments, I request that they only submit write-ups of work that has not been presented. This forces them to constantly work ahead in the notes and accelerates the class considerably. In this clip you can also see how I draw out one student who has not presented yet and how I yield authority to the student at the board by telling him he may present in any fashion he likes. In fact, this argument for P19 P19
Show that if \(\begin{equation}M\end{equation}\) is a point set and there is a point \(\begin{equation}p\end{equation}\) which is the first point to the right of \(\begin{equation}M\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\).
becomes a class effort. Often when a student is presenting for the first time, I allow such collaboration. If a student is more confident or has presented multiple times and there is a mistake, then I'll allow the students to make a decision: fix the problem at the board, yield the problem to another student, or hold the problem until the next class period. Hence, I become more strict in my policy of presenting individually and without help from the class as the student's confidence and number of presentations grow.

Diary 9.25.11

All day went to a very nice presentation of P13 P13
For each positive integer \(\begin{equation}n\end{equation}\),
let \(\begin{equation}p_{2n} = 1/(2n-1)\end{equation}\), and
let \(\begin{equation}p_{2n-1}= 1/2n\end{equation}\). Show that the
sequence \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) converges to \(\begin{equation}0\end{equation}\).
by Han. He definitely had good intuition and much of the algebra handled, but did not have a complete proof when he went to the board. With just a tad of guidance, I think he has a much better understanding and can write a nice proof of it. And I think the class got their arms wrapped around convergence today, so I expect good success for the next few classes.

Diary 9.28.11

I gently chastised the members of the class who have not presented. And I've noted on my web site that those who have been to see me in my office are having success. I complimented Han’s nice work on P13 P13
For each positive integer \(\begin{equation}n\end{equation}\),
let \(\begin{equation}p_{2n} = 1/(2n-1)\end{equation}\), and
let \(\begin{equation}p_{2n-1}= 1/2n\end{equation}\). Show that the
sequence \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) converges to \(\begin{equation}0\end{equation}\).
and addressed a minor question about it from Chris. Rabbi then put up a very nice start to P9 P9
If \(\begin{equation}H\end{equation}\) is a point set and \(\begin{equation}K\end{equation}\) is a point set and \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H \cup K\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}H\end{equation}\) or \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\).
. Jessica made a very astute observation, pointing out a flaw in his argument. He was so close, that I made a few summary statements (leading too much?) and then Amber broke the back of it. Amber then sketched her argument for P12 P12
For each positive integer \(\begin{equation}n\end{equation}\),
let \(\begin{equation}p_{2n-1}=1/(2n-1)\end{equation}\)
and let \(\begin{equation}p_{2n}=1+1/2n\end{equation}\).
Does the sequence \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\)
converge to \(\begin{equation}0\end{equation}\)?
, but it was rushed so I wrote a summary and a statement of what she must conclude on Friday in order to complete the problem.

Diary 9.30.11

Amber completed P12 P12
For each positive integer \(\begin{equation}n\end{equation}\),
let \(\begin{equation}p_{2n-1}=1/(2n-1)\end{equation}\)
and let \(\begin{equation}p_{2n}=1+1/2n\end{equation}\).
Does the sequence \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\)
converge to \(\begin{equation}0\end{equation}\)?
, although it was a rough proof and not quite complete because she had failed to show that for any natural number \(\begin{equation}N\end{equation}\) there was an \(\begin{equation}m>N\end{equation}\) so that \(\begin{equation}p_m\end{equation}\) was not in her chosen \(\begin{equation}(a,b)\end{equation}\). Still a good attempt. As is typical of bright students who can often make large intuitive leaps, Jacob has good intuition but does not believe he needs to write things carefully. When he tries, he often makes logical flaws in his writing and thus does not have a proof. We'll need to start with his P15 P15
Show that if \(\begin{equation}p \ne 0\end{equation}\), then \(\begin{equation}p\end{equation}\) is not a limit point of the set \(\begin{equation}\{ 1 , {1 \over 2} , {1 \over 3} , \dots \}\end{equation}\).
on Monday.

Diary 10.3.11

Barron asked a question about the definition of open set OPEN SET
The statement that the point set \(\begin{equation}M\end{equation}\) is an open point set means that for every point \(\begin{equation}p\end{equation}\) of \(\begin{equation}M\end{equation}\) there is an open interval which contains \(\begin{equation}p\end{equation}\) and is a subset of \(\begin{equation}M\end{equation}\).
and continues to struggle with the writing, attempting to write completely in symbols rather than English. This adds a layer of difficulty that is hindering him, but I respect his perseverance. Not properly defining the variables is a common problem in all students’ proofs. There are many times when a set or point is not defined and their write-ups become confused as a result. I then asked Jessica to put up what she had done on T2 T2
If \(\begin{equation}M\end{equation}\) is a finite point set, then \(\begin{equation}M\end{equation}\) has a right-most point and a left-most point.
. To prove this she was assuming that a finite set could be ordered which is equivalent. Even though trained in induction in the previous class, I never have seen a student recognize that induction would solve this problem. Then I discussed T2 T2
If \(\begin{equation}M\end{equation}\) is a finite point set, then \(\begin{equation}M\end{equation}\) has a right-most point and a left-most point.
, open and closed sets, and continuity. A much better discussion about continuity occurred after class!

Diary 10.5.11

Jessica made another attempt at T2 T2
If \(\begin{equation}M\end{equation}\) is a finite point set, then \(\begin{equation}M\end{equation}\) has a right-most point and a left-most point.
, but failed to understand the necessary inductive step. She believed that simply defining \(\begin{equation}k!\end{equation}\) was sufficient to showing that there were in fact \(\begin{equation}k!\end{equation}\) orderings of a finite set of \(\begin{equation}k\end{equation}\) elements. I attempted to clarify the problem. Then Lucas gave a nice argument for P16 P16
Show that if \(\begin{equation}c\end{equation}\) is a number and \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a sequence which converges to the point \(\begin{equation}x\end{equation}\), then the sequence \(\begin{equation}c \cdot p_1,c \cdot p_2,c \cdot p_3,\dots\end{equation}\)
converges to \(\begin{equation}c \cdot x\end{equation}\).
He recognized that it is not self-evident that if \(\begin{equation}(a,b)\end{equation}\) contains \(\begin{equation}x\end{equation}\) then \(\begin{equation}(ca, cb)\end{equation}\) represents an arbitrary interval containing \(\begin{equation}cx\end{equation}\), but he also proved a lemma, which he presented to me in my office, showing that every open interval containing \(\begin{equation}cx\end{equation}\) could be represented as \(\begin{equation}(ca,cb)\end{equation}\) for some \(\begin{equation}(a,b)\end{equation}\) containing \(\begin{equation}x\end{equation}\). I showed how one could start with an open interval about \(\begin{equation}cx\end{equation}\) and create one around \(\begin{equation}x\end{equation}\) in the standard way.

Diary 10.7.11

Today Barron informed me that he has been moved from nights to days at the post office, so will no longer be able to attend classes. I gave a lecture on continuity CONTINUITY
The statement that the function \(\begin{equation}f\end{equation}\) is continuous at the point \(\begin{equation}p=(x,y)\end{equation}\) means that:

1. \(\begin{equation}p\end{equation}\) is a point on \(\begin{equation}f\end{equation}\), and

2. if \(\begin{equation}H\end{equation}\) and \(\begin{equation}K\end{equation}\) are any two horizontal lines with \(\begin{equation}p\end{equation}\) between them, then there are two vertical lines, \(\begin{equation}h\end{equation}\) and \(\begin{equation}k\end{equation}\) with \(\begin{equation}p\end{equation}\) between them so that if \(\begin{equation}t\end{equation}\) is any point in the domain of \(\begin{equation}f\end{equation}\) between \(\begin{equation}h\end{equation}\) and \(\begin{equation}k,\end{equation}\) then \(\begin{equation}(t,f(t))\end{equation}\) is in the rectangle bounded by \(\begin{equation}h, k, H,\end{equation}\) and \(\begin{equation}K.\end{equation}\)
. Jessica gave a very nice proof to T2 T2
If \(\begin{equation}M\end{equation}\) is a finite point set, then \(\begin{equation}M\end{equation}\) has a right-most point and a left-most point.
, after some guidance in my office, so now we know that all finite sets can be ordered. Then I tossed out a question, LP1 LP1
Does there exist a set such that every point is a limit point, no point outside the set is a limit point, and the set contains no closed interval?
.

Commentary 10.7.11

This clip demonstrates how one can control the speed of a class. My preference is to lecture minimally, but if I want to accelerate a class, it is easy to do so by giving an introductory lecture on a new definition. Here I lecture briefly over continuity. Between the 7th and the 10th we see our first transition, which may well have been assisted by this lecture. See Item 5 under Course Observations which discusses the transitions for the course. Sometimes the right amount of precision at just the right time can really help. Over the next three days, we put up six well-written mathematical arguments. Even though some were not completely perfect or missed cases, they were still well-written mathematics. They now have the language down which will certainly aid the proving of the harder theorems.

Diary 10.10.11

Today Amber presented a very nice argument for P15 P15
Show that if \(\begin{equation}p \ne 0\end{equation}\), then \(\begin{equation}p\end{equation}\) is not a limit point of the set \(\begin{equation}\{ 1 , {1 \over 2} , {1 \over 3} , \dots \}\end{equation}\).
and Jacob broke new ground in using his ability to write correct mathematics, something he has struggled with all semester. Because he does not often write things out carefully, he makes leaps which are not correct. Today, however, he gave a very nice argument for P17 P17
Show that if the sequence \(\begin{equation}p_1,p_2,p_3,\end{equation}\)... converges to \(\begin{equation}x\end{equation}\) and the sequence \(\begin{equation}q_1,q_2,q_3,\dots\end{equation}\) converges to \(\begin{equation}y\end{equation}\), then the sequence \(\begin{equation}p_1+q_1,p_2+q_2,p_3+q_3,\end{equation}\)... converges to \(\begin{equation}x+y\end{equation}\).
, missing only the fact that he had not proved that the interval he constructed around \(\begin{equation}x+y\end{equation}\) was arbitrary. A very nice class with two very nice proofs.

Diary 10.12.11

Clint W. put up T1 T3
If the sequence \(\begin{equation}p_1 ,p_2 ,p_3 \dots\end{equation}\) converges to the point \(\begin{equation}x\end{equation}\) and \(\begin{equation}y\end{equation}\) is a point different from \(\begin{equation}x\end{equation}\), then \(\begin{equation}p_1 , p_2 , p_3 , \dots\end{equation}\) does not converge to \(\begin{equation}y\end{equation}\).
and Milagro put up T3 T3
If the point \(\begin{equation}p\end{equation}\) is a limit point of the point set \(\begin{equation}M\end{equation}\) and \(\begin{equation}S\end{equation}\) is an open interval containing \(\begin{equation}p\end{equation}\), then \(\begin{equation}S \cap M\end{equation}\) is infinite.
, both nice proofs, even as Milagro works on his writing.

Diary 10.14.11

Anthony went to the board for the first time and gave a good partial argument for T4 T4
If the sequence \(\begin{equation}p_1, p_2, p_3, \ldots\end{equation}\) converges to the point \(\begin{equation}x\end{equation}\) and \(\begin{equation}y\end{equation}\) is a point different from \(\begin{equation}x\end{equation}\), then \(\begin{equation}y\end{equation}\) is not a limit point of
\(\begin{equation}\{p_i:i=1,2,3,\dots \},\end{equation}\) the range of the sequence.
, assuming that the sequence must be increasing or decreasing. Thus, he has more work. Aaron, who recently told me he dropped a different class to find time for this one, gave a flawless argument for T6 T6
If \(\begin{equation}M\end{equation}\) is an open point set and \(\begin{equation}M\end{equation}\) is not all points, then the set of all points not in \(\begin{equation}M\end{equation}\) is a closed point set.
. He works full time as a manager at a coffee shop.

Diary 10.17.11

During the last class I attempted to convince Anthony that he had handled only a special case of T4 T4
If the sequence \(\begin{equation}p_1, p_2, p_3, \ldots\end{equation}\) converges to the point \(\begin{equation}x\end{equation}\) and \(\begin{equation}y\end{equation}\) is a point different from \(\begin{equation}x\end{equation}\), then \(\begin{equation}y\end{equation}\) is not a limit point of
\(\begin{equation}\{p_i:i=1,2,3,\dots \},\end{equation}\) the range of the sequence.
, so today when he had a question, I tried to provide enough guidance to increase his odds for success. Then Amber presented T6 T6
If \(\begin{equation}M\end{equation}\) is a closed point set and \(\begin{equation}M\end{equation}\) is not all points, then the set of all points not in \(\begin{equation}M\end{equation}\) is an open point set.
smoothly. From there I gave a quick overview of the completeness axiom, the idea behind the nested interval problem and a quick note about what to expect on the midterm.

Diary 10.19.11

Midterm

Diary 10.21.11

Han showed T9 T9
If \(\begin{equation}M\end{equation}\) is a closed and bounded point set, then there is a left-most point of \(\begin{equation}M\end{equation}\) and a right-most point of \(\begin{equation}M.\end{equation}\)
, with the help of the class and me. Weston presented P25 P25
Let \(\begin{equation}f\end{equation}\) be the function such that \(\begin{equation}f(x)=x^2\end{equation}\) for all numbers \(\begin{equation}x\end{equation}\). Show that \(\begin{equation}f\end{equation}\) is continuous at the point \(\begin{equation}(2,4)\end{equation}\).
and is approaching it in an unusual way. Rather than applying the definition of continuity at the point, he showed that \(\begin{equation}g(x)=x\end{equation}\) is continuous at any point \(\begin{equation}(c,g(c))\end{equation}\) and now is attempting to show that the product of two continuous functions is continuous.

Diary 10.24.11

I discussed a favorite quote attributed to many individuals, among them Von Goethe: "Treat an individual as he is and he will remain as he is. But if you treat him as if he were what he ought to be and could be, he will become what he ought to be and could be." I related this to my grading where a student I had been working with for years (since Calculus a year or so back) had turned in a beautiful proof this weekend. Following this, Jacob gave a nice argument for T8 T8
If the sequence \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\)
converges to the point \(\begin{equation}x\end{equation}\), then
\(\begin{equation}M=\{p_1,p_2,p_3\dots\}\end{equation}\) is bounded.
and Chris showed P24 P24
Show that if \(\begin{equation}f\end{equation}\) is a function and \(\begin{equation}(x,f(x))\end{equation}\) is a point on \(\begin{equation}f\end{equation}\), and \(\begin{equation}x\end{equation}\) is not a limit point of the domain of \(\begin{equation}f\end{equation}\), then \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}(x,f(x)).\end{equation}\)
.

Diary 10.26.11

Aaron made a good attempt at T7 T7
If \(\begin{equation}p\end{equation}\) is a point, there is a sequence of open intervals \(\begin{equation}S_1,S_2,S_3,\dots\end{equation}\) each containing \(\begin{equation}p\end{equation}\) such that for each positive integer \(\begin{equation}n\end{equation}\),
\(\begin{equation}S_{n+1} \subseteq S_n,\end{equation}\) and \(\begin{equation}p\end{equation}\) is the only point that is in every open interval in the sequence.
and John made a good attempt at T11 T11
If \(\begin{equation}M\end{equation}\) has \(\begin{equation}p\end{equation}\) as a limit point, then there exists either an increasing or a decreasing sequence of points of \(\begin{equation}M\end{equation}\) converging to \(\begin{equation}p.\end{equation}\)
. Both created nested sequences of open intervals in an attempt to solve their problem. Neither realized that the lengths of intervals must tend to zero in order to resolve the problem. It was nice to see two problems on the same day reinforcing the same technique. It is perhaps worth noting that I have given a "gag order" to Lucas, who sits next to me. He has almost every problem and I have him come to my office to show me solutions. I offered to allow him to quit coming to class, but he says he learns a lot from what goes on the board and chose to continue coming. Ironically, this removes a shining star from my video project, but I'm confident that working with him outside of class is in his best interest as he prepares for graduate school.

Diary 10.28.11

Just after midterm, of the twenty students in the class, the averages are:

- two "Fs" (one took the pre-requisite course fifteen years ago and was advised against taking the course, the other is
  working full time as a coach and taking eighteen hours of mathematics)
- two "Ds" (one of whom did not meet the minimum grade in the prerequisite course and requested that I allow her to take   the course, which I did, explaining that I recommended against it)
- four "C/D" borderline
- six "C"
- one "B/C"
- three "B"
- two "A"

Today I again discussed the quotes of von Goethe and the comment of a former student, who wrote that the best part of the class was "the teaching method (the instructor treating us as if we could prove theorems before we could)." We also closed the class with a brief discussion of a set that a student thought might answer question LP1 LP1
Does there exist a set such that every point is a limit point, no point outside the set is a limit point, and the set contains no closed interval?
. Two more nice problems went on the board. Aaron presented T7 T7
If \(\begin{equation}p\end{equation}\) is a point, there is a sequence of open intervals \(\begin{equation}S_1,S_2,S_3,\dots\end{equation}\) each containing \(\begin{equation}p\end{equation}\) such that for each positive integer \(\begin{equation}n\end{equation}\),
\(\begin{equation}S_{n+1} \subseteq S_n,\end{equation}\) and \(\begin{equation}p\end{equation}\) is the only point that is in every open interval in the sequence.
which he attempted yesterday and Clint G. presented P20 P20
Show that if \(\begin{equation}M\end{equation}\) is a point set and \(\begin{equation}p\end{equation}\) is a point and every closed interval containing \(\begin{equation}p\end{equation}\) contains a point of \(\begin{equation}M\end{equation}\) different from \(\begin{equation}p\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\).
, his first presentation since the first day.

Commentary 10.28.11

All three grades (presentation, written and midterm/final average) tend to increase as the semester progresses since students become better at proving, writing and presenting. Thus, these grades aren't as bad as they appear at first glance. Even though I grade by the method described on the syllabus, grading could be done in another way that would result in the same grades. If they develop good intuition on a number of problems, they earn a "C". If they develop good intuition on a number of problems and present some of these with very well-written mathematics, they earn a "B". If they break the back of some difficult problems, have numerous successes and present well-written mathematics, they earn an "A".

In this clip I discuss the afore mentioned quotes because I liked the parallel, but also because most of my students will teach at some point, whether in high school, community colleges, as graduate students, as professors or as parents. Thus I always want to sprinkle a bit of my philosophy on instruction in order that they understand why I teach as I do and for the benefit of these future teachers. I do treat my students as if they can do mathematics and can present well, because I believe that they can. Year after year, the majority prove me right.

From here we see a nice presentation of the nested interval theorem. Look at this proof and the writing and then look back at a clip a few weeks into the class. This is beautifully written mathematics we are now seeing. And we see an answer to the question, "Do Moore Method instructors lecture?" Almost all my lectures are reactive, and I "lecture" over two things in this class. In the first case, a proof is on the board but not all students believe the theorem, much less the proof. After we refine the proof based on my "lecture," we look at specific examples to help illustrate the theorem. From here, I move on to a problem that I want my best students working on which has been outstanding for quite some time. The problem is not in the notes, but I toss it out every semester. Lucas has thought about LP1 LP1
Does there exist a set such that every point is a limit point, no point outside the set is a limit point, and the set contains no closed interval?
and in order to keep it fresh in the minds of the students, I discuss an attempt that he brought to my office.

Diary 10.31.11

Katie had been working on P17 P17
Show that if the sequence \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) converges to \(\begin{equation}x\end{equation}\) and the sequence \(\begin{equation}q_1,q_2,q_3,\dots\end{equation}\) converges to \(\begin{equation}y\end{equation}\), then the sequence \(\begin{equation}p_1+q_1,p_2+q_2,p_3+q_3,\dots\end{equation}\) converges to \(\begin{equation}x+y\end{equation}\).
for weeks when Jacob beat her to it last week. Katie was starting the problem with an open interval around \(\begin{equation}x+y\end{equation}\) and trying to find intervals around \(\begin{equation}x\end{equation}\) and \(\begin{equation}y\end{equation}\) so that she could use the definition of convergence for each of these intervals. Jacob started with intervals around \(\begin{equation}x\end{equation}\) and \(\begin{equation}y\end{equation}\) so that when he finished, he still needed to prove a lemma that every open interval around \(\begin{equation}x+y\end{equation}\) could be constructed as the sum of an open interval around \(\begin{equation}x\end{equation}\) and a second open interval around \(\begin{equation}y\end{equation}\). Neither Jacob, nor Rabbi, nor Katie had gotten it and today Katie said she had progress and went to present it. When she got stuck in the middle of her presentation, Jessica jumped up and helped her out. Katie does not have the pre-requisite to the course, so is handicapped with regards to the rest of the students and this was her first presentation. I found the support touching. While rushed, I gave a summary of how one can combine the work of Katie and Jacob to create one elegant proof.

Anthony then presented P21 P21
Show that it is not true that if p is a limit point of a point set \(\begin{equation}M\end{equation}\), then every closed interval containing p must contain a point of \(\begin{equation}M\end{equation}\) different from p.
, demonstrating that there is a reason why we use open intervals in the definition of limit point and not closed intervals.

Commentary 10.31.11

It's a two-fer! In this first clip you'll first see what I do every Monday -- pass out the papers one at a time to each student before class starts. This allows me to make individual encouraging comments as often as possible. Sometimes they are just "good job" or "come see me," but sometimes I am truly impressed by a nice write-up by some student and can discuss some specific technique they developed that is important and tie it to some mathematician who made use of that historically. After this, you'll see see what I consider to be a touching episode that dispels the myth that there is some competition in Moore Method courses that is somehow detrimental to the students. This is not deep mathematics, but every student who goes out to teach should go out having proved and written up some mathematics on their own and now Katie has done this. Even though Jacob beat her to the problem, her contribution was necessary to make his argument complete. Thus the competition of two students working on the same problem ended up with each of them (a future teacher and a future electrical engineer) contributing to the solution. Perhaps I am an outlier, but it seems so very important to me that those who teach high school mathematics actually have done some mathematics independently so that they know what mathematics is! A second clip from this day shows what I do when I have a bit of time left and there is not enough time for another presentation or there are no presentations. Rabbi does have something ready to go, but I field a request for help on the definition of continuity. Rabbi will present on continuity tomorrow, so perhaps more will understand the proof if I can lay the groundwork. I am willing to help when students struggle with a new definition, but even here I am building off of the success of other students and I discuss what Weston and Rabbi have done to put what I am doing in the context of what the students have done, reinforcing that I know what each has done and that their work is what is important in this class.

Diary 11.2.11

Clint W. put up a very nice argument for P20 P20
Show that if \(\begin{equation}M\end{equation}\) is a point set and \(\begin{equation}p\end{equation}\) is a point and every closed interval containing \(\begin{equation}p\end{equation}\) contains a point of \(\begin{equation}M\end{equation}\) different from \(\begin{equation}p\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\).
. John attempted T11 T11
If \(\begin{equation}M\end{equation}\) has \(\begin{equation}p\end{equation}\) as a limit point, then there exists either an increasing or a decreasing sequence of points of \(\begin{equation}M\end{equation}\) converging to \(\begin{equation}p.\end{equation}\)
and had an idea, but it was not developed well enough to generate a proof. I drew some examples and created a conjecture and thought he left with an idea, but he did not have it the next day, so perhaps I need to talk more about the concept required.

Diary 11.4.11

Before class started, I discussed our Fast Track BS/MS and our MS Programs as there are several students in the class well-prepared for one or both. Filming started a bit late, just as I ended this discussion and was talking about enjoying mathematics and the importance of enjoying the process of struggling with a problem. From here I discussed the second semester of this course, encouraging students to take it. Milagro spent most of the period working problem T4 T4
If the sequence \(\begin{equation}p_1, p_2, p_3, \ldots\end{equation}\) converges to the point \(\begin{equation}x\end{equation}\) and \(\begin{equation}y\end{equation}\) is a point different from \(\begin{equation}x\end{equation}\), then \(\begin{equation}y\end{equation}\) is not a limit point of
\(\begin{equation}\{p_i:i=1,2,3,\dots \},\end{equation}\) the range of the sequence.
, which he has been working on for weeks. He has visited my office many times and invested a lot of time on the problem. This was Milagro’s third trip to the board and one can see improved confidence, even as he struggles with both notation and defending his work. I concluded class with a five minute lecture tying our notion of "closeness" to the definitions of "limits" that a teacher might see in high school or calculus texts.

Diary 11.7.11

Lucas attempted P22 P22
True or false? If \(\begin{equation}[a,b]\end{equation}\) is a closed interval and \(\begin{equation}G\end{equation}\) is a collection of open intervals with the property that every point in \(\begin{equation}[a,b]\end{equation}\) is in some open interval in \(\begin{equation}G\end{equation}\) then there is a finite subcollection of \(\begin{equation}G\end{equation}\) with the same property.
. He has been to the board multiple times. He is a senior and has had multiple classes with me. He has also done an REU. Thus, I informed him early on that I would not let him present much and he shows me several proofs outside of class for each one he is allowed to present inside class. Because he had not shown me P22 P22
True or false? If \(\begin{equation}[a,b]\end{equation}\) is a closed interval and \(\begin{equation}G\end{equation}\) is a collection of open intervals with the property that every point in \(\begin{equation}[a,b]\end{equation}\) is in some open interval in \(\begin{equation}G\end{equation}\) then there is a finite subcollection of \(\begin{equation}G\end{equation}\) with the same property.
outside of class and because I know that a mistake won't hurt his confidence and may well spur him to work even harder than his self-driven work ethic already requires, I let him put up P22 P22
True or false? If \(\begin{equation}[a,b]\end{equation}\) is a closed interval and \(\begin{equation}G\end{equation}\) is a collection of open intervals with the property that every point in \(\begin{equation}[a,b]\end{equation}\) is in some open interval in \(\begin{equation}G\end{equation}\) then there is a finite subcollection of \(\begin{equation}G\end{equation}\) with the same property.
even though I strongly suspected from a question he asked earlier that it was incorrect. Then Rabbi put up a nice argument for P25 P25
Let \(\begin{equation}f\end{equation}\) be the function such that \(\begin{equation}f(x)=x^2\end{equation}\) for all numbers \(\begin{equation}x\end{equation}\). Show that \(\begin{equation}f\end{equation}\) is continuous at the point \(\begin{equation}(2,4)\end{equation}\).
. I had a very difficult decision to make here. Kim had a nice geometric argument for P25 P25
Let \(\begin{equation}f\end{equation}\) be the function such that \(\begin{equation}f(x)=x^2\end{equation}\) for all numbers \(\begin{equation}x\end{equation}\). Show that \(\begin{equation}f\end{equation}\) is continuous at the point \(\begin{equation}(2,4)\end{equation}\).
which had a mistake and would really have helped the class understand the geometric definition of continuity. I still want to see it presented and will tell her that when I advise her today. I concluded class with a brief introduction to T15 T15
If \(\begin{equation}I_1,I_2,I_3, \dots\end{equation}\) is a sequence of closed intervals such that for each positive integer \(\begin{equation}n\end{equation}\), \(\begin{equation}I_{n+1}\subseteq I_n,\end{equation}\) then there is a point \(\begin{equation}p\end{equation}\) such that if \(\begin{equation}n\end{equation}\) is any positive integer, then \(\begin{equation}p\end{equation}\) is in \(\begin{equation}I_n\end{equation}\). In other words, there is a point \(\begin{equation}p\end{equation}\) which is in all the closed intervals of the sequence \(\begin{equation}I_1,I_2,I_3 , \dots\end{equation}\).
and T16 T16
If \(\begin{equation}I_1,I_2,I_3 , \dots\end{equation}\) is a sequence of closed intervals so that for each positive integer n, \(\begin{equation}I_{n+1} \subseteq I_n,\end{equation}\) and the length of \(\begin{equation}I_n\end{equation}\) is less than \(\begin{equation}1 \over n\end{equation}\), then there is only one point p such that for each positive integer n, \(\begin{equation}p \in I_n\end{equation}\).
.

Diary 11.9.11

Michael put up P23 P23
Let f be the function such that \(\begin{equation}f(x)=2\end{equation}\) for all numbers \(\begin{equation}x>5\end{equation}\), and \(\begin{equation}f(x)=1\end{equation}\) for all numbers \(\begin{equation}x \leq 5\end{equation}\).

1. Show that \(\begin{equation}f\end{equation}\) is not continuous at the point \(\begin{equation}(5,1)\end{equation}\).

2. Show that if \(\begin{equation}t\end{equation}\) is a number and \(\begin{equation}t>5\end{equation}\), then \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}(t,2)\end{equation}\).
and Weston put up P26 P26
If \(\begin{equation}f\end{equation}\) is a function which is continuous on \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}x \in (a,b)\end{equation}\) such that \(\begin{equation}f(x) > 0\end{equation}\) then there exists an open interval, \(\begin{equation}T,\end{equation}\) containing \(\begin{equation}x\end{equation}\) such that \(\begin{equation}f(t) > 0\end{equation}\) for all \(\begin{equation}t \in T\end{equation}\).
. This problem was interesting because he had received a grade of "A" on homework and I was convinced that he fully understood the problem. He had turned in two proofs, one a "D" and one an "A" for the same problem. He believed that he did not have a proof, even though I had given him an "A". As is typical of Weston, who has not had the pre-requisite course, he had a proof in his mind. He also had written statements that were not valid. Through careful questioning, I was able to convince him that he had a proof.

Diary 11.11.11

Today I started passing out some of the many articles I have saved over the years. Lucas discussed P22 P22
True or false? If \(\begin{equation}[a,b]\end{equation}\) is a closed interval and \(\begin{equation}G\end{equation}\) is a collection of open intervals with the property that every point in \(\begin{equation}[a,b]\end{equation}\) is in some open interval in \(\begin{equation}G\end{equation}\) then there is a finite subcollection of \(\begin{equation}G\end{equation}\) with the same property.
which he believes is true, although he does not have a proof yet. I encouraged him to put up what he had to spark interest in this problem. So many nice tools can be discovered working on this problem. He gave a good presentation of where he is on the problem. Milagro then gave a nice presentation on T12 T12
If \(\begin{equation}f\end{equation}\) is a function and \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a sequence of points in the domain of \(\begin{equation}f\end{equation}\) converging to the number \(\begin{equation}x\end{equation}\) in the domain of \(\begin{equation}f\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}(x,f(x))\end{equation}\), then \(\begin{equation}f(p_1),f(p_2 ), \dots\end{equation}\) converges to \(\begin{equation}f(x)\end{equation}\).
. I concluded class by explaining that T12 T12
If \(\begin{equation}f\end{equation}\) is a function and \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a sequence of points in the domain of \(\begin{equation}f\end{equation}\) converging to the number \(\begin{equation}x\end{equation}\) in the domain of \(\begin{equation}f\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}(x,f(x))\end{equation}\), then \(\begin{equation}f(p_1),f(p_2 ), \dots\end{equation}\) converges to \(\begin{equation}f(x)\end{equation}\).
provides another equivalent definition for continuity.

Commentary 11.11.11

This clip shows a nice proof from a student who was somewhat unsure at the beginning of the course and has not presented a lot, but has really given an almost complete argument at the board. And it shows another example of how I lecture reactively to tie together the multiple definitions of continuity once individual students have mastered several of them.

Diary 11.14.11

I passed out an internship opportunity to ORNL and some math-ed literature. Anthony’s write up this weekend was very close to a proof for T14 T14
Suppose \(\begin{equation}f\end{equation}\) and \(\begin{equation}g\end{equation}\) are functions having domain \(\begin{equation}M\end{equation}\) and each is continuous at the point \(\begin{equation}p\end{equation}\) in \(\begin{equation}M\end{equation}\). Suppose that \(\begin{equation}h\end{equation}\) is a function with domain \(\begin{equation}M\end{equation}\) such that
\(\begin{equation}f(p)=h(p)=g(p)\end{equation}\)
and for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}M\end{equation}\),
\(\begin{equation}f(x) \le h(x) \le g(x).\end{equation}\)
Prove \(\begin{equation}h\end{equation}\) is continuous at \(\begin{equation}p\end{equation}\).
. He jumped up today and put a nice argument on the board for it. Then I encouraged Kim to put up what she had so that we could talk about how one applies the completeness axiom. I was at the board more than usual, so this one might show how I lecture, if one calls it that.

Diary 11.16.11

I started before class with a discussion of the importance of College Algebra on campus, because it is the last mathematics course taken by many students. Thus, on every campus, this is the course that should be considered the most important to our profession, as many politicians may have seen this as their last mathematics class and if they left with a good impression of what mathematics might do for them, we might get more support for this all-important discipline. Lucas puts up a flawless proof of T14 T14
Suppose \(\begin{equation}f\end{equation}\) and \(\begin{equation}g\end{equation}\) are functions having domain \(\begin{equation}M\end{equation}\) and each is continuous at the point \(\begin{equation}p\end{equation}\) in \(\begin{equation}M\end{equation}\). Suppose that \(\begin{equation}h\end{equation}\) is a function with domain \(\begin{equation}M\end{equation}\) such that
\(\begin{equation}f(p)=h(p)=g(p)\end{equation}\)
and for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}M\end{equation}\),
\(\begin{equation}f(x) \le h(x) \le g(x).\end{equation}\)
Prove \(\begin{equation}h\end{equation}\) is continuous at \(\begin{equation}p\end{equation}\).
. Several students have begun working on differentiability so I put up the definition of tangent line and attempted to query the class about it. There wasn't a lot of discussion, but hopefully it still serves to give them a better understanding of the definitions.

Diary 11.18.11

Video probably won't be good today; Brandy was ill and camera was stationary. But Kim did an interesting attempt at T20 T20
Let \(\begin{equation}f\end{equation}\) be a continuous function whose domain includes the closed interval \(\begin{equation}[a,b]\end{equation}\). If \(\begin{equation}f(a)<0\end{equation}\) and \(\begin{equation}f(b)>0\end{equation}\), then there is a number \(\begin{equation}x\end{equation}\) between \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\) such that \(\begin{equation}f(x)=0\end{equation}\).
and the second video shows it well. John attempted T11 T11
If \(\begin{equation}M\end{equation}\) has \(\begin{equation}p\end{equation}\) as a limit point, then there exists either an increasing or a decreasing sequence of points of \(\begin{equation}M\end{equation}\) converging to \(\begin{equation}p.\end{equation}\)
again, but was still not there.

Commentary 11.18.11

I was very disappointed that they did not have T11 T11
If \(\begin{equation}M\end{equation}\) has \(\begin{equation}p\end{equation}\) as a limit point, then there exists either an increasing or a decreasing sequence of points of \(\begin{equation}M\end{equation}\) converging to \(\begin{equation}p.\end{equation}\)
and T20 T20
Let \(\begin{equation}f\end{equation}\) be a continuous function whose domain includes the closed interval \(\begin{equation}[a,b]\end{equation}\). If \(\begin{equation}f(a)<0\end{equation}\) and \(\begin{equation}f(b)>0\end{equation}\), then there is a number \(\begin{equation}x\end{equation}\) between \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\) such that \(\begin{equation}f(x)=0\end{equation}\).
on Monday. When I watched all the video prior to building the webinar of my father's teaching, I observed the incredible percentage of time that a student struggled on one day and completed the problem the next day. That not only accelerated the class, it also gives a measure of the effectiveness of the teacher. While I'd like to attribute this to the fact that Emory students live on campus, rarely work and are rarely first generation students, I'd be remiss if I did not observe that Dad seemed particularly adept at patiently questioning the students at the board. In so doing, he was able to alert the student as to what was missing without lecturing and without giving the student the solution. The students seemed always to understand the problem well enough after the questioning to be able to work through the problem the next class period.

Diary 11.21.11

Today Han attempted a proof of T18 T18
If \(\begin{equation}f\end{equation}\) is a continuous function whose domain includes a closed interval \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}p \in [a,b]\end{equation}\), then the set of all numbers \(\begin{equation}x \in [a,b]\end{equation}\) such that \(\begin{equation}f(x)=f(p)\end{equation}\) is a closed point set.
at the board. He had a decent proof on his paper, but after questioning him at the board he did not seem to have a good understanding of the problem. As if to offset this disappointment, Jessica presented a beautiful argument for T19 T19
No closed interval is the union of two mutually exclusive closed point sets.
. She had asked questions in the office to clarify points, but the heart of the idea was all hers.

Diary 11.23.11

Today demonstrates how to generate progress, even on the Wednesday before Thanksgiving. Most students, due to exams, work schedules and the difficulty of the latest problems have hit a wall and we don't have a back log of problems ready to present. Three students have been to my office struggling with problems requiring the Completeness Axiom. Thus, today I encouraged Clint G. to present what he had on the Intermediate Value Theorem, T21 T21
If \(\begin{equation}f\end{equation}\) is a continuous function whose domain includes a closed interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}L\end{equation}\) is a horizontal line, and \(\begin{equation}(a,f(a))\end{equation}\) is below \(\begin{equation}L\end{equation}\), and \(\begin{equation}(b,f(b))\end{equation}\) is above \(\begin{equation}L\end{equation}\), then there is a number \(\begin{equation}x\end{equation}\) between \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\) such that \(\begin{equation}(x,f(x))\end{equation}\) is on \(\begin{equation}L\end{equation}\).
, in order to encourage success with the problems requiring completeness. While a successful proof did not go on the board, we demonstrated two sets, either of which would yield success once the completeness axiom and continuity are applied. Shaymal then presented P28 P28
Use any of the definitions of derivative to show that if \(\begin{equation}f(x)=x^2 + 1\end{equation}\) then \(\begin{equation}f\end{equation}\) has
derivative \(\begin{equation}6\end{equation}\) at \(\begin{equation}3\end{equation}\).
, the first simple example for differentiability. Finally, I attempted, rather poorly, to tie the \(\begin{equation}\epsilon-\delta\end{equation}\) definition of differentiability to the geometric definition. In the last two weeks of the semester after Thanksgiving, we'll see enough derivative problems to make the final cover limit points, convergence, continuity and derivatives. Sadly, this semester we did not reach integration. On the bright side, when I took on this course at Lamar ten years ago and asked advice, the former instructor told me he'd never made it past the Extreme Value Theorem! Also, on the bright side at this point ten students of the twenty have registered for the second semester so we will have a good class going forward and should see some measure theory, integration, Cauchy sequences, subsequences and other important concepts for their future studies.

Diary 11.28.11

Today, no one had anything to present. It is rare that we make it this far into the semester before this happens. I also knew from office visits that there are people working hard on upcoming problems and so I wasn't sure I wanted to give hints on upcoming problems that might take away the students’ chances of getting the problems on their own. And I wasn't sure that the attempts people had at this point were worth spending board time on. If the class had regularly not had problems to present, I would not have lectured, but would have pushed forward in the notes or encouraged a student to go to the board, but since this was the first time in the entire semester that they were unprepared, I gave a full hour lecture. I should note that Lucas did have material ready to present, so I could have let him have the floor and will on Wednesday if no one has anything, but I never let a bright student run away with the class. I only let bright students put up quite difficult problems. The lecture was on the numbers, starting with Peano’s Postulates, we talked about the natural, rational and real numbers from an axiomatic approach with a bit of intuition thrown in. Because the axiomatic treatment of the real numbers provides poor intuition, we showed that \(\begin{equation}\sqrt{2}\end{equation}\) is irrational. Then we talked about how to show that the natural numbers and rational numbers are countable and how to show that the real numbers are not. My goal was to tease them with some elementary number theory questions, but I didn't make it that far!

Commentary 11.28.11

This clip is the only full hour of lecture given during the two semester course.

Diary 11.30.11

Entire day went to Weston putting up that the product of two continuous functions is continuous, but he really had the entire idea. He needed one lemma that I put on the board.

Commentary 11.30.11

Note that it has been months since he decided to prove that \(\begin{equation}f(x)=x^2\end{equation}\) was continuous by showing that \(\begin{equation}g(x)=x\end{equation}\) was continuous and then proving that the product of two continuous functions was continuous. I have continually questioned him as to whether he has this proof in order to encourage him to continue working on it because the result he finally obtained was so much stronger than the result I had originally asked for.

Diary 12.2.11

Kimberly put up a nice problem, T20 T20
Let \(\begin{equation}f\end{equation}\) be a continuous function whose domain includes the closed interval \(\begin{equation}[a,b]\end{equation}\). If \(\begin{equation}f(a)<0\end{equation}\) and \(\begin{equation}f(b)>0\end{equation}\), then there is a number \(\begin{equation}x\end{equation}\) between \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\) such that \(\begin{equation}f(x)=0\end{equation}\).
, that she had worked on for some time. Milagro then put up T15 T15
If \(\begin{equation}I_1,I_2,I_3, \dots\end{equation}\) is a sequence of closed intervals such that for each positive integer \(\begin{equation}n\end{equation}\), \(\begin{equation}I_{n+1}\subseteq I_n,\end{equation}\) then there is a point \(\begin{equation}p\end{equation}\) such that if \(\begin{equation}n\end{equation}\) is any positive integer, then \(\begin{equation}p\end{equation}\) is in \(\begin{equation}I_n\end{equation}\). In other words, there is a point \(\begin{equation}p\end{equation}\) which is in all the closed intervals of the sequence \(\begin{equation}I_1,I_2,I_3 , \dots\end{equation}\).
, so they should be masters at the completeness axiom by now!

Commentary 12.2.11

This clip illustrates three things that I think are important. First, we see the oft supported theme that not only does the Moore Method not work only for the brightest students, but under the method both minorities and females tend to perform better. For evidence of this, see for example the study by Karen Laursen et al. This video also demonstrates how I carefully question, without leading, to determine if a student has a proof or in order to demonstrate why they don't have a proof. I don't give them a proof, I simply force them to carefully articulate that which they intuitively understand already. Kim really does have this problem well in mind and can write and produce good mathematics when questioned. Milagro has all the right pieces as well, even if it takes a bit of work to finish it. He uses the acronym "N.T.S." that we used in class to denote "Need To Show."

Second, I hope this demonstrates why I support pushing the students to work outside of class rather than using collaborative work in class. The first reason is simple, I believe that homework should be done outside of class where the individual learning comes. Secondly, while a group effort working on these problems might have produced a proof during class, I think that these two students spent many hours on these problems outside of class and now know that they can do mathematics independently. My goal is for my students to graduate having proved some non-trivial mathematics independently in order to build their confidence in their ability. Having worked in industry, I have seen too many workers and students who feel that they can do something if someone helps or someone shows them how. I want to propagate the belief not that a student can learn something if it is well taught, but that every student can learn something independently through effort and time.

Finally, I conclude the class by discussing how this definition gives an alternate defintion for completeness. We have not discussed completeness in this class, but those who stay for the second semester will see Cauchy sequences and a proof that the reals are complete. Thus, this lecture introduces them to completeness and hints at something that is important that will happen next semester.

Diary 12.5.11

Amber went to the board on T21 T21
If \(\begin{equation}f\end{equation}\) is a continuous function whose domain includes a closed interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}L\end{equation}\) is a horizontal line, and \(\begin{equation}(a,f(a))\end{equation}\) is below \(\begin{equation}L\end{equation}\), and \(\begin{equation}(b,f(b))\end{equation}\) is above \(\begin{equation}L\end{equation}\), then there is a number \(\begin{equation}x\end{equation}\) between \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\) such that \(\begin{equation}(x,f(x))\end{equation}\) is on \(\begin{equation}L\end{equation}\).
admitting that she was close. She had the intuition and had previously done problems where she used the technique she needed. Still, she struggled with seeing how to use continuity to find the contradiction she needed. I'm sure it cleared the proof technique up for both her and Clint G. who had worked on this one and T20 T20
Let \(\begin{equation}f\end{equation}\) be a continuous function whose domain includes the closed interval \(\begin{equation}[a,b]\end{equation}\). If \(\begin{equation}f(a)<0\end{equation}\) and \(\begin{equation}f(b)>0\end{equation}\), then there is a number \(\begin{equation}x\end{equation}\) between \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\) such that \(\begin{equation}f(x)=0\end{equation}\).
.

Diary 12.7.11

Last day of class. Rabbi presented T25 T25
Suppose that each of \(\begin{equation}f\end{equation}\) and \(\begin{equation}g\end{equation}\) are functions that are differentiable at the point \(\begin{equation}p\end{equation}\) and that \(\begin{equation}h\end{equation}\) is the function defined by \(\begin{equation}h(x) = f(x) + g(x)\end{equation}\) for all \(\begin{equation}x \in D_f.\end{equation}\) Show that \(\begin{equation}h\end{equation}\) is also differentiable at the point \(\begin{equation}p\end{equation}\).
and had T24 T24
Suppose that \(\begin{equation}f\end{equation}\) is a function that is differentiable at the point \(\begin{equation}p\end{equation}\) and that \(\begin{equation}c \in \mathbb{R}.\end{equation}\) Show that the function \(\begin{equation}g\end{equation}\) defined by \(\begin{equation}g(x) = cf(x)\end{equation}\) for all \(\begin{equation}x \in D_f\end{equation}\) is also differentiable at the point \(\begin{equation}p\end{equation}\).
ready. Lucas presented T16 T16
If \(\begin{equation}I_1,I_2,I_3 , \dots\end{equation}\) is a sequence of closed intervals so that for each positive integer n, \(\begin{equation}I_{n+1} \subseteq I_n,\end{equation}\) and the length of \(\begin{equation}I_n\end{equation}\) is less than \(\begin{equation}1 \over n\end{equation}\), then there is only one point p such that for each positive integer n, \(\begin{equation}p \in I_n\end{equation}\).
, at my request, because I wanted another example of a nested interval theorem given that there is one on the final.

End-of-Course Commentary

By some measures, this class has been more successful than previous classes. There was never a day when students did not have material ready to present and of the twenty students who started,

- all but one presented and that one was working as a full-time coach and taking eighteen hours of mathematics,
- all improved their ability to write mathematics as demonstrated on the weekly written homework,
- all earned a grade of "C" or better, except two, the coach and one who had a legitimate reason for an incomplete,
- ten signed up to take the optional second semester.

Be other measures, the class has been less successful. We did not cover as much material as we usually do. Previous classes usually make it to integration. Because students always had material to present, I only squeezed in abbreviated supplementary lectures that I often give. I suppose that is a good thing. And often I give a proof of one or more theorems, which I think can help them see how I figure proofs out as I write them. Normally, I give several more detailed mini-lectures over topics such as:

- the axioms that give us the natural and real numbers,
- countability and cardinality,
- measure theory (an ad for next semester),
- relating our definitions of continuity to how continuity might be taught in high school or freshman calculus, and
- relating our definitions of differentiable to how differentiability might be taught in high school or freshman calculus.

Due to several factors, the student presentations were less interactive than I've seen in the past. My belief is that there were multiple reasons for this. My own personality was modified by the presence of the camera. The presence of the camera also forced me to sit near the front, when I usually sit to the side and near the back where it is easier to interact with the whole class and where my role is diminished. The class started very slowly and numerous early classes were unproductive due to a few students who did not have the correct background but were placed in the class. This set a slower pace for the class than for previous classes. This was the largest analysis class I've had. When I taught my first analysis course here, there were only five students in the class. I teach the course only every other year, because I feel that a student who has not succeeded in the course under my instruction should not be subjected to the same instructor again and we only offer one section per year. As we have quadrupled the number of majors over the past decade, this course has grown from five students to more than twenty students. When I wrote this class summary at the end of the semester, I did not know of all the personal problems that students were just starting to have and which escalated during the second semester, as discussed in Course Observations.

Spring Semester, Math 4326

As is typical, about half of last semester's students continued on and we added one new one, Tri.

Diary 1.18.12

I surprised the students who knew me by starting out with a lecture! I wanted to encourage the students to try to give some intuition for a proof before writing a formal proof on the board. Thus I did just this on two problems. For the first problem, I gave intuition via verbal and graphical arguments before providing a complete proof. For the second problem, I gave the intuition, but not a proof, in order to set them up to give a proof.

Diary 1.20.12

Before I got to class Jessica had an attempt on the board for T23 T23
If \(\begin{equation}M\end{equation}\) is a point set then \(\begin{equation}Cl(M)\end{equation}\) is a closed point set.
, but does not have the problem yet. Good class discussion, although Lucas tried to give her a proof and I eventually stopped it. She is a strong student and doesn't need hints, she just needs to go work on it. Then Tri made a good start on P30 P30
Use any of the definitions of derivative to show that if \(\begin{equation}f\end{equation}\) is a function whose domain includes \(\begin{equation}(-1,1)\end{equation}\) and for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}(-1,1)\end{equation}\), \(\begin{equation}{-x^2 \le f(x) \le x^2}\end{equation}\), then the derivative of \(\begin{equation}f\end{equation}\) at the point \(\begin{equation}(0,0)\end{equation}\) is \(\begin{equation}0.\end{equation}\)
. As the only student who did not have the first semester, he made a good start. I hope that I pointed him in the right direction. Lastly, Lucas put up a perfect proof of T24 T24
Suppose that \(\begin{equation}f\end{equation}\) is a function that is differentiable at the point \(\begin{equation}p\end{equation}\) and that \(\begin{equation}c \in \mathbb{R}.\end{equation}\) Show that the function \(\begin{equation}g\end{equation}\) defined by \(\begin{equation}g(x) = cf(x)\end{equation}\) for all \(\begin{equation}x \in D_f\end{equation}\) is also differentiable at the point \(\begin{equation}p\end{equation}\).
.

Diary 1.23.12

Tri was not ready with P30 P30
Use any of the definitions of derivative to show that if \(\begin{equation}f\end{equation}\) is a function whose domain includes \(\begin{equation}(-1,1)\end{equation}\) and for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}(-1,1)\end{equation}\), \(\begin{equation}{-x^2 \le f(x) \le x^2}\end{equation}\), then the derivative of \(\begin{equation}f\end{equation}\) at the point \(\begin{equation}(0,0)\end{equation}\) is \(\begin{equation}0.\end{equation}\)
. He is missing class because of the Lunar New Year. Sudents missing clases for Weed Day and the Lunar New Year are firsts for me in my teaching career! Jessica does not have T23 T23
If \(\begin{equation}M\end{equation}\) is a point set then \(\begin{equation}Cl(M)\end{equation}\) is a closed point set.
ready, although she says she understands the problem and based on previous experience, if she works on it, she'll get it. Lucas did a nice job of P27 P27
If \(\begin{equation}f\end{equation}\) is a function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\) and the range of \(\begin{equation}f\end{equation}\) is \(\begin{equation}\{-1,1\}\end{equation}\), then there is a number \(\begin{equation}x\end{equation}\) in \(\begin{equation}[a,b]\end{equation}\) at which \(\begin{equation}f\end{equation}\) is not continuous.
and there was some discussion which was good. John has returned to T11 T11
If \(\begin{equation}M\end{equation}\) has \(\begin{equation}p\end{equation}\) as a limit point, then there exists either an increasing or a decreasing sequence of points of \(\begin{equation}M\end{equation}\) converging to \(\begin{equation}p.\end{equation}\)
which he never resolved last semester and would have made the difference between a "C" and a "B" for his final grade. I hadn't looked over the notes closely enough and did not realize the humdinger I had in P31 P31
Does there exist a function \(\begin{equation}f\end{equation}\) defined and continuous on \(\begin{equation}[0,1]\end{equation}\) such that \(\begin{equation}f(0)=0\end{equation}\) and \(\begin{equation}f(1)=1\end{equation}\) and \(\begin{equation}f'(x)=0\end{equation}\) at all but countably many points of \(\begin{equation}[0,1]\end{equation}\)?
, so may not have done a good job of introducing this problem to the class. So, it was a good example of a video that shows what happens when the entire class, myself included, are unprepared!

Diary 1.25.12

Tri worked on P30 P30
Use any of the definitions of derivative to show that if \(\begin{equation}f\end{equation}\) is a function whose domain includes \(\begin{equation}(-1,1)\end{equation}\) and for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}(-1,1)\end{equation}\), \(\begin{equation}{-x^2 \le f(x) \le x^2}\end{equation}\), then the derivative of \(\begin{equation}f\end{equation}\) at the point \(\begin{equation}(0,0)\end{equation}\) is \(\begin{equation}0.\end{equation}\)
for the first half of class. I believe he has a proof, but I did not ask him to write it up. I'm tempted to. Kim attempted P18 P18
Show that if \(\begin{equation}M\end{equation}\) is a point set, then there cannot be both a right-most point of \(\begin{equation}M\end{equation}\) and a first point to the right of \(\begin{equation}M.\end{equation}\)
for the second half of class, but did not have a proof, although she had the seed of an idea.

Diary 1.27.12

John worked on T11 T11
If \(\begin{equation}M\end{equation}\) has \(\begin{equation}p\end{equation}\) as a limit point, then there exists either an increasing or a decreasing sequence of points of \(\begin{equation}M\end{equation}\) converging to \(\begin{equation}p.\end{equation}\)
and has a better understanding. I'm not succeeding as my father did at seeding the students perfectly and neither Jessica nor Kim had their problems. I'm hopeful after today that John will resolve T11 T11
If \(\begin{equation}M\end{equation}\) has \(\begin{equation}p\end{equation}\) as a limit point, then there exists either an increasing or a decreasing sequence of points of \(\begin{equation}M\end{equation}\) converging to \(\begin{equation}p.\end{equation}\)
. And Katie said she has an argument for T24 T24
Suppose that \(\begin{equation}f\end{equation}\) is a function that is differentiable at the point \(\begin{equation}p\end{equation}\) and that \(\begin{equation}c \in \mathbb{R}.\end{equation}\) Show that the function \(\begin{equation}g\end{equation}\) defined by \(\begin{equation}g(x) = cf(x)\end{equation}\) for all \(\begin{equation}x \in D_f\end{equation}\) is also differentiable at the point \(\begin{equation}p\end{equation}\).
which Lucas already proved, but I note that she did not realize he had proved it and she used a different definition to prove it.

Diary 1.30.12

Katie presented a proof of T24 T24
Suppose that \(\begin{equation}f\end{equation}\) is a function that is differentiable at the point \(\begin{equation}p\end{equation}\) and that \(\begin{equation}c \in \mathbb{R}.\end{equation}\) Show that the function \(\begin{equation}g\end{equation}\) defined by \(\begin{equation}g(x) = cf(x)\end{equation}\) for all \(\begin{equation}x \in D_f\end{equation}\) is also differentiable at the point \(\begin{equation}p\end{equation}\).
using a different definition than the one that Lucas had used. She had the heart of the argument, but needed details. I think that in the future when I do not give exams, I need to make them turn in and post final write-ups of proofs and give them a grade for them. Aaron made a good attempt at T25 T25
Suppose that each of \(\begin{equation}f\end{equation}\) and \(\begin{equation}g\end{equation}\) are functions that are differentiable at the point \(\begin{equation}p\end{equation}\) and that \(\begin{equation}h\end{equation}\) is the function defined by \(\begin{equation}h(x) = f(x) + g(x)\end{equation}\) for all \(\begin{equation}x \in D_f.\end{equation}\) Show that \(\begin{equation}h\end{equation}\) is also differentiable at the point \(\begin{equation}p\end{equation}\).
. I think both were helped by going to the board and I demonstrated how gentle I try to be with the students who appear less confident of their mathematical abilities.

Diary 2.1.12

What to do when nothing to present. Weston talks about P31 P31
Does there exist a function \(\begin{equation}f\end{equation}\) defined and continuous on \(\begin{equation}[0,1]\end{equation}\) such that \(\begin{equation}f(0)=0\end{equation}\) and \(\begin{equation}f(1)=1\end{equation}\) and \(\begin{equation}f'(x)=0\end{equation}\) at all but countably many points of \(\begin{equation}[0,1]\end{equation}\)?
and I give some ideas. Showing that the interval is not countable and introducing convergence of sequences. Lucas talks about T26 T26
If \(\begin{equation}f\end{equation}\) is a function, \(\begin{equation}x\end{equation}\) is in the domain of \(\begin{equation}f\end{equation}\), and \(\begin{equation}f\end{equation}\) has a derivative at \(\begin{equation}(x,f(x))\end{equation}\), then \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}(x,f(x))\end{equation}\).
to show where he is. I try to help the class understand differentiability better. It doesn't break the back of T26 T26
If \(\begin{equation}f\end{equation}\) is a function, \(\begin{equation}x\end{equation}\) is in the domain of \(\begin{equation}f\end{equation}\), and \(\begin{equation}f\end{equation}\) has a derivative at \(\begin{equation}(x,f(x))\end{equation}\), then \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}(x,f(x))\end{equation}\).
for Lucas, but it does set up success the next day for Aaron for T25 T25
Suppose that each of \(\begin{equation}f\end{equation}\) and \(\begin{equation}g\end{equation}\) are functions that are differentiable at the point \(\begin{equation}p\end{equation}\) and that \(\begin{equation}h\end{equation}\) is the function defined by \(\begin{equation}h(x) = f(x) + g(x)\end{equation}\) for all \(\begin{equation}x \in D_f.\end{equation}\) Show that \(\begin{equation}h\end{equation}\) is also differentiable at the point \(\begin{equation}p\end{equation}\).
.

Commentary 2.1.12

Here we have another day when no student has a solution to present. This clip and the one on the subsequent day, show how one can seed ideas on one day in order to have success on the next. My discussion with Lucas, which appears completely focused on him, is sufficient that Aaron resolves T25 T25
Suppose that each of \(\begin{equation}f\end{equation}\) and \(\begin{equation}g\end{equation}\) are functions that are differentiable at the point \(\begin{equation}p\end{equation}\) and that \(\begin{equation}h\end{equation}\) is the function defined by \(\begin{equation}h(x) = f(x) + g(x)\end{equation}\) for all \(\begin{equation}x \in D_f.\end{equation}\) Show that \(\begin{equation}h\end{equation}\) is also differentiable at the point \(\begin{equation}p\end{equation}\).
, without help from me in my office, on the next day. This day also shows how to use a day when we don't have something to present to look at what attempts we have and springboard from these attempts to new material. I used Weston's approach to P31 P31
Does there exist a function \(\begin{equation}f\end{equation}\) defined and continuous on \(\begin{equation}[0,1]\end{equation}\) such that \(\begin{equation}f(0)=0\end{equation}\) and \(\begin{equation}f(1)=1\end{equation}\) and \(\begin{equation}f'(x)=0\end{equation}\) at all but countably many points of \(\begin{equation}[0,1]\end{equation}\)?
to introduce pointwise functional convergence. Watching the video, I wish I had stated the examples I used as problems for presentation to follow this thread. If I had done so, we could have started a thread on functional convergence that would have quickly led to uniform convergence of sequences of functions. What can I say, hind sight is 20/20.

Diary 2.3.12

Kim T18 T18
If \(\begin{equation}f\end{equation}\) is a continuous function whose domain includes a closed interval \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}p \in [a,b]\end{equation}\), then the set of all numbers \(\begin{equation}x \in [a,b]\end{equation}\) such that \(\begin{equation}f(x)=f(p)\end{equation}\) is a closed point set.
, Aaron T25 T25
Suppose that each of \(\begin{equation}f\end{equation}\) and \(\begin{equation}g\end{equation}\) are functions that are differentiable at the point \(\begin{equation}p\end{equation}\) and that \(\begin{equation}h\end{equation}\) is the function defined by \(\begin{equation}h(x) = f(x) + g(x)\end{equation}\) for all \(\begin{equation}x \in D_f.\end{equation}\) Show that \(\begin{equation}h\end{equation}\) is also differentiable at the point \(\begin{equation}p\end{equation}\).
, Tri P34 part 1 P34 part 1
Let \(\begin{equation}f(x)=0\end{equation}\) for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}[0,1]\end{equation}\) except \(\begin{equation}x=0\end{equation}\), and let \(\begin{equation}f(0)=1\end{equation}\). Show that:

1. if \(\begin{equation}P\end{equation}\) is a partition of \(\begin{equation}[0,1]\end{equation}\), then \(\begin{equation}0 < U_P f,\end{equation}\)

2. if \(\begin{equation}\epsilon > 0,\end{equation}\) then there is a partition \(\begin{equation}P\end{equation}\) of \(\begin{equation}[0,1]\end{equation}\) such that \(\begin{equation}U_P f < \epsilon\end{equation}\),

3. zero is the only lower Riemann sum for \(\begin{equation}f\end{equation}\) on \(\begin{equation}[0,1]\end{equation}\).
.

Commentary 2.3.12

I open this clip with perhaps the best motivational argument I have ever given for teaching the way I do. I talk about my respect for all good teaching, lectured included, and discuss some courses I took. Then I discuss a fascinating educational paper by J. Boaler, Open and closed mathematics approaches: Student experiences and understandings, Journal for Research in Mathematics Education, Vol. 29, No. 1,January 1998, pp. 41-62., which may be the best argument ever for Moore Method teaching. When I finally shut up, we see three nice proofs!

Diary 2.6.12

Tri presents P34 part 2 P34 part 2
Let \(\begin{equation}f(x)=0\end{equation}\) for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}[0,1]\end{equation}\) except \(\begin{equation}x=0\end{equation}\), and let \(\begin{equation}f(0)=1\end{equation}\). Show that:

1. if \(\begin{equation}P\end{equation}\) is a partition of \(\begin{equation}[0,1]\end{equation}\), then \(\begin{equation}0 < U_P f,\end{equation}\)

2. if \(\begin{equation}\epsilon > 0,\end{equation}\) then there is a partition \(\begin{equation}P\end{equation}\) of \(\begin{equation}[0,1]\end{equation}\) such that \(\begin{equation}U_P f < \epsilon\end{equation}\),

3. zero is the only lower Riemann sum for \(\begin{equation}f\end{equation}\) on \(\begin{equation}[0,1]\end{equation}\).
. with help from class. Part 3 is never addressed. Lucas presents P29 P29
Use the definition of tangent to show that if \(\begin{equation}f\end{equation}\) is a function whose domain includes \(\begin{equation}(-1,1),\end{equation}\) and for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}(-1,1)\end{equation}\), \(\begin{equation}{-x^2 \le f(x) \le x^2}\end{equation}\), then the x-axis is tangent to \(\begin{equation}f\end{equation}\) at the point \(\begin{equation}(0,0)\end{equation}\).
I comment on P29 P29
Use the definition of tangent to show that if \(\begin{equation}f\end{equation}\) is a function whose domain includes \(\begin{equation}(-1,1),\end{equation}\) and for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}(-1,1)\end{equation}\), \(\begin{equation}{-x^2 \le f(x) \le x^2}\end{equation}\), then the x-axis is tangent to \(\begin{equation}f\end{equation}\) at the point \(\begin{equation}(0,0)\end{equation}\).
, but probably didn't need to and should have foreshadowed upcoming work.

Diary 2.8.12

Tri tries T34 T34
. Let \(\begin{equation}f(x)=0\end{equation}\) for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}[0,1]\end{equation}\) except \(\begin{equation}x=0\end{equation}\), and let \(\begin{equation}f(0)=1\end{equation}\). Show that:

1. if \(\begin{equation}P\end{equation}\) is a partition of \(\begin{equation}[0,1]\end{equation}\), then \(\begin{equation}0 < U_P f,\end{equation}\)

2. if \(\begin{equation}\epsilon > 0,\end{equation}\) then there is a partition \(\begin{equation}P\end{equation}\) of \(\begin{equation}[0,1]\end{equation}\) such that \(\begin{equation}U_P f < \epsilon\end{equation}\),

3. zero is the only lower Riemann sum for \(\begin{equation}f\end{equation}\) on \(\begin{equation}[0,1]\end{equation}\).
. There are not enough volunteers for problems and I let them know. The disadvantage of allowing only board work is that they simply won't do enough and the only consequence is low grades at the end of the semester. In the past with only a small group of five-to-eight in the second semester this has worked well, but with larger classes, I will need to continue with midterm, final and write-ups. After making clear my displeasure, I passed out MathNerds pencils to use only on my class and told them that part of their grade would be how much of the pencil they had worn down at the end of the semester. (Commentary: Sure enough, two of them turned them in taped to their portfolios!) If there is not progress at a faster pace next week, I'll make them come in for grade reports. Next class we need to address T34 T34
. Let \(\begin{equation}f(x)=0\end{equation}\) for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}[0,1]\end{equation}\) except \(\begin{equation}x=0\end{equation}\), and let \(\begin{equation}f(0)=1\end{equation}\). Show that:

1. if \(\begin{equation}P\end{equation}\) is a partition of \(\begin{equation}[0,1]\end{equation}\), then \(\begin{equation}0 < U_P f,\end{equation}\)

2. if \(\begin{equation}\epsilon > 0,\end{equation}\) then there is a partition \(\begin{equation}P\end{equation}\) of \(\begin{equation}[0,1]\end{equation}\) such that \(\begin{equation}U_P f < \epsilon\end{equation}\),

3. zero is the only lower Riemann sum for \(\begin{equation}f\end{equation}\) on \(\begin{equation}[0,1]\end{equation}\).
and I need to talk about T31 T31
If \(\begin{equation}f\end{equation}\) is a function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is a closed point set.
, T32 T32
If \(\begin{equation}f\end{equation}\) is a function with domain \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is bounded.
, T33 T33
If \(\begin{equation}f\end{equation}\) is a continuous function with domain \(\begin{equation}[a,b]\end{equation}\), then there is a number \(\begin{equation}x \in [a,b]\end{equation}\) such that if \(\begin{equation}t \in [a,b]\end{equation}\), then \(\begin{equation}f(t) \le f(x)\end{equation}\).
to foreshadow these important results.

Diary 2.10.12

Tri is given help at the board and completes T34 T34
. Let \(\begin{equation}f(x)=0\end{equation}\) for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}[0,1]\end{equation}\) except \(\begin{equation}x=0\end{equation}\), and let \(\begin{equation}f(0)=1\end{equation}\). Show that:

1. if \(\begin{equation}P\end{equation}\) is a partition of \(\begin{equation}[0,1]\end{equation}\), then \(\begin{equation}0 < U_P f,\end{equation}\)

2. if \(\begin{equation}\epsilon > 0,\end{equation}\) then there is a partition \(\begin{equation}P\end{equation}\) of \(\begin{equation}[0,1]\end{equation}\) such that \(\begin{equation}U_P f < \epsilon\end{equation}\),

3. zero is the only lower Riemann sum for \(\begin{equation}f\end{equation}\) on \(\begin{equation}[0,1]\end{equation}\).
. Weston discusses P31 P31
Does there exist a function \(\begin{equation}f\end{equation}\) defined and continuous on \(\begin{equation}[0,1]\end{equation}\) such that \(\begin{equation}f(0)=0\end{equation}\) and \(\begin{equation}f(1)=1\end{equation}\) and \(\begin{equation}f'(x)=0\end{equation}\) at all but countably many points of \(\begin{equation}[0,1]\end{equation}\)?
, which he is making nice progress on having programmed his Ipad(!) to show limits of sequences of functions! Aaron takes a look at T27 T27
If \(\begin{equation}f\end{equation}\) is a function having domain \(\begin{equation}[a,b]\end{equation}\), \(\begin{equation}x \in (a,b)\end{equation}\), \(\begin{equation}f(x) \ge f(t)\end{equation}\) for all \(\begin{equation}t \in (a,b)\end{equation}\), and \(\begin{equation}f\end{equation}\) has a derivative at \(\begin{equation}x\end{equation}\), then \(\begin{equation}f'(x)=0\end{equation}\).
which he has a good understanding of and I made a mess of guiding him in the right direction, not having looked carefully at the problem before I went to the board to help.

Diary 2.13.12

I was at a medical appointment and class worked on T26 T26
If \(\begin{equation}f\end{equation}\) is a function, \(\begin{equation}x\end{equation}\) is in the domain of \(\begin{equation}f\end{equation}\), and \(\begin{equation}f\end{equation}\) has a derivative at \(\begin{equation}(x,f(x))\end{equation}\), then \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}(x,f(x))\end{equation}\).
and T27 T27
If \(\begin{equation}f\end{equation}\) is a function having domain \(\begin{equation}[a,b]\end{equation}\), \(\begin{equation}x \in (a,b)\end{equation}\), \(\begin{equation}f(x) \ge f(t)\end{equation}\) for all \(\begin{equation}t \in (a,b)\end{equation}\), and \(\begin{equation}f\end{equation}\) has a derivative at \(\begin{equation}x\end{equation}\), then \(\begin{equation}f'(x)=0\end{equation}\).
.

Diary 2.15.12

All day went to one problem with Katie trying to prove T28 T28
If \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a sequence of points in the closed interval \(\begin{equation}[a,b]\end{equation}\), then there is a point in \(\begin{equation}[a,b]\end{equation}\) which is not in the sequence
\(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\).
. We discussed and I gave a hint.

Diary 2.17.12

Lucas gave a very nice, partial argument for T26 T26
If \(\begin{equation}f\end{equation}\) is a function, \(\begin{equation}x\end{equation}\) is in the domain of \(\begin{equation}f\end{equation}\), and \(\begin{equation}f\end{equation}\) has a derivative at \(\begin{equation}(x,f(x))\end{equation}\), then \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}(x,f(x))\end{equation}\).
which definitely can be turned into a complete argument. I then talked about the tools that might help with T31 T31
If \(\begin{equation}f\end{equation}\) is a function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is a closed point set.
and T32 T32
If \(\begin{equation}f\end{equation}\) is a function with domain \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is bounded.
with the intent of introducing subsequences. However, after an introduction to the result that the Intermediate Value Theorem along with T31 T31
If \(\begin{equation}f\end{equation}\) is a function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is a closed point set.
and T32 T32
If \(\begin{equation}f\end{equation}\) is a function with domain \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is bounded.
can yield, I was out of time!

Diary 2.20.12

Rabbi attempted T26 T26
If \(\begin{equation}f\end{equation}\) is a function, \(\begin{equation}x\end{equation}\) is in the domain of \(\begin{equation}f\end{equation}\), and \(\begin{equation}f\end{equation}\) has a derivative at \(\begin{equation}(x,f(x))\end{equation}\), then \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}(x,f(x))\end{equation}\).
which Lucas had given a good argument for previously using the geometric definition of derivative. Lucas wanted to see an argument using the epsilon-delta definition, and I knew Rabbi was attempting that approach, so I thought it worth seeing the same theorem twice. Rabbi is close and I allow a lot of time, hoping for class discussion or for him to make progress at the board. When that does not occur, I offer a suggestion on how I would attempt to find the delta that will work. Students had been struggling with T11 T11
If \(\begin{equation}M\end{equation}\) has \(\begin{equation}p\end{equation}\) as a limit point, then there exists either an increasing or a decreasing sequence of points of \(\begin{equation}M\end{equation}\) converging to \(\begin{equation}p.\end{equation}\)
and T30 T30
If \(\begin{equation}f\end{equation}\) is a function with domain \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is a closed point set.
, so I introduced subsequences as a tool that might help and got good interaction. In fact, Kim raised a question that I was planning to answer. I wanted them to know that if a sequence converges, then any subsequence also must converge to that same number. This theorem helps a lot with T31 T31
If \(\begin{equation}f\end{equation}\) is a function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is a closed point set.
and T32 T32
If \(\begin{equation}f\end{equation}\) is a function with domain \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is bounded.
. Thus, even though we weren't to sequences, I wanted to introduce them to this group. Also, we needed some easier problems!

Commentary 2.20.12

This class really started off slowly in the second semester. Not only did I lose two strong students, but I made the mistake of not doing weekly write-ups and eliminating exams. Lastly, we had ended the last semester stuck on T28 T28
If \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a sequence of points in the closed interval \(\begin{equation}[a,b]\end{equation}\), then there is a point in \(\begin{equation}[a,b]\end{equation}\) which is not in the sequence
\(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\).
, T29 T29
If \(\begin{equation}x_1, x_2, x_3, \dots\end{equation}\) is a sequence of distinct points in the closed interval \(\begin{equation}[a,b],\end{equation}\) then the range of the sequence has a limit point.
and T30 T30
If \(\begin{equation}f\end{equation}\) is a function with domain \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is a closed point set.
all difficult applications of continuity. In the past, with smaller classes, no exams or written work had worked fine. However, as I mention in Course Observations many of my best students this year had serious personal life-issues to deal with. Thus, this clip marks the beginning of my attempt to accelerate the class. What I did was to move away from several hard problems on continuity and move toward some easier material on sequences. Over the next few days when this was not immediately successful, I passed out focused assignments to get everyone working on the exact same material. See the Additional Resources worksheets for the materials that I handed out at regularly for the next few weeks to re-focus everyone on the material that I felt we really needed to cover. If there is a lesson here, it is that teaching is fluid and what works in one class will not always work in the next. The more autonomy one gives the students, the more possibilities there are and the more we need to be ready to adapt to the possibilities!

Diary 2.22.12

Weston attempted P31 P31
Does there exist a function \(\begin{equation}f\end{equation}\) defined and continuous on \(\begin{equation}[0,1]\end{equation}\) such that \(\begin{equation}f(0)=0\end{equation}\) and \(\begin{equation}f(1)=1\end{equation}\) and \(\begin{equation}f'(x)=0\end{equation}\) at all but countably many points of \(\begin{equation}[0,1]\end{equation}\)?
and basically can prove pointwise convergence of a sequence of functions that he created, \(\begin{equation}f_n(x) = \lfloor{nx}\rfloor / n\end{equation}\) converges to \(\begin{equation}g(x)=x\end{equation}\). Then Tri attempted T32 T32
If \(\begin{equation}f\end{equation}\) is a function with domain \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is bounded.
and hopefully has a better understanding of continuity and unbounded functions. I tried to point them in the right direction. I proved that the subsequence of the odd subscripted terms of a convergent sequence converges.

Diary 2.24.12

Weston again discussed P31 P31
Does there exist a function \(\begin{equation}f\end{equation}\) defined and continuous on \(\begin{equation}[0,1]\end{equation}\) such that \(\begin{equation}f(0)=0\end{equation}\) and \(\begin{equation}f(1)=1\end{equation}\) and \(\begin{equation}f'(x)=0\end{equation}\) at all but countably many points of \(\begin{equation}[0,1]\end{equation}\)?
and I introduced uniform convergence, because Weston essentially discovered the notion of uniform convergence and, without realizing it, was trying to prove uniform convergence, not point-wise convergence. On Monday, I'm going to pull Lucas out of the class and make him visit me separately. Today I'm going to put on the web site the following message. "Last semester, each of you demonstrated to me that you have mathematical talent and it disappoints me greatly not to see the fruits of your work last semester developing into successful proofs this semester. I believe it will help for the class to focus on fewer problems at a time. Therefore, until they are resolved, I want everyone working on T28 T28
If \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a sequence of points in the closed interval \(\begin{equation}[a,b]\end{equation}\), then there is a point in \(\begin{equation}[a,b]\end{equation}\) which is not in the sequence
\(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\).
, T29 T29
If \(\begin{equation}x_1, x_2, x_3, \dots\end{equation}\) is a sequence of distinct points in the closed interval \(\begin{equation}[a,b],\end{equation}\) then the range of the sequence has a limit point.
& T30 T30
If \(\begin{equation}f\end{equation}\) is a function with domain \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is a closed point set.
. I want everyone to stop by my office next week to get their current grade." This type of statement is antithetical to Moore method teaching, but I feel I have no choice; one-third of the way through the semester, we appear to be making minimal progress, even as I know that some individuals are working hard.

Commentary 2.24.12

Over the next year, I hope to experiment with a wiki that will replace both my blog and my facebook page. My goal is to create a digital, social classroom where three things can happen. (1) Each student can have an individual workspace where s/he does all work electronically in something like scribtex so I can track and edit each students progress. This will enable me to see the progress of each student individually and do early interventions if needed. (2) I will have the ability to push one student's attempted solution out to all the students on the day before it is presented, so students have the opportunity to view the work and prepare questions. (3) All students will have a common area where questions about past material may be discussed.

Diary 2.27.12

Katie attempted T28 T28
If \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a sequence of points in the closed interval \(\begin{equation}[a,b]\end{equation}\), then there is a point in \(\begin{equation}[a,b]\end{equation}\) which is not in the sequence
\(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\).
, but has a ways to go. Tri attempted T28 T28
If \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a sequence of points in the closed interval \(\begin{equation}[a,b]\end{equation}\), then there is a point in \(\begin{equation}[a,b]\end{equation}\) which is not in the sequence
\(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\).
, and now fully understands the intuitive statement of the problem. I went to the board and gave them everything they needed except one piece. I told them how to construct the nested sequence of intervals, how to make their lengths tend to zero and only left out that the first one needed to exclude \(\begin{equation}x_1\end{equation}\), the second one needed to exclude \(\begin{equation}x_1\end{equation}\) and \(\begin{equation}x_2\end{equation}\), etc. I even hinted at this. Hopefully we'll have a proof by Wednesday.

Diary 2.29.12

John attempted T28 T28
If \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a sequence of points in the closed interval \(\begin{equation}[a,b]\end{equation}\), then there is a point in \(\begin{equation}[a,b]\end{equation}\) which is not in the sequence
\(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\).
, yet assumed that in his construction, one of the open intervals must contain only one point. Weston attempted T28 T28
If \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a sequence of points in the closed interval \(\begin{equation}[a,b]\end{equation}\), then there is a point in \(\begin{equation}[a,b]\end{equation}\) which is not in the sequence
\(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\).
, but his language was not precise and I had to work to get him to precisely define a sequence of intervals. Now, I believe that the class understands what must be shown and will do so Friday. Therefore, rather than waste the time or introduce something that might take away their focus, I tied the definitions of continuity and differentiability from calculus to our notation and showed how one might show in calculus our T26 T26
If \(\begin{equation}f\end{equation}\) is a function, \(\begin{equation}x\end{equation}\) is in the domain of \(\begin{equation}f\end{equation}\), and \(\begin{equation}f\end{equation}\) has a derivative at \(\begin{equation}(x,f(x))\end{equation}\), then \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}(x,f(x))\end{equation}\).
, that functions differentiable at a point are continuous at that point.

Diary 3.2.12

Tri answered Kimberly’s Question QUESTION
If \(\begin{equation}(q_n)\end{equation}\) is subsequence of \(\begin{equation}(p_n)\end{equation}\) and \(\begin{equation}(p_n) \rightarrow x\end{equation}\), does \(\begin{equation}(q_n) \rightarrow x\end{equation}\)?
. I spent considerable time making sure the class understood the proof. Then Kimberly attempted T30 T30
If \(\begin{equation}f\end{equation}\) is a function with domain \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is a closed point set.
again and I spent considerable time trying to show how to use the completeness axiom to finish it.

Diary 3.5.12

Aaron presented T29 T29
If \(\begin{equation}x_1, x_2, x_3, \dots\end{equation}\) is a sequence of distinct points in the closed interval \(\begin{equation}[a,b],\end{equation}\) then the range of the sequence has a limit point.
. Although not complete, I allowed him to simply write up what was left and put it up on our Facebook page. Then Rabbi returned to T26 T26
If \(\begin{equation}f\end{equation}\) is a function, \(\begin{equation}x\end{equation}\) is in the domain of \(\begin{equation}f\end{equation}\), and \(\begin{equation}f\end{equation}\) has a derivative at \(\begin{equation}(x,f(x))\end{equation}\), then \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}(x,f(x))\end{equation}\).
, and gave a valid \(\begin{equation}\epsilon - \delta\end{equation}\) proof. I challenged him on one step, but with the help of Weston, he was able to convince me after class that it was correct. Again, I asked for the final proof to go on Facebook after he clarified the last step.

Diary 3.7.12

No video today; videographer Brandy is sick. Shaymal attempted T30 T30
If \(\begin{equation}f\end{equation}\) is a function with domain \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is a closed point set.
, but has a long way to go. John put up SS2 SS2
Suppose that \(\begin{equation}q_1, q_2, q_3, \dots\end{equation}\) is a subsequence of \(\begin{equation}p_1, p_2, p_3, \dots\end{equation}\) and there is a number \(\begin{equation}x\end{equation}\) so that \(\begin{equation}q_1, q_2, q_3, \dots\end{equation}\) converges to \(\begin{equation}x.\end{equation}\) Is it true that \(\begin{equation}p_1, p_2, p_3, \dots\end{equation}\) converges to \(\begin{equation}x\end{equation}\)?
by giving an example of a sequence that did not converge, but had convergent subsequences.

Diary 3.9.12

No one has had success with T31 T31
If \(\begin{equation}f\end{equation}\) is a function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is a closed point set.
, T32 T32
If \(\begin{equation}f\end{equation}\) is a function with domain \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is bounded.
, & T33 T33
If \(\begin{equation}f\end{equation}\) is a continuous function with domain \(\begin{equation}[a,b]\end{equation}\), then there is a number \(\begin{equation}x \in [a,b]\end{equation}\) such that if \(\begin{equation}t \in [a,b]\end{equation}\), then \(\begin{equation}f(t) \le f(x)\end{equation}\).
so I gave a sketch of T31 T31
If \(\begin{equation}f\end{equation}\) is a function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is a closed point set.
after Weston wrote his idea on the board. Watching my father, I realized how good he was at leading students at the board, but I don't seem able to do this as well. And I won't give hints or lectures on T32 T32
If \(\begin{equation}f\end{equation}\) is a function with domain \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is bounded.
& T33 T33
If \(\begin{equation}f\end{equation}\) is a continuous function with domain \(\begin{equation}[a,b]\end{equation}\), then there is a number \(\begin{equation}x \in [a,b]\end{equation}\) such that if \(\begin{equation}t \in [a,b]\end{equation}\), then \(\begin{equation}f(t) \le f(x)\end{equation}\).
, but rather, will just wait and see if they can get them. Lucas, who I pulled out of the class because he has already proved all of these, all the basic set theory and is moving through the integration. While he could present these to the class, I don't think that it would be as valuable to them as having them continue to work. And of course, I am paying the price by watching the class move more slowly.

Spring Break

Diary 3.19.12

Work got very busy, no entry.

Diary 3.21.12

On this day, Shaymal put up T32 T32
If \(\begin{equation}f\end{equation}\) is a function with domain \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is bounded.
using a standard argument. To earn credit for the problem, all aspects must be presented fully, so I simply helped state the two lemmas that are needed to complete the problem. If in a future class meeting, he resolves these, then he will get full credit for the problem.

Diary 3.23.12

I missed class to deliver talk and class conducts itself in my absence, probably better than when I attend. Aaron does a good job on T35 T35
If f is a bounded function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\), and P is a partition of \(\begin{equation}[a,b]\end{equation}\), then \(\begin{equation}L_ P (f) \le U_P (f)\end{equation}\).
and Tri makes a good start on SS4 SS4
The sequence \(\begin{equation}p_1, p_2, p_3, \dots\end{equation}\) is a Cauchy sequence if and only if it is true that for each positive number \(\begin{equation}\epsilon\end{equation}\), there is a positive integer \(\begin{equation}N\end{equation}\) such that if \(\begin{equation}n\end{equation}\) is a positive integer and \(\begin{equation}n \ge N\end{equation}\),
then \(\begin{equation}|p_n-p_N| < \epsilon\end{equation}\).
.

Commentary 3.23.12

This clip shows a good example of how a class can be productive without the professor present. When I watched this clip in preparing this web site I thought, "wow, I'm doing a really good job of keeping my mouth shut and they are doing great." Then I realized that I was absent! One nice thing about putting the responsibility of presentation on the students is that when I am unable to attend class, they conduct it productively. I'm particularly impressed about how they force a very carefully written argument of T35 T35
If f is a bounded function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\), and P is a partition of \(\begin{equation}[a,b]\end{equation}\), then \(\begin{equation}L_ P (f) \le U_P (f)\end{equation}\).
in my absence making him clearly state his hypothesis. And he fields all these questions gracefully and without frustration, simply improving the writing throughout the discussion. Aaron even steps back to the board to help erase it before Tri attempts SS4 SS4
The sequence \(\begin{equation}p_1, p_2, p_3, \dots\end{equation}\) is a Cauchy sequence if and only if it is true that for each positive number \(\begin{equation}\epsilon\end{equation}\), there is a positive integer \(\begin{equation}N\end{equation}\) such that if \(\begin{equation}n\end{equation}\) is a positive integer and \(\begin{equation}n \ge N\end{equation}\),
then \(\begin{equation}|p_n-p_N| < \epsilon\end{equation}\).
, a job I often do to show respect and support for the presenter. While the remainder of the class goes to them working jointly on SS4 SS4
The sequence \(\begin{equation}p_1, p_2, p_3, \dots\end{equation}\) is a Cauchy sequence if and only if it is true that for each positive number \(\begin{equation}\epsilon\end{equation}\), there is a positive integer \(\begin{equation}N\end{equation}\) such that if \(\begin{equation}n\end{equation}\) is a positive integer and \(\begin{equation}n \ge N\end{equation}\),
then \(\begin{equation}|p_n-p_N| < \epsilon\end{equation}\).
without success, I believe that they do nail down the issues before they leave. Tri did not take the first semester, having just joined Lamar after earning a BS in math from a nationally known R1 university. While we may not cover as much as R1 universities, Lamar's undergraduates can prove theorems independently before they graduate. If you were admitting students to a graduate program, would you rather they have seen all the theorems in Rudin's "Principles of Analysis" or that they could prove half the theorems in it independently? Two class periods later, Tri does successfully present SS4 SS4
The sequence \(\begin{equation}p_1, p_2, p_3, \dots\end{equation}\) is a Cauchy sequence if and only if it is true that for each positive number \(\begin{equation}\epsilon\end{equation}\), there is a positive integer \(\begin{equation}N\end{equation}\) such that if \(\begin{equation}n\end{equation}\) is a positive integer and \(\begin{equation}n \ge N\end{equation}\),
then \(\begin{equation}|p_n-p_N| < \epsilon\end{equation}\).
, so this class was successful, if we measure success based on the final outcome of student understanding. I can always gauge the success of a Moore Method course by whether I feel confident that leaving them in charge of the class, mathematics will be presented, discussed and the validity determined. Maria Montessori wrote, "The greatest sign of success for a teacher is to be able to say, 'The children are now working as if I did not exist.'" If we accept this as valid, then I can say we are having a successful class. Except for the first few weeks of a course, I leave all my classes to work independently when I travel. I request only that each student send me an email report of what was presented and whether they consider it resolved. Some emails coming in are simply a list of problems, while others are essays! They are always enjoyable to read and always leave me with a sense of what was accomplished in class the day I was gone.

Diary 3.26.12

Mostly me lecturing and clearing up SS4 SS4
The sequence \(\begin{equation}p_1, p_2, p_3, \dots\end{equation}\) is a Cauchy sequence if and only if it is true that for each positive number \(\begin{equation}\epsilon\end{equation}\), there is a positive integer \(\begin{equation}N\end{equation}\) such that if \(\begin{equation}n\end{equation}\) is a positive integer and \(\begin{equation}n \ge N\end{equation}\),
then \(\begin{equation}|p_n-p_N| < \epsilon\end{equation}\).
which I had missed while delivering a talk. Then we discussed T33 T33
If \(\begin{equation}f\end{equation}\) is a continuous function with domain \(\begin{equation}[a,b]\end{equation}\), then there is a number \(\begin{equation}x \in [a,b]\end{equation}\) such that if \(\begin{equation}t \in [a,b]\end{equation}\), then \(\begin{equation}f(t) \le f(x)\end{equation}\).
which Aaron probably has, but left at home. Then I talked about the next upcoming integration problem T37 T37
If \(\begin{equation}f\end{equation}\) is a bounded function with domain \(\begin{equation}[a,b]\end{equation}\), and for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}[a,b]\end{equation}\), \(\begin{equation}f(x) \ge 0\end{equation}\), and for some number \(\begin{equation}z\end{equation}\) in \(\begin{equation}[a,b]\end{equation}\), \(\begin{equation}f(z) > 0\end{equation}\) and \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}z\end{equation}\), then \(\begin{equation}\displaystyle _U \int_a^b f>0\end{equation}\).
. I misstated the hypothesis, then caught my mistake and corrected it at the board, offering up a lemma lemma
If \(\begin{equation}f\end{equation}\) is a function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\), and there is a sequence \(\begin{equation}x_1, x_2, \dots\end{equation}\) of points in \(\begin{equation}[a,b]\end{equation}\) converging to the point, \(\begin{equation}c\end{equation}\), and for each positive integer \(\begin{equation}n\end{equation}\) we have that \(\begin{equation}f(x)=n\end{equation}\), then \(\begin{equation}f\end{equation}\) is not continuous at \(\begin{equation}(c,f(c))\end{equation}\).
that will be helpful in proving it. The lemma lemma
If \(\begin{equation}f\end{equation}\) is a function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\), and there is a sequence \(\begin{equation}x_1, x_2, \dots\end{equation}\) of points in \(\begin{equation}[a,b]\end{equation}\) converging to the point, \(\begin{equation}c\end{equation}\), and for each positive integer \(\begin{equation}n\end{equation}\) we have that \(\begin{equation}f(x)=n\end{equation}\), then \(\begin{equation}f\end{equation}\) is not continuous at \(\begin{equation}(c,f(c))\end{equation}\).
should probably be in the notes before this theorem.

Diary 3.28.12

I overstepped my speaking bounds! Tri presented SS4 SS4
The sequence \(\begin{equation}p_1, p_2, p_3, \dots\end{equation}\) is a Cauchy sequence if and only if it is true that for each positive number \(\begin{equation}\epsilon\end{equation}\), there is a positive integer \(\begin{equation}N\end{equation}\) such that if \(\begin{equation}n\end{equation}\) is a positive integer and \(\begin{equation}n \ge N\end{equation}\),
then \(\begin{equation}|p_n-p_N| < \epsilon\end{equation}\).
and then Katie does part of P87.1 P87.1
Assume that \(\begin{equation}\Lambda\end{equation}\) is a set and that \(\begin{equation}A_{\lambda}\end{equation}\) is a set for each \(\begin{equation}\lambda \in \Lambda\end{equation}\). Show that:

1. \(\begin{equation}\left( \bigcup_{\lambda \in \Lambda} A_\lambda \right)^c = \bigcap_{\lambda \in \Lambda} \left(A_\lambda\right)^c\end{equation}\),

2. \(\begin{equation}\left( \bigcap_{\lambda \in \Lambda} A_\lambda \right)^c= \bigcup_{\lambda \in \Lambda} \left( A_\lambda \right)^c\end{equation}\),

3. \(\begin{equation}A \cap (\bigcup_{\lambda \in \Lambda} A_\lambda) = \bigcup_{\lambda \in \Lambda} (A \cap A_\lambda)\end{equation}\),

4. \(\begin{equation}A \cup (\bigcap_{\lambda \in \Lambda} A_\lambda) = \bigcap_{\lambda \in \Lambda} (A \cup A_\lambda)\end{equation}\).
. This shows the importance of having problems at every level.

Diary 3.30.12

Good class! Three complete proofs, discussion before class. Katie completes P87.1 P87.1
Assume that \(\begin{equation}\Lambda\end{equation}\) is a set and that \(\begin{equation}A_{\lambda}\end{equation}\) is a set for each \(\begin{equation}\lambda \in \Lambda\end{equation}\). Show that:

1. \(\begin{equation}\left( \bigcup_{\lambda \in \Lambda} A_\lambda \right)^c = \bigcap_{\lambda \in \Lambda} \left(A_\lambda\right)^c\end{equation}\),

2. \(\begin{equation}\left( \bigcap_{\lambda \in \Lambda} A_\lambda \right)^c= \bigcup_{\lambda \in \Lambda} \left( A_\lambda \right)^c\end{equation}\),

3. \(\begin{equation}A \cap (\bigcup_{\lambda \in \Lambda} A_\lambda) = \bigcup_{\lambda \in \Lambda} (A \cap A_\lambda)\end{equation}\),

4. \(\begin{equation}A \cup (\bigcap_{\lambda \in \Lambda} A_\lambda) = \bigcap_{\lambda \in \Lambda} (A \cup A_\lambda)\end{equation}\).
. Shaymal presents T36 T36
If \(\begin{equation}f\end{equation}\) is bounded on \(\begin{equation}[a,b]\end{equation}\) then the set of all Riemann sums of \(\begin{equation}f\end{equation}\) is bounded.
. Rabbi presents SS6 SS6
If the sequence \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) converges to a point \(\begin{equation}x\end{equation}\), then \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a Cauchy
sequence.
.

Commentary 3.30.12

Finally, half way through the semester, I have this class working in good order and productively, as this clip shows. Additionally, this clip illustrates how important it is to have problems at every level of difficulty. The majority of problems should be at or just above the average ability of the class. Still, it is good to have some problems that reinforce previous concepts or introduce new, but elementary concepts. In this class, Katie presents P87.1 P87.1
Assume that \(\begin{equation}\Lambda\end{equation}\) is a set and that \(\begin{equation}A_{\lambda}\end{equation}\) is a set for each \(\begin{equation}\lambda \in \Lambda\end{equation}\). Show that:

1. \(\begin{equation}\left( \bigcup_{\lambda \in \Lambda} A_\lambda \right)^c = \bigcap_{\lambda \in \Lambda} \left(A_\lambda\right)^c\end{equation}\),

2. \(\begin{equation}\left( \bigcap_{\lambda \in \Lambda} A_\lambda \right)^c= \bigcup_{\lambda \in \Lambda} \left( A_\lambda \right)^c\end{equation}\),

3. \(\begin{equation}A \cap (\bigcup_{\lambda \in \Lambda} A_\lambda) = \bigcup_{\lambda \in \Lambda} (A \cap A_\lambda)\end{equation}\),

4. \(\begin{equation}A \cup (\bigcap_{\lambda \in \Lambda} A_\lambda) = \bigcap_{\lambda \in \Lambda} (A \cup A_\lambda)\end{equation}\).
. which are essentially the basic set theory of DeMorgan's Laws for uncountable unions and intersections of sets, along with the interesting properties of inverse functions. Even though many students have seen these, they are nice to have refreshed before graduate school.

Diary 4.2.12

One presentation by Shaymal which used a definition of continuity that we didn't have for uniform continuity on the interval. And had some notational difficulties. Then John attempted SS9 SS9
If \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a Cauchy sequence, then the sequence \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) converges to some point.
, but without success. Then Weston looked at P31 P31
Does there exist a function \(\begin{equation}f\end{equation}\) defined and continuous on \(\begin{equation}[0,1]\end{equation}\) such that \(\begin{equation}f(0)=0\end{equation}\) and \(\begin{equation}f(1)=1\end{equation}\) and \(\begin{equation}f'(x)=0\end{equation}\) at all but countably many points of \(\begin{equation}[0,1]\end{equation}\)?
again, but still has not handle on how to tackle the problem. Some days are diamonds, some are stones.

Diary 4.6.12, no class on Good Friday

Work remains very busy, no entry.

Diary 4.9.12

Katie presents more of P87 P87
Assume that \(\begin{equation}\Lambda\end{equation}\) is a set and that \(\begin{equation}A_{\lambda}\end{equation}\) is a set for each \(\begin{equation}\lambda \in \Lambda\end{equation}\). Show that:

1. \(\begin{equation}\left( \bigcup_{\lambda \in \Lambda} A_\lambda \right)^c = \bigcap_{\lambda \in \Lambda} \left(A_\lambda\right)^c\end{equation}\),

2. \(\begin{equation}\left( \bigcap_{\lambda \in \Lambda} A_\lambda \right)^c= \bigcup_{\lambda \in \Lambda} \left( A_\lambda \right)^c\end{equation}\),

3. \(\begin{equation}A \cap (\bigcup_{\lambda \in \Lambda} A_\lambda) = \bigcup_{\lambda \in \Lambda} (A \cap A_\lambda)\end{equation}\),

4. \(\begin{equation}A \cup (\bigcap_{\lambda \in \Lambda} A_\lambda) = \bigcap_{\lambda \in \Lambda} (A \cup A_\lambda)\end{equation}\).
. Rabbi attempts SS8 SS8
If \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a Cauchy sequence, then the set \(\begin{equation}\{p_1,p_2,p_3,\dots\}\end{equation}\) does not have two limit points.
. He proceeds by contradiction, but rather than assuming it does not have two limit points, he assumes it does not converge to two different points. He gives a nice proof for what he claims and this is very close to the statement of the problem, but not equivalent. I attempt to see how to help him out and suggest a few ideas, but I'm afraid I did not show a clear path.

Diary 4.11.12

Shaymal presented T48 T48
If \(\begin{equation}f\end{equation}\) is a continuous function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}F\end{equation}\) is the function such that for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}[a,b]\end{equation}\), \(\begin{equation}\displaystyle F(x)=\int_a^x f\end{equation}\), then for each number \(\begin{equation}c \in [a,b]\end{equation}\), \(\begin{equation}F\end{equation}\) has a derivative at \(\begin{equation}c\end{equation}\) and \(\begin{equation}F'(c)~=~f(c)\end{equation}\).
and is quite close, but does not fully understand how to use T47 T47
If f is a continuous function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\), then there is a number \(\begin{equation}c\end{equation}\) in \(\begin{equation}[a,b]\end{equation}\) such that
\(\begin{equation}\displaystyle \int_a^b f = f(c) (b-a)\end{equation}\).
. Aaron presented T33 T33
If \(\begin{equation}f\end{equation}\) is a continuous function with domain \(\begin{equation}[a,b]\end{equation}\), then there is a number \(\begin{equation}x \in [a,b]\end{equation}\) such that if \(\begin{equation}t \in [a,b]\end{equation}\), then \(\begin{equation}f(t) \le f(x)\end{equation}\).
and is quite close, but it still needs work. Thus Rabbi’s version of SS8, Aaron’s T33 T33
If \(\begin{equation}f\end{equation}\) is a continuous function with domain \(\begin{equation}[a,b]\end{equation}\), then there is a number \(\begin{equation}x \in [a,b]\end{equation}\) such that if \(\begin{equation}t \in [a,b]\end{equation}\), then \(\begin{equation}f(t) \le f(x)\end{equation}\).
and Shaymal’s T48 T48
If \(\begin{equation}f\end{equation}\) is a continuous function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}F\end{equation}\) is the function such that for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}[a,b]\end{equation}\), \(\begin{equation}\displaystyle F(x)=\int_a^x f\end{equation}\), then for each number \(\begin{equation}c \in [a,b]\end{equation}\), \(\begin{equation}F\end{equation}\) has a derivative at \(\begin{equation}c\end{equation}\) and \(\begin{equation}F'(c)~=~f(c)\end{equation}\).
are all within epsilon of complete.

Diary 4.13.12

My videographer and I were at the Sectional MAA meeting, so we have no film. Weston worked on T8 T8
If the sequence \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\)
converges to the point \(\begin{equation}x\end{equation}\), then
\(\begin{equation}M=\{p_1,p_2,p_3\dots\}\end{equation}\) is bounded.
and Katie worked on T28 T28
If \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a sequence of points in the closed interval \(\begin{equation}[a,b]\end{equation}\), then there is a point in \(\begin{equation}[a,b]\end{equation}\) which is not in the sequence
\(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\).
.

Diary 4.16.12

Katie presented T28 T28
If \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a sequence of points in the closed interval \(\begin{equation}[a,b]\end{equation}\), then there is a point in \(\begin{equation}[a,b]\end{equation}\) which is not in the sequence
\(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\).
saying that she was not sure of it. But she has a very nice argument, albeit a somewhat burdensome notation. She certainly had a proof which was nice because she has not had a lot of success in the course to date. I think this day shows the need for both easy and hard problems. She has been presenting the easier ones on basic set theory and this may have relaxed her about her grade, enabling her to spend more time on a harder one. Regardless of the reason, she succeeded in showing that the interval is not countable to my satisfaction today. Rabbi and I then co-presented SS8 SS8
If \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a Cauchy sequence, then the set \(\begin{equation}\{p_1,p_2,p_3,\dots\}\end{equation}\) does not have two limit points.
.

Commentary 4.16.12

Today's clip shows how patient I will be in trying to understand a student's proof. Katie supposes that there is a sequence that "fills" the interval and proceeds to create a nested sequence of closed intervals, each excluding one element of a the sequence so that the intersection of them will be a point not in the sequence. Her notation is quite complex, but I take the time to review it carefully and am able to conclude that she does have a valid proof. I ask her to write it up for her portfolio, in the hopes that when she writes it again, she will see a simpler notation. Either way, she has a nice proof and it is one of the more difficult problems that she has presented in the two semesters, showing just how much a student can grow in developing their creativity. She still needs work in her writing, but it is easier to develop writing skills than creativity and confidence in problem solving.

Diary 4.18.12

Work got very busy, no entry.

Diary 4.20.12

Kimberly presented the lemma to T32 T32
If \(\begin{equation}f\end{equation}\) is a function with domain \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is bounded.
, so should have T32 T32
If \(\begin{equation}f\end{equation}\) is a function with domain \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\), then the range of \(\begin{equation}f\end{equation}\) is bounded.
soon. This was a major success for her. Very few people attended (How depressing -- a student actually told me that it’s weed day!), so I introduced measure theory and summarized countability. This day illustrates how I try to tell the "story" of analysis by recapping what they have accomplished and using that to lead into another forthcoming topic.

Commentary 4.20.12

This clip simply shows another student success and how I tend to lecture on "big" ideas, leaving detailed proofs to the students. They discover the trees, I try to show them the forest. I first summarize what we have learned about countability and then try to give them a good grounding in the motivation behind measure theory, including a definition for the outer measure of a set.

Diary 4.23.12

Tri presents his work on T39 T39
If \(\begin{equation}f\end{equation}\) is a bounded function with domain the closed interval \(\begin{equation}[a,b],\end{equation}\) \(\begin{equation}P\end{equation}\) is a partition of \(\begin{equation}[a,b]\end{equation}\), \(\begin{equation}Q\end{equation}\) is a partition of \(\begin{equation}[a,b],\end{equation}\) and \(\begin{equation}Q\end{equation}\) is a refinement of \(\begin{equation}P\end{equation}\),
then \(\begin{equation}L_P(f) \le L_Q(f)\end{equation}\)
and \(\begin{equation}U_P(f) \ge U_Q(f)\end{equation}\).
and has a special case, where the function is increasing. I point out that his argument works for decreasing and functions that are neither increasing nor decreasing and I believe he sees it now. I discuss uniform continuity and continuity, which both Weston and Tri have asked about. Hopefully that will point them in the right direction. I dismiss class early, first time in two semesters, and tell them to get to work.

Diary 4.25.12

Work got very busy, no entry.

Diary 4.27.12

Shaymal presents T49 T49
If \(\begin{equation}f\end{equation}\) is a function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}f\end{equation}\) has a derivative at each point of \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}f'\end{equation}\) is continuous at each point in \(\begin{equation}[a,b]\end{equation}\),
then \(\begin{equation}\displaystyle \int_a^b f' = f(b)-f(a)\end{equation}\).
. Tri presents again for T39 T39
If \(\begin{equation}f\end{equation}\) is a bounded function with domain the closed interval \(\begin{equation}[a,b],\end{equation}\) \(\begin{equation}P\end{equation}\) is a partition of \(\begin{equation}[a,b]\end{equation}\), \(\begin{equation}Q\end{equation}\) is a partition of \(\begin{equation}[a,b],\end{equation}\) and \(\begin{equation}Q\end{equation}\) is a refinement of \(\begin{equation}P\end{equation}\),
then \(\begin{equation}L_P(f) \le L_Q(f)\end{equation}\)
and \(\begin{equation}U_P(f) \ge U_Q(f)\end{equation}\).
. He is still struggling with what constitutes a proof for T39 T39
If \(\begin{equation}f\end{equation}\) is a bounded function with domain the closed interval \(\begin{equation}[a,b],\end{equation}\) \(\begin{equation}P\end{equation}\) is a partition of \(\begin{equation}[a,b]\end{equation}\), \(\begin{equation}Q\end{equation}\) is a partition of \(\begin{equation}[a,b],\end{equation}\) and \(\begin{equation}Q\end{equation}\) is a refinement of \(\begin{equation}P\end{equation}\),
then \(\begin{equation}L_P(f) \le L_Q(f)\end{equation}\)
and \(\begin{equation}U_P(f) \ge U_Q(f)\end{equation}\).
, but he has the idea down now.

Diary 5.2.12

Shaymal puts T37 T37
If \(\begin{equation}f\end{equation}\) is a bounded function with domain \(\begin{equation}[a,b]\end{equation}\), and for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}[a,b]\end{equation}\), \(\begin{equation}f(x) \ge 0\end{equation}\), and or some number \(\begin{equation}z\end{equation}\) in \(\begin{equation}[a,b]\end{equation}\), \(\begin{equation}f(z) > 0\end{equation}\) and \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}z\end{equation}\), then \(\begin{equation}\displaystyle _U \int_a^b f>0\end{equation}\).
using a technique that I did not understand, using some notation that we had not developed, the \(\begin{equation}<<\end{equation}\) notation. When a student uses some notation that I did not understand, I simply ask them to defend their work. In this case he could not and I sat him down.

Diary 5.4.12

This was a very nice class. First we saw a very nice proof of SS9 SS9
If \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) is a Cauchy sequence, then the sequence \(\begin{equation}p_1,p_2,p_3,\dots\end{equation}\) converges to some point.
that every Cauchy sequence converges by Rabbi, who had made several attempts at this previously. Then Weston presented a nice proof of CNT2 CNT2
Suppose that \(\begin{equation}M_1\end{equation}\) and \(\begin{equation}M_2\end{equation}\) are two countable sets. Show that
\(\begin{equation}M_1 \cup M_2\end{equation}\) is a countable set.
that the finite union of countable sets is countable. Finally, Shaymal showed at least one case of the problem that he attempted at the board yesterday using notation that he could not defend.

Commentary 5.4.12

So, here we are at the end of the semester. I'm at the board a lot in this class because I want to be sure that we have these topics wrapped up and understand the statements and proofs of the theorems, as well as the implications of the theorems and how they might see them in the future. Hence, now that we have that the real line is complete, I talk about completeness in Banach spaces, although I don't use that word. This clip is a very nice class. While Rabbi's proof looks like a near picture-perfect proof, this was the result of multiple attempts.

Diary 5.7.12

I pass out evaluations, Weston presents CNT2 CNT2
Suppose that \(\begin{equation}M_1\end{equation}\) and \(\begin{equation}M_2\end{equation}\) are two countable sets. Show that
\(\begin{equation}M_1 \cup M_2\end{equation}\) is a countable set.
and concludes it. Then Shaymal attempts UC3 UC3
Show that there is a set \(\begin{equation}M\end{equation}\) of the reals and a function \(\begin{equation}f\end{equation}\) defined on \(\begin{equation}M\end{equation}\) so that \(\begin{equation}f\end{equation}\) is continuous on \(\begin{equation}M\end{equation}\) but \(\begin{equation}f\end{equation}\) is not uniformly continuous on \(\begin{equation}M\end{equation}\).
again, but fails. Then Shaymal shows T48 T48
If \(\begin{equation}f\end{equation}\) is a continuous function with domain the closed interval \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}F\end{equation}\) is the function such that for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}[a,b]\end{equation}\), \(\begin{equation}\displaystyle F(x)=\int_a^x f\end{equation}\), then for each number \(\begin{equation}c \in [a,b]\end{equation}\), \(\begin{equation}F\end{equation}\) has a derivative at \(\begin{equation}c\end{equation}\) and \(\begin{equation}F'(c)~=~f(c)\end{equation}\).
and succeeds, albeit with a few last minute corrections by me.

Diary 5.9.12, last day of class

Only one student has something to present. John presents the answer to the question of whether there exists a set with the properties that every point is a limit point, the set is closed, and the set contains no interval. While not all the details are there, he does have a good intuition for the problem and fields my questions nicely.

End-of-Course Commentary

On the bright side,

- this was the largest number of students I have ever had in the second semester,
- we completed the basics of analysis including limit points, sequences, Cauchy sequences, subsequences, compactness,
  continuity, differentiability, integrability, uniform continuity and uniform convergence, and
- one student, Lucas, completed almost all the notes, only missing out on the existence and uniqueness section on
  differential equations and the measure theory.

On the other hand,

- we did not cover as much material as previous classes have,
- I spent more time at the board than in previous classes, and
- in general student discussion was less than normal.

I'm quite confident that the same reasons I outlined at the end of the first semester and under Observations explain this. The unusual number of individual student's personal hardships, the camera in the classroom, and the high number of students lacking pre-requisites account for the less productive than usual class. On the other hand, if the goal of the course was to have every student do some mathematics independently, learn to write mathematics, learn to critique the mathematics of others and see a large number of topics in analysis, then I would argue that the course has been successful.