Foundations of Mathematics

W. S. (Bill) Mahavier, Spring 2008, Emory 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.

The Students

His Last Two Honors Students
 His Last Two Honors Students

When viewing the early clips, it is often the case that the name on the board is not the name of the student at the board because the first student going to the board writes her or his name and the second student does not erase that name, but simply starts to work with what is already on the board. The following stills show the students by row on the fourth day of filming after they have settled into the seats they will remain in more-or-less for the rest of the semester. We refer to the students by last name, as Mahavier calls them by last name.



Here are the students in the front row labeled from left to right:
1. Zahn, Joseph Alan
2. Sayanii, Sameer
3. Rasmussen, Tyler Scott
4. DeMitchell, Isabelle
5. Norman, Michael







Here are the students in the back row labeled from left to right:
1. Phelan, Preston Daniel
2. Valero, Alina
3. Herring, James
4. Famber, Wayne
5. Winterhalter, Paul








Videos


You may wish to view the problem sequence or syllabus before venturing into the videos. Hovering your mouse over any problem will show you a statement of that problem. These resources, and more, may be found via the link Additional Resources. We do not have video for the first two days of class and summarize these from Mahavier's diary.

Day 1, Wednesday 1/16/08, (no video available).

Observe how low-level the first mathematics is and how slowly Mahavier starts each class. His goal in the first days of any course is not coverage, but always to create active participants in the classroom by having them attempt to present real mathematics. At the same time, he is creating an environment whereby students are at ease going to the board, asking and answering questions.

After letting the students read the syllabus and discussing the goals of the class, Mahavier discusses Properties P1-P5 PROPERTIES P1-P5
Properties of the number line
P1. If each of \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\) is a number then exactly one of the following is true: \(\begin{equation}a\end{equation}\) is \(\begin{equation}b\end{equation}\), \(\begin{equation}a\end{equation}\) is to the left of \(\begin{equation}b\end{equation}\), or \(\begin{equation}a\end{equation}\) is to the right of \(\begin{equation}b\end{equation}\).
P2. If each of \(\begin{equation}a\end{equation}\), \(\begin{equation}b\end{equation}\), and \(\begin{equation}c\end{equation}\) is a number, and \(\begin{equation}a\end{equation}\) is to the left of \(\begin{equation}b\end{equation}\), and \(\begin{equation}b\end{equation}\) is to the left of \(\begin{equation}c\end{equation}\), then \(\begin{equation}a\end{equation}\) is to the left of \(\begin{equation}c\end{equation}\).
P3. If \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\) are two points on the number line there is a point between them.
P4. If \(\begin{equation}a\end{equation}\) is a point, there is a smallest integer to the right of \(\begin{equation}a\end{equation}\) and a largest integer to the left of \(\begin{equation}a\end{equation}\).
P5 If \(\begin{equation}n\end{equation}\) is an integer then \(\begin{equation}n-1\end{equation}\) and \(\begin{equation}n+1\end{equation}\) are integers and \(\begin{equation}n\end{equation}\) is the only integer between \(\begin{equation}n-1\end{equation}\) and \(\begin{equation}n+1\end{equation}\).
.

He gives the definitions for point set A point set is a set of one or more points. , segment The statement that the point set \(\begin{equation}S\end{equation}\) is a segment means that there are two (different) points \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\), called the endpoints of \(\begin{equation}S\end{equation}\) such that \(\begin{equation}S\end{equation}\) is the set of all points between \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\). and 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 segment that contains \(\begin{equation}p\end{equation}\) contains a point of \(\begin{equation}M\end{equation}\) different from \(\begin{equation}p\end{equation}\). and asks "Is \(\begin{equation}1\end{equation}\) is a limit point of \(\begin{equation}M=\{0\}\end{equation}\)?" which is quickly resolved. He then asks, "If \(\begin{equation}p>0\end{equation}\) can \(\begin{equation}p\end{equation}\) be a limit point of this set?"

This line of inquiry leads to a solution to Problem 1 PROBLEM 1
Show that if \(\begin{equation}M\end{equation}\) is the set which contains only the number \(\begin{equation}0\end{equation}\), and \(\begin{equation}p\end{equation}\) is a point, then \(\begin{equation}p\end{equation}\) is not a limit point of \(\begin{equation}M\end{equation}\).
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At the end of class Mahavier hands out the first two pages of the problem sequence.

Day 2, Friday 1/18/08, (no video available).

The class presents Problem 2 PROBLEM 2
Show that if \(\begin{equation}M\end{equation}\) is the point set that contains only the two points \(\begin{equation}0\end{equation}\) and \(\begin{equation}1\end{equation}\), then no point is a limit point of \(\begin{equation}M\end{equation}\).
, Problem 3 PROBLEM 3
Show that \(\begin{equation}1\end{equation}\) is a limit point of the segment \(\begin{equation}(0,1)\end{equation}\).
, Problem 4 PROBLEM 4
Show that \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\) are limit points of the segment \(\begin{equation}(a,b)\end{equation}\).
,and Problem 5 PROBLEM 5
If \(\begin{equation}S\end{equation}\) is the segment \(\begin{equation}(a,b)\end{equation}\), show that every point of \(\begin{equation}S\end{equation}\) is a limit point of \(\begin{equation}S\end{equation}\).
.

Mahavier dismisses class early.

Monday 1.21.08, Martin Luther King Day, no class

Day 3, Wednesday 1/23/08.

The first few days demonstrate several features that will remain constant throughout the semester. Mahavier is not asking for volunteers, but rather calling on them by name to present the problems, so the presumption is that every student will have something to present every day, hopefully at least the first few outstanding problems. Mahavier calls first on those students who have presented the least. He regularly asks the students if there are any previously presented problems that they wish to see presented again. While he is complimentary of the class as a whole, he challenges any statement at the board that is not clear. When a student puts up only a special case of a problem, he allows the full presentation of this special case and then allows another student to present the full case. Thus he is supportive of any correctly written and stated mathematics, even if it is not exactly the mathematics he expected to see, but demands precise statements of all students.

Mahavier asks if anyone wants to see any previously presented problems and DeMitchell asks to see Problem 4 PROBLEM 4
Show that \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\) are limit points of the segment \(\begin{equation}(a,b)\end{equation}\).
a second time. Mahavier makes every effort at the beginning to assure that no diligent student is left behind. He knows from experience that a slower starting student may turn out to be a high-performing student. Valero re-presents Problem 4 PROBLEM 4
Show that \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\) are limit points of the segment \(\begin{equation}(a,b)\end{equation}\).
. Perhaps because DeMitchell requested to see this problem again, after the presentation Mahavier asks her to negate the definition of 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 segment that contains \(\begin{equation}p\end{equation}\) contains a point of \(\begin{equation}M\end{equation}\) different from \(\begin{equation}p\end{equation}\).. He is gauging DeMitchell's understanding of the definition, reinforcing the importance of negation (in preparation for the negation of the definitions of convergence and continuity), and recapping an important definition. It is often the case that what appears to be a simple, off-the-cuff remark is carefully constructed to be a highly efficient communique addressing the entire class and not simply the individual student.

Presentations thus far have been primarily pictures and some notation. Mahavier emphasizes that when writing up the problems for the weekly written grade, the only rule he'll give is to be sure to define all the letters you use. He is gently nudging them toward writing full proofs at the board, as we will see in the later clips.

The next clip shows how delicately he will work with a student who is struggling. DeMitchell attempts Problem 7 PROBLEM 7
Show that if \(\begin{equation}H\end{equation}\) is a point set, and \(\begin{equation}K\end{equation}\) is a point set, and \(\begin{equation}H\subseteq K\end{equation}\), 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 \(\begin{equation}K\end{equation}\).
. She mistakenly assumes that the sets must be segments The statement that the point set \(\begin{equation}S\end{equation}\) is a segment means that there are two (different) points \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\), called the endpoints of \(\begin{equation}S\end{equation}\) such that \(\begin{equation}S\end{equation}\) is the set of all points between \(\begin{equation}a\end{equation}\) and \(\begin{equation}b\end{equation}\). and shows a valid argument for this special case. Mahavier carefully questions, but does not embarrass her at the board. He allows her to present the special case and then questions the class regarding whether the question was resolved. He then allows another student to give a proof, carefully pointing out exactly when and where DeMitchell used the fact that the sets were intervals.

Day 4, Friday 1.25.08.

The class votes on what day of the week to turn in weekly assignments. Mahavier reminds them to make sure and define any symbols they use before they appear in the write-ups and allows them to write up anything that has been presented in class.

Valero presents Problem 9 PROBLEM 9
Show that if \(\begin{equation}p\end{equation}\) is a limit point of the point set \(\begin{equation}M\end{equation}\) and \(\begin{equation}S\end{equation}\) is a segment containing \(\begin{equation}p\end{equation}\), then \(\begin{equation}S\end{equation}\) contains 2 points of \(\begin{equation}M\end{equation}\).
and is clearly trying hard at the board while being questioned by Mahavier. Eventually, Valero takes a seat and Herring concludes the argument.

A student, Norman, who previously gave a flawless verbal argument of Problem 7 PROBLEM 7
Show that if \(\begin{equation}H\end{equation}\) is a point set, and \(\begin{equation}K\end{equation}\) is a point set, and \(\begin{equation}H\subseteq K\end{equation}\), 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 \(\begin{equation}K\end{equation}\).
struggles to show Problem 10 PROBLEM 10
Let \(\begin{equation}H\end{equation}\) be a point set which has a limit point, and let \(\begin{equation}K\end{equation}\) be the set of all limit points of \(\begin{equation}H\end{equation}\) . Show that if \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\) , then \(\begin{equation}p\end{equation}\) is also a limit point of \(\begin{equation}H\end{equation}\).
.

Day 5, Monday 1.28.08.

We can observe four themes from this day. First we see Mahavier's emphasis on precise language. Second we see that his notes are not static, he adds problems as necessary to facilitate progress. Third we see that students may appear bewildered early on in the course. They don't know exactly what he wants or where he is guiding them, and they are trying to feel their way. Still, they are working. Even if it is not clear how much students at their seats understand of the sketches presented at the board, this will be determined by their weekly written work which will be their first graded feedback. Fourth, we see that seemingly weak students may turn out to be strong. Norman's makes a third unsuccessful attempt at a problem, but becomes a strong student as the course progresses.

Norman makes his third attempt at Problem 10 PROBLEM 10
Let \(\begin{equation}H\end{equation}\) be a point set which has a limit point, and let \(\begin{equation}K\end{equation}\) be the set of all limit points of \(\begin{equation}H\end{equation}\) . Show that if \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\) , then \(\begin{equation}p\end{equation}\) is also a limit point of \(\begin{equation}H\end{equation}\).
and succeeds. During his presentation he states "every segment containing p contains a point z" and Mahavier tackles the imprecise use of language. Certainly there is no point, other than \(\begin{equation}p\end{equation}\) that is in every segment containing \(\begin{equation}p\end{equation}\). What the student means is that "this interval, \(\begin{equation}(x,y)\end{equation}\) contains a point of \(\begin{equation}H\end{equation}\) other than \(\begin{equation}p\end{equation}\)."

Winterhalter presents Problem 12 PROBLEM 12
Show that if \(\begin{equation}p \ne 0\end{equation}\), then \(\begin{equation}p\end{equation}\) is not a limit point of the set of all reciprocals of positive integers.
which states that zero is not a limit point of the set of reciprocals of natural numbers. Notice how Mahavier takes an attempted statement about the reciprocals of consecutive integers and guides Winterhalter to write a precise mathematical statement, encouraging such statements in the forthcoming written work. Note also his willingness to allow Winterhalter at the board to add forgotten cases and modify his work to make it precise and complete. Then, even as the class believes the proof is complete, Mahavier adds a problem, Problem 10.5 (although he probably meant 12.5) , to the notes and wants Winterhalter to complete this problem during the next class period, rather than allow another student, who has already had success, resolve the problem today. Thus, he reserves this problem for Winterhalter for the next class. This shows how Mahavier tailor's the class to individual students. Depending on the level of preparation, what has happened in his office, the apparent confidence of the student, and the student's performance in past classes, Mahavier may allow other students to give guidance, or he may not. He may allow another student to take over the board, or he may not. Finally, he may allow a student to reserve a piece of a problem for the next day as he does in this example.

Even while students are heading for the board, Mahavier is using the class time efficiently by letting students know that he knows who has been to the board and that he wants to get them to the board. After Zahn presents Problem 13 PROBLEM 13
If \(\begin{equation}H\end{equation}\) and \(\begin{equation}K\end{equation}\) are two point sets having a common point, 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 \(\begin{equation}K\end{equation}\).
efficiently, Mahavier recaps and complements the argument. You can see Mahavier's pleasure in this clean, well-prepared presentation, as he recaps it for the benefit of the class.

Day 6, Wednesday 1.30.08.

Winterhalter completes Problem 12 PROBLEM 12
Show that if \(\begin{equation}p \ne 0\end{equation}\), then \(\begin{equation}p\end{equation}\) is not a limit point of the set of all reciprocals of positive integers.
by completing the problem Mahavier stated last class, Problem 10.5. Mahavier observes that Problem 10.5 was false as stated because while the case where \(\begin{equation}p\end{equation}\) is a positive integer was handled during the last class, he did not state that \(\begin{equation}p\end{equation}\) needed to be a positive number that was not a positive integer.

DeMitchell gives a flawless presentation of Problem 15 PROBLEM 15
Suppose that \(\begin{equation}M\end{equation}\) is a point set and every interval containing \(\begin{equation}p\end{equation}\) contains a point of \(\begin{equation}M\end{equation}\) different from \(\begin{equation}p\end{equation}\). Must it be true that \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\)?
. The clear smile on her face after what will turn out to be her first success of many successes in the course may be the best argument for the method. Mahavier's careful concern in his diary at the beginning and special attention to this student begins to pay dividends. In his diary early on he stated that she "needed help." Mahavier uses a powerful combination of asking direct questions to see that each student can defend his or her thoughts, while complimenting successes so that students want to be prepared when they go to the board, but don't appear fearful to go to the board despite his direct questioning.

With only a few minutes left in the class, Mahavier again uses time wisely by working to motivate the students. He passes out a sheet of student responses to the question "if you were going to offer advice to people taking this class for the first time, what would you say to them" and reads his favorite, which was "The best advice I could give would be to be willing to embarrass yourself by presenting work at the board. Often I presented problems I was not terribly confident about and even though I often had a flaw in my proof, I always came away with a better understanding of the problem and I never felt badly or belittled by other students or Dr. Mahavier for making a mistake..." This handout and reading is clearly done in order to motivate the students who are going to the board and failing. He even says that one student previously went to the board ten times before getting a problem right and after that never got one wrong.

Day 7, Friday 2.1.2008.

Problems 17 and 18 demonstrate effectively how Mahavier stair steps difficult problems. From experience, I know that Problem 18 often leads to difficulty as students attempt to prove it directly. Problem 17 sets the stage for an indirect argument of Problem 18 by handling the heart of the matter and planting the seed for an indirect proof.

In Phelan's attempted solution to Problem 18 PROBLEM 18
Show that if \(\begin{equation}H\end{equation}\) and \(\begin{equation}K\end{equation}\) are two point sets 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 \(\begin{equation}K\end{equation}\).
, we see how even when students believe they have a solution, forcing them to write it on the board helps them determine if they have a solution. Mahavier says absolutely nothing and yet the student, who has a neat idea of nesting intervals and infinite sequences, realizes he does not have a solution. Mahavier complements him (both in class and in his diary) and the class moves on.

Norman fails to show Problem 20 PROBLEM 20
Show that if \(\begin{equation}M\end{equation}\) is a point set and \(\begin{equation}M\end{equation}\) is finite then there is a smallest (and a largest) number in \(\begin{equation}M\end{equation}\).
that every finite set has a least element. After he sits down, a long discussion ensues. Mahavier is in no hurry and spends as much time as is needed until the students make progress. The student who finally makes the breakthrough, Sayanii, is struggling in the class. He has been to the board multiple times in the past and has failed to prove anything yet. Here he states several equivalent statements about being able to order the set, which of course one can't do if one can't show there is a least element. The moment he gets it, he rushes to the board to write it down and succeeds. Mathematicians often worry about coverage in Moore Method courses, but if our students can't prove elementary problems then how can we assume they can understand our polished proofs of more technical arguments in advanced undergraduate courses?

Day 8, Monday 2.4.2008.

Today we see the first transition for the class. Because Mahavier has never written more than a few words on the board at anytime, students are now able to write and speak more clearly at the board and from their seats. Herring attempts to present an alternative argument to Problem 21 PROBLEM 21
Show that if \(\begin{equation}M\end{equation}\) is a point set, and \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\), and \(\begin{equation}S\end{equation}\) is a segment containing \(\begin{equation}p\end{equation}\), and \(\begin{equation}n\end{equation}\) is a positive integer, then \(\begin{equation}S\end{equation}\) contains \(\begin{equation}n\end{equation}\) points so that \(\begin{equation}S\cap M\end{equation}\) is infinite.
which Valero has just presented. He starts defining carefully all the variables and making precise mathematical statements. Valero then sees a problem and points out that his argument is not valid. Even though Herring does not see how to correct his proof at the board, he clearly is on the cusp of a complete argument and the ideas he has are sufficient to conclude the theorem.

A lesson in negation... Mahavier is willing to spend 14 minutes of class time on one problem to facilitate understanding negation. The clip starts with Zahn showing Problem 22 PROBLEM 22
Show that the sequence \(\begin{equation}0,1,0,1,...\end{equation}\) does not converge to \(\begin{equation}0\end{equation}\).
which leads to a discussion of bare denial. Valero and Zahn attempt to write a precise statement for "the sequence \(\begin{equation}x_1, x_2, x_3, \dots\end{equation}\) does not converge to \(\begin{equation}p\end{equation}\)". As far as they get is that "the sequence \(\begin{equation}x_1, x_2, x_3 \dots\end{equation}\) does not converge to \(\begin{equation}p\end{equation}\) if there is a segment so that it is not true that there is a positive integer \(\begin{equation}n\end{equation}\) so that \(\begin{equation}x_m\end{equation}\) is in the segment for all \(\begin{equation}m>n\end{equation}\)."

Day 9, Wednesday 2.6.2008.

The transformation from confusion to correct mathematics continues. Phelan has been to the board once, two class periods ago. Even though Mahavier said little, Phelan realized he could not write down a proof of Problem 18 PROBLEM 18
Show that if \(\begin{equation}H\end{equation}\) and \(\begin{equation}K\end{equation}\) are two point sets 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 tossed the chalk into the tray, frustrated. Sometimes failure is a necessary precursor to success. This time, he has everything well in mind and presents a nice argument for Problem 24 PROBLEM 24
Show that if the sequence \(\begin{equation}x_1, x_2, x_3 \dots\end{equation}\) converges to the number \(\begin{equation}c\end{equation}\) and \(\begin{equation}(x,y)\end{equation}\) is a segment containing \(\begin{equation}c\end{equation}\) then there are not infinitely many terms of the sequence \(\begin{equation}x_1, x_2, x_3 \dots\end{equation}\) that are not in \(\begin{equation}(x,y)\end{equation}\).
. He is asked only one question, which he immediately fields.

Zahn's attempt of Problem 25 PROBLEM 25
Does the sequence
\(\begin{equation}1/2,1/1,1/4,1/3,1/6,1/5,\ldots\end{equation}\) converge to \(\begin{equation}0\end{equation}\)? Note that this is the sequence \(\begin{equation}x\end{equation}\) where \(\begin{equation}x_n=\frac{1}{n-1}\end{equation}\) if \(\begin{equation}n\end{equation}\) is an even positive integer and \(\begin{equation}x_n=\frac{1}{n+1}\end{equation}\) if \(\begin{equation}n\end{equation}\) is an odd positive integer.
shows us both good collaboration within the classroom and the importance of posing good problems. Mahavier asks for a vote and the class is split, proof that this is a good problem for this audience. Zahn admits he's not sure, but goes to the board and attempts to prove that it converges. There is good class discussion when he has difficulty. Several students have a good understanding of the problem at hand, as evidenced from comments from Valero, Herring and Winterhalter. Each has clearly worked the problem. Norman steps up and gives an argument, defending questions from Mahavier, and we see an instance when Mahavier is willing to go to the board and write a full mathematical statement. In nine classes, this is perhaps the third or fourth time he has written anything on the board - the definition of limit point, one lemma that a student needed, and this clarifying statement of what must be shown to conclude the argument. In short, the students learn to write and speak correct mathematics precisely because Mahavier does not do it for them.

Day 10, Friday 2.8.2008.

The second transition has occurred. Not only are they writing and speaking correct mathematics, in 10 days, they are now putting nice proofs on the board.

Winterhalter attempts Problem 26 PROBLEM 26
Assume that the sequence \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) converges to the point \(\begin{equation}c\end{equation}\) and \(\begin{equation}d\end{equation}\) is a point different from \(\begin{equation}c\end{equation}\). Show that \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) does not converge to \(\begin{equation}d\end{equation}\).
, but assumes that every sequence is either increasing or decreasing. Mahavier lets him hold the problem till Monday and then gives one of his longest discussions to date, over four minutes. He first states a new problem NEW PROBLEM
Let \(\begin{equation}S\end{equation}\) be a segment, let \(\begin{equation}p\end{equation}\) be a point of \(\begin{equation}S\end{equation}\), and let \(\begin{equation}M\end{equation}\) be a finite subset of \(\begin{equation}S\end{equation}\). Show that there is a segment that contains \(\begin{equation}p\end{equation}\) but no point of \(\begin{equation}M\end{equation}\) different from \(\begin{equation}p\end{equation}\)
verbally, allowing the students to write it as he speaks. He then reminds them that Problem 18 PROBLEM 18
Show that if \(\begin{equation}H\end{equation}\) and \(\begin{equation}K\end{equation}\) are two point sets 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 \(\begin{equation}K\end{equation}\).
is still outstanding. Even though this seems an arbitrary new problem and an arbitrary chat, the proof technique required for this problem will break the back of both Problem 18 and Problem 26, thus is directly related to the problem that Winterhalter has just attempted and the outstanding Problem 18. Winterhalter has not had much success yet at the board and with his lecture, Mahavier is maximizing the odds of success for this student on both the long-standing problem and the current one. Winterhalter will turn out to be a very successful student in the course.

The next two clips demonstrate how quickly the students grow. Watching clips day-by-day, progress can seem painfully slow. Watching a pair of clips separated by five or ten days can show the progress more clearly.

This video would be valuable to watch DeMitchell's first attempt at the board on Day 3. In this clip, DeMitchell gives a flawless argument to Problem 27 PROBLEM 27
Show that if the sequence \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) converges to the point \(\begin{equation}c\end{equation}\), and, for each positive integer \(\begin{equation}n\end{equation}\), \(\begin{equation}x_n \ne x_{n+1}\end{equation}\), then \(\begin{equation}c\end{equation}\) is a limit point of the range of the sequence.

This video would be valuable to watch back-to-back with Winterhalter's first attempt at the board on Day 5. In this clip, Winterhalter gives flawless argument to Problem 28 PROBLEM 28
Find a sequence \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) such that the point \(\begin{equation}c\end{equation}\) is a limit point of the range \(\begin{equation}x_1,x_2,...\end{equation}\) of the sequence but
\(\begin{equation}x_1,x_2,x_3,...\end{equation}\) does not converge to \(\begin{equation}c\end{equation}\).

Mahavier reminds them of a question he posed several classes ago as to whether every sequence with range the reciprocals of the natural numbers converges.

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.'" This clip demonstrates the success of the method in developing students who can work independently and without a teacher to guide them. Mahavier discovers that the students don't have enough problems to work on because he failed to pass out the next sheet of problems during the last class period. He leaves class to make copies and there is a stretch of silence in the room before a discussion between students leads to an argument for Problem 29 PROBLEM 29
Show that if \(\begin{equation}d\end{equation}\) is a number and \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is a sequence which converges to the point \(\begin{equation}c\end{equation}\), then the sequence \(\begin{equation}d \cdot x_1,d \cdot x_2,d \cdot x_3,...\end{equation}\) converges to \(\begin{equation}d \cdot c\end{equation}\).
which is completed before Mahavier returns.

The longest lecture Mahavier has given up to this point in the course is this lecture on Axiom 1: The Completeness Property AXIOM 1:
THE COMPLETENESS PROPERTY

If \(\begin{equation}M\end{equation}\) is a point set which is bounded above (below) , then there is a point \(\begin{equation}p\end{equation}\) such that either

1. \(\begin{equation}p \in M\end{equation}\) and \(\begin{equation}p\end{equation}\) is the largest (smallest) number in \(\begin{equation}M\end{equation}\), or

2. \(\begin{equation}p \notin M\end{equation}\) and \(\begin{equation}p\end{equation}\) is the smallest (largest) number such that each point of \(\begin{equation}M\end{equation}\) is to the left (right) of \(\begin{equation}p\end{equation}\).

Day 11, Monday 2.11.2008.

Mahavier asks about Problem 31 PROBLEM 31
Assume that \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is a point sequence that converges to the point \(\begin{equation}c\end{equation}\) and \(\begin{equation}d\end{equation}\) is a point different from \(\begin{equation}c\end{equation}\). Show that \(\begin{equation}d\end{equation}\) is not a limit point of the range of the sequence.
. You can see in this clip how the students are taking ownership of their own learning and are comfortable enough in the class to redirect what is about to be presented. Winterhalter reminds Mahavier that he was working on Problem 26 PROBLEM 26
Assume that the sequence \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) converges to the point \(\begin{equation}c\end{equation}\) and \(\begin{equation}d\end{equation}\) is a point different from \(\begin{equation}c\end{equation}\). Show that \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) does not converge to \(\begin{equation}d\end{equation}\).
last time and did not resolve it. He asks Mahavier if he can present it to be sure that he has it right. Mahavier apologizes and sends him to the board to show it. Winterhalter shows a nice argument and Mahavier is very complimentary.

Class time has become more efficient. In less than the 20 minutes of remaining class time, three students attempt problems without success. DeMitchell tries Problem 31 PROBLEM 31
Assume that \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is a point sequence that converges to the point \(\begin{equation}c\end{equation}\) and \(\begin{equation}d\end{equation}\) is a point different from \(\begin{equation}c\end{equation}\). Show that \(\begin{equation}d\end{equation}\) is not a limit point of the range of the sequence.
, Zahn tries Problem 32 PROBLEM 32
Show that if \(\begin{equation}M\end{equation}\) is a point set, there cannot be both a rightmost point of \(\begin{equation}M\end{equation}\) and a smallest number which is larger than each number in \(\begin{equation}M\end{equation}\).
, and Valero tries Problem 33 PROBLEM 33
If \(\begin{equation}M\end{equation}\) is a point set and each point of \(\begin{equation}M\end{equation}\) is to the left of \(\begin{equation}p\end{equation}\) and \(\begin{equation}p\end{equation}\) is the smallest number such that each point of \(\begin{equation}M\end{equation}\) is to the left of \(\begin{equation}p\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\).
.

Day 12, Wednesday 2.13.2008.

This is representative of what I consider an optimal day. Five problems are presented, three successfully and two unsuccessfully. Even the unsuccessful attempts generate understanding. Mahavier's method of calling on the unsuccessful students earlier is slowly assuring the success of most students. In my experience, there is one such pivotal day in every class, usually around 20-33% through the class when the students are writing and speaking correct mathematics and the speed of the class increases substantially. At this point, when problems are presented correctly, they are likely presented as quickly as a faculty lecturing would present. It might be worth watching the full day to observe how efficient a Moore Method class can be once the students become proficient at producing, writing and explaining their mathematics. Of course, if there are multiple Moore Method courses in a department, then the initial lag time exists only in the first course and all other courses move quickly from the start. This is the point in the course when students who have not put sufficient work in to have success find that the pace now leaves them behind. These are not necessarily weak or strong students, simply students who did not commit time early in the course.

Norman answers the question as to whether there is a sequence with range the reciprocals of the natural numbers which does not converge to zero with a nice example. His example is not one-to-one and did not need to be to answer the original question, so Mahavier then seeds him with another question, asking, "Is there a one-to-one function that defines a sequence, with range the reciprocals of the natural numbers, which does not converge to zero?", and Norman shows that there is not.

DeMitchell attempts Problem 31 PROBLEM 31
Assume that \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is a point sequence that converges to the point \(\begin{equation}c\end{equation}\) and \(\begin{equation}d\end{equation}\) is a point different from \(\begin{equation}c\end{equation}\). Show that \(\begin{equation}d\end{equation}\) is not a limit point of the range of the sequence.
. Zahn presents Problem 32 PROBLEM 32
Show that if \(\begin{equation}M\end{equation}\) is a point set, there cannot be both a rightmost point of \(\begin{equation}M\end{equation}\) and a smallest number which is larger than each number in \(\begin{equation}M\end{equation}\).
. Valero completes Problem 33 PROBLEM 33
If \(\begin{equation}M\end{equation}\) is a point set and each point of \(\begin{equation}M\end{equation}\) is to the left of \(\begin{equation}p\end{equation}\) and \(\begin{equation}p\end{equation}\) is the smallest number such that each point of \(\begin{equation}M\end{equation}\) is to the left of \(\begin{equation}p\end{equation}\), then \(\begin{equation}p\end{equation}\) is a limit point of \(\begin{equation}M\end{equation}\).
. DeMitchell attempts Problem 34 PROBLEM 34
If the range of the increasing sequence \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is bounded above, then \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) converges to some point.
, but is not successful before the class period ends.

Day 13, Friday 2.15.2008.

It is often asked whether the Moore Method works better for the more confident or stronger student. The answer lies in the instructor. In this clip you can see how Mahavier is patient in allowing the student to answer the questions, but does not feed him answers. Zahn presents Problem 34 PROBLEM 34
If the range of the increasing sequence \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is bounded above, then \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) converges to some point.
which is a nice presentation by a student who clearly is appears not to be confident at the board, but has had a string of successes in the course.

Two observations can be made from this clip. First, Mahavier often uses non-standard definitions and later includes problems that show that these definitions are equivalent to the standard definitions found in books. Second, Mahavier treats students differently depending on their level of success to date and their confidence at the board. Winterhalter, who has had numerous successes, attempts a problem and takes a seat. We see the seamless transition as Winterhalter fails, but is not humiliated or embarrassed, and Herring successfully presents Problem 35 PROBLEM 35
Assume that \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is a point sequence and if \(\begin{equation}S\end{equation}\) is a segment centered at the point \(\begin{equation}c\end{equation}\), then there is a positive integer \(\begin{equation}n\end{equation}\) such that \(\begin{equation}\{x_n,x_{n+1},x_{n+2}\ldots\} \subseteq S\end{equation}\). Show that \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) converges to \(\begin{equation}c\end{equation}\).

Mahavier gives his second lecture in which he briefly discusses the definition of continuity. As soon as he gives the definition and an example, he asks if the function which consists of the single point \(\begin{equation}(1,2)\end{equation}\) is continuous, illustrating his belief that much can be learned by allowing students to investigate first the absolute simplest of examples. Sayanii suggests that this is not a function, showing just what misconceptions students have about functions after several semesters of Calculus.

Day 14, Monday 2.18.2008.

Rasmussen goes to the board for the first time. He attempts Problem 37 PROBLEM 37
If the range of the non decreasing sequence \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is bounded above, then \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) converges to some point.
. He has a good start, but can't accurately write the cases he has in his mind. Herring claims he can finish it and Mahavier asks Rasmussen if he wishes to allow Herring to finish it. (Editor's Note: It's not clear to me why Mahavier did not encourage Rasmussen to save it in order to maximize the opportunity for him to have success at the board, since Herring has had one recent success. I believe that Rasmussen indicates that he thinks he can get it for the next class period and that Mahavier does not hear him.) Herring attempts Problem 37 PROBLEM 37
If the range of the non decreasing sequence \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is bounded above, then \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) converges to some point.
and is very close to a solution, but also gets stuck on a very technical point when he makes a statement that Mahavier says is not true. In fact, it is true. Mahavier tells him to finish it tomorrow.

Day 15, Wednesday 2.20.2008.

We are now one-third of the way through the course and more problems are being covered each day. Five problems are presented successfully in class and others are discussed. First Mahavier collects papers and offers to represent Problem 31 PROBLEM 31
Assume that \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is a point sequence that converges to the point \(\begin{equation}c\end{equation}\) and \(\begin{equation}d\end{equation}\) is a point different from \(\begin{equation}c\end{equation}\). Show that \(\begin{equation}d\end{equation}\) is not a limit point of the range of the sequence.
, which went on the board during the last class and perhaps was not clear to everyone. There are no takers, but there is a brief discussion and then DeMitchell asks to see Problem 36 again. Sayanii and Norman had previously attempted Problem 36 PROBLEM 36
Show that if the sequence \(\begin{equation}p_1,p_2,p_3,...\end{equation}\) converges to \(\begin{equation}c\end{equation}\) and the sequence \(\begin{equation}q_1,q_2,q_3,...\end{equation}\) converges to \(\begin{equation}d\end{equation}\), then the sequence \(\begin{equation}p_1+q_1,p_2+q_2,p_3+q_3,\dots\end{equation}\) converges to \(\begin{equation}c+d\end{equation}\).
. Norman presents Problem 36 PROBLEM 36
Show that if the sequence \(\begin{equation}p_1,p_2,p_3,...\end{equation}\) converges to \(\begin{equation}c\end{equation}\) and the sequence \(\begin{equation}q_1,q_2,q_3,...\end{equation}\) converges to \(\begin{equation}d\end{equation}\), then the sequence \(\begin{equation}p_1+q_1,p_2+q_2,p_3+q_3,\dots\end{equation}\) converges to \(\begin{equation}c+d\end{equation}\).
nicely. Mahavier compliments the class and offers to present one of a few problems that were recently presented, stating that this class is ten problems further along than his last two sections of this course were at this point. A students asks and, after first offering to allow the student who presented it to show it a second time, Mahavier presents Problem 31 PROBLEM 31
Assume that \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is a point sequence that converges to the point \(\begin{equation}c\end{equation}\) and \(\begin{equation}d\end{equation}\) is a point different from \(\begin{equation}c\end{equation}\). Show that \(\begin{equation}d\end{equation}\) is not a limit point of the range of the sequence.
. Valero presents the last piece of the puzzle that completes Herring's previous argument of Problem 37 PROBLEM 37
If the range of the non decreasing sequence \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is bounded above, then \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) converges to some point.
, concluding that problem. Phelan presents Problem 38 PROBLEM 38
If \(\begin{equation}M\end{equation}\) is a bounded point set then the \(\begin{equation}lub(M)\end{equation}\) (and the \(\begin{equation}glb(M)\end{equation}\)) is either a point of \(\begin{equation}M\end{equation}\) or a limit point of \(\begin{equation}M\end{equation}\).
and Rasmussen presents Problem 39 PROBLEM 39
Give an example of a bounded point set such that lub(M) is both a point of \(\begin{equation}M\end{equation}\) and a limit point of \(\begin{equation}M\end{equation}\).
and Problem 40 PROBLEM 40
If the sequence \(\begin{equation}x_1,x_2,x_3,...\end{equation}\)
converges to the point \(\begin{equation}c\end{equation}\), then
\(\begin{equation}M=\{x_1,x_2,x_3...\}\end{equation}\)
is bounded.
.

Day 16, Friday 2.22.2008.

Mahavier spends a full fifteen minutes helping a student create an accurate conjecture. Winterhalter attempts to present Problem 43 PROBLEM 43
Let \(\begin{equation}f\end{equation}\) be the graph such that \(\begin{equation}f(x)=x^2\end{equation}\) for each number \(\begin{equation}x\end{equation}\). Show that \(\begin{equation}f\end{equation}\) is continuous at the point \(\begin{equation}(2,4)\end{equation}\).
. He has an idea and believes that if the length of the interval around 2 (twice delta in standard language) is less than the length of the interval about 4 (twice epsilon), then all points in the interval about 2 will map into the interval about 4. At the end, Winterhalter sees right into the heart of the issue - that the rate of change of the graph affects the problem. And Mahavier has helped him state clearly his conjecture, albeit a false one, on which he can work. This is perhaps why the Moore Method is so powerful. Even though he is working on a relatively simple example, he does not have the benefit of examples so is following his own creativity. Because he is struggling without examples to guide him, he sees deeply what is going on. In the next class, he shows that his conjecture is false.

Day 17, Monday 2.25.2008.

Mahavier encourages Valero to attempt Problem 41 PROBLEM 41
If \(\begin{equation}M\end{equation}\) is an infinite and bounded point set, then \(\begin{equation}M\end{equation}\) has a limit point.
. She does not have it, but he allows an eight minute exploration while two students think and talk about the problem, without success.

Norman presents Problem 45 PROBLEM 45
If \(\begin{equation}f\end{equation}\) is a simple graph and \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is a sequence of points in the domain of \(\begin{equation}f\end{equation}\) converging to the number \(\begin{equation}c\end{equation}\) in the domain of \(\begin{equation}f\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}(c,f(c))\end{equation}\), then \(\begin{equation}f(x_1),f(x_2 ), ...\end{equation}\) converges to \(\begin{equation}f(c)\end{equation}\).
and clearly has a handle on it. This is an example where Mahavier treats a student quite differently than he might treat another. Norman is now a strong student who has had many successes. Although he is quite likely to be able to write this problem up correctly with only a few changes to the imprecise notation he has used at the board, Mahavier makes him "start from scratch" and reproduce the entire argument a second time using accurate notation. Mahavier's reasoning is two-fold. First, Norman will be forced to correct the error in his notation. Second, Norman is strong enough at the board that the class can benefit from this lesson as he presents it a second time. Mahavier is using Norman as an instrument to instruct the class.

Day 18, Wednesday 2.27.2008.

Now halfway through the course Mahavier has started to sprinkle more difficult problems into the notes which remain outstanding for several class periods while simpler problems are resolved. The question of whether there is a sequence with range the interval [0,1] is one such unresolved problem. The fact that every infinite bounded set has a limit point is another. As Norman and Valero work on these, we can see how they both speak very clear mathematics and when confronted will either defend their mathematics or recognize their error.

Norman exhibits a sequence that he believes includes all the numbers between 0 and 1. The class discusses it and Mahavier encourages the students to work on this particular problem, even as he points out that \(\begin{equation}\frac{1}{3}\end{equation}\) is not in the list of numbers that Norman creates.

Valero attempts Problem 41 PROBLEM 41
If \(\begin{equation}M\end{equation}\) is an infinite and bounded point set, then \(\begin{equation}M\end{equation}\) has a limit point.
which states that every infinite bounded set has a limit point. Mahavier makes his only reference to Moore at this point when he debates whether to provide any guidance. Clearly, Valero has worked hard on the problem and he decides against offering a hint, quoting Moore who said in the MAA film titled Challenge in the Classroom, "The student is taught best who is told the least."

Following these failures, DeMitchell gives a nice proof that a constant multiple of a continuous function is continuous and Zahn gives a nice proof that a function is continuous at every point in the domain that is not a limit point of the domain.

Day 19, Friday 2.29.2008.

Valero resolves Problem 41 PROBLEM 41
If \(\begin{equation}M\end{equation}\) is an infinite and bounded point set, then \(\begin{equation}M\end{equation}\) has a limit point.
that she attempted last class, with a tad of help from the class.

Norman resolves Question 1 QUESTION 1
Suppose \(\begin{equation}f\end{equation}\) is a simple graph with domain the interval \(\begin{equation}[-1,1]\end{equation}\) and \(\begin{equation}f(0)=0\end{equation}\). Suppose also that \(\begin{equation}f\end{equation}\) is continuous at the point \(\begin{equation}(0,0)\end{equation}\). Must there be two vertical lines with \(\begin{equation}(0,0)\end{equation}\) between them such that \(\begin{equation}f\end{equation}\) is continuous at each point on \(\begin{equation}f\end{equation}\) between the two vertical lines?
by exhibiting a function that is continuous at (0,0) but which is not continuous on any interval containing (0,0). Mahavier then gives his longest lecture to date, pointing out two alternative ways one might do this problem, foreshadowing the squeeze theorem and giving his philosophy on the upcoming midterm.

More leading the students... Zahn gets very close to Problem 49 PROBLEM 49
If \(\begin{equation}f\end{equation}\) and \(\begin{equation}g\end{equation}\) are simple graphs having domain the interval \(\begin{equation}I\end{equation}\), \(\begin{equation}a \in I\end{equation}\),
\(\begin{equation}f(a)=g(a)\end{equation}\), and each of \(\begin{equation}f\end{equation}\) and \(\begin{equation}g\end{equation}\) is continuous at the point \(\begin{equation}(a,f(a))\end{equation}\) and \(\begin{equation}h\end{equation}\) is a simple graph with domain \(\begin{equation}I\end{equation}\),
and for each number \(\begin{equation}x\end{equation}\) in \(\begin{equation}I\end{equation}\),
\(\begin{equation}f(x) \le h(x) \le g(x)\end{equation}\) then \(\begin{equation}h\end{equation}\) is
continuous at \(\begin{equation}(a,h(a))\end{equation}\).
, the squeeze theorem. Mahavier definitely guides him closer to understanding the issue that Zahn is missing and the odds for success next time are quite good. The amount of guidance he gives is intended to be enough that the student has a fair chance at success, but not so much that the student does not feel he has solved it independently. Zahn completes this problem efficiently the next day.

Day 20, Monday 3.3.2008.

Zahn completes the squeeze theorem efficiently. Herring and Phelan have been quiet for some time. Today they produce nice arguments. This is one of the benefits of the method - students who are working consistently are likely to have success, even if it does not come quickly. Both conclude the course with many successful presentations.

It has been five class periods since Herring has been to the board. Today he presents Problem 50 PROBLEM 50
Let \(\begin{equation}f\end{equation}\) be a graph whose domain is an interval \(\begin{equation}[a,b]\end{equation}\), and which has points both above and below the \(\begin{equation}x-axis\end{equation}\). Let \(\begin{equation}c\end{equation}\) be a number in \(\begin{equation}[a,b]\end{equation}\) such that \(\begin{equation}f(c)> 0\end{equation}\) and let \(\begin{equation}H\end{equation}\) denote the set of all numbers \(\begin{equation}x\end{equation}\) in \(\begin{equation}[a,b]\end{equation}\) such that \(\begin{equation}f(x)\le 0\end{equation}\). Show that if \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}(c,f(c))\end{equation}\), then \(\begin{equation}c\end{equation}\) is not a limit point of \(\begin{equation}H\end{equation}\).
very nicely.

Phelan attempts Problem 51 PROBLEM 51
Let \(\begin{equation}f\end{equation}\) be a graph whose domain is an interval \(\begin{equation}[a,b]\end{equation}\), let \(\begin{equation}c\end{equation}\) be a number in \(\begin{equation}[a,b]\end{equation}\) such that \(\begin{equation}f(c)>0\end{equation}\) and let \(\begin{equation}K\end{equation}\) denote the set of all numbers \(\begin{equation}x\end{equation}\) in \(\begin{equation}[a,b]\end{equation}\) such that \(\begin{equation}f(x)>0\end{equation}\). Show that if \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}(c,f(c))\end{equation}\), then \(\begin{equation}c\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\).
, but is winging it, with nothing written down. This clip shows just how patient Mahavier is with a student who is attempting a problem, but has not yet had regular success at the board. He is hoping that some seed will come from this that will allow him to give some guidance that will lead the student to success on the problem, both to build up the student's confidence and to resolve the problem for the class so that the progress continues. On the other hand, I think Mahavier believes that this student is capable and has not spent sufficient time outside of class and Mahavier suggests at the end that the student write it down before he attempts to present it at the next class.

The class discusses Question 2 QUESTION 2
If \(\begin{equation}M_1,M_2,M_3,\dots\end{equation}\) is a sequence of point sets such that for each positive integer \(\begin{equation}n\end{equation}\), \(\begin{equation}M_{n+1} \subseteq M_n\end{equation}\), must there be a point that is common to all the sets of the sequence \(\begin{equation}M_1,M_2,M_3,\ldots\end{equation}\)?
. This is a long discussion, but much progress is made.

Winterhalter begins Problem 52 PROBLEM 52
Let \(\begin{equation}p\end{equation}\) be a point and for each positive integer \(\begin{equation}n\end{equation}\), let
\(\begin{equation}S_n=(p-1/n,p+1/n)\end{equation}\).
Show that every segment containing \(\begin{equation}p\end{equation}\) contains one of the segments in the sequence \(\begin{equation}S_1,S_2,S_3,\ldots\end{equation}\).
, clearly has the idea, but runs out of time.

Day 21, Wednesday 3.5.2008.

Mahavier's patience with Phelan yesterday pays off as Phelan presents Problem 51 PROBLEM 51
Let \(\begin{equation}f\end{equation}\) be a graph whose domain is an interval \(\begin{equation}[a,b]\end{equation}\), let \(\begin{equation}c\end{equation}\) be a number in \(\begin{equation}[a,b]\end{equation}\) such that \(\begin{equation}f(c)>0\end{equation}\) and let \(\begin{equation}K\end{equation}\) denote the set of all numbers \(\begin{equation}x\end{equation}\) in \(\begin{equation}[a,b]\end{equation}\) such that \(\begin{equation}f(x)>0\end{equation}\). Show that if \(\begin{equation}f\end{equation}\) is continuous at \(\begin{equation}(c,f(c))\end{equation}\), then \(\begin{equation}c\end{equation}\) is a limit point of \(\begin{equation}K\end{equation}\).
. His first difficult theorem, this is a rather long technical proof, but a good one. He clearly has created the proof himself and he has worked on a lot of problems in the past without writing them up well and without this level of polished success.

As Winterhalter presents Problem 52 PROBLEM 52
Let \(\begin{equation}p\end{equation}\) be a point and for each positive integer \(\begin{equation}n\end{equation}\), let
\(\begin{equation}S_n=(p-1/n,p+1/n)\end{equation}\).
Show that every segment containing \(\begin{equation}p\end{equation}\) contains one of the segments in the sequence \(\begin{equation}S_1,S_2,S_3,\ldots\end{equation}\).
Mahavier points out when they use previous theorems.

There is a good class discussion about Question 2 QUESTION 2
If \(\begin{equation}M_1,M_2,M_3,\dots\end{equation}\) is a sequence of point sets such that for each positive integer \(\begin{equation}n\end{equation}\), \(\begin{equation}M_{n+1} \subseteq M_n\end{equation}\), must there be a point that is common to all the sets of the sequence \(\begin{equation}M_1,M_2,M_3,\ldots\end{equation}\)?
where all the students agree that there is a point common to any nested sequence of sets, but no one can prove it. Mahavier uses this example to point out how much harder it is to do research, that is, to solve a problem when one does not know what is true. He tells them that their intuition is wrong and lets them work on it again.

Phelan has now hit his stride which is impressive, as he is the only freshman in the course. He presents Problem 53 PROBLEM 53
Let \(\begin{equation}p\end{equation}\) be a point and for each positive integer \(\begin{equation}n\end{equation}\), let
\(\begin{equation}S_n=(p-1/n,p+1/n)\end{equation}\).
Show that \(\begin{equation}p\end{equation}\) is the only point common to all the segments in the sequence \(\begin{equation}S_1,S_2,S_3,\ldots\end{equation}\).
flawlessly and efficiently.

Mahavier lectures on induction. Mahavier uses negation to teach induction, which is nice and keeps notation to a minimum. If a statement is not true for all natural numbers then there must be a smallest natural number for which it is not true. If we call this number \(\begin{equation}n+1\end{equation}\), then the statement is true for \(\begin{equation}n\end{equation}\). By using the fact that it is true for \(\begin{equation}n\end{equation}\) to show that it is also true for \(\begin{equation}n+1,\end{equation}\) we produce a contradiction to the fact that it is not true for \(\begin{equation}n+1\end{equation}\).

Day 22, Friday 3.7.2008.

Mahavier passes out more problems with one comment encouraging them to be clear about the difference between the domain of a sequence, the natural numbers, and the range of the sequence, a possibly finite set. Such very brief lectures demonstrate how most of Mahavier's lecturing is in response to student work, whether at the board or in the written homework.

Mahavier asks for solutions to Problem 54 or induction.

Sayanii asks to see an example of induction again and Mahavier presents for a second time the induction example that he presented when he introduced induction. Mahavier uses negation to do inductive arguments, which is nice and keeps the notation to a minimum.

Monday 3.10.2008 - Friday 3.14.2008, Spring Break, no class

Day 23, Monday 3.17.2008.

Zahn presented Problem 55 PROBLEM 55
Show that if \(\begin{equation}n\end{equation}\) is a positive integer, then \(\begin{equation}2^n>n\end{equation}\).
.

Valero presents Problem 56 PROBLEM 56
Show that the sequence \(\begin{equation}1,1/2,1/4,1/8\ldots\end{equation}\) converges to \(\begin{equation}0\end{equation}\).
.

Day 24, Wednesday 3.19.2008.

While it may not be worth watching the entire clip, which spans the whole class period, several observations can be made from this clip. First, it demonstrates how patiently Mahavier questions and guides students at the board. Second, it demonstrates that, as long as a student is making progress, Mahavier will continue this process even if it takes the entire class. Third, it demonstrates how Mahavier is almost exclusively focused on maximizing the learning of the student at the board. Lastly, this clip illustrates something significant about course note development. Mahavier has been intertwining multiple concepts: limit points, convergence and continuity. Rasmussen has not had a lot of success in the class to date, but when Mahavier adds the new concept of nested intervals, Rasmussen has success. Writing notes so that there are several threads occurring simultaneously increases the odds of each student succeeding and decreases the odds of a small number of students getting a bit ahead and running away with the course.

Rasmussen attempts to presents Problem 57 PROBLEM 57
Assume that \(\begin{equation}I_1,I_2,I_3,...\end{equation}\) is a sequence of intervals such that for each positive integer \(\begin{equation}n\end{equation}\), \(\begin{equation}I_{n+1} \subseteq I_n\end{equation}\). Show that there is a point that is common to all the intervals of the sequence \(\begin{equation}I_1,I_2,I_3,...\end{equation}\).
which states that a nested sequence of intervals has a point common to all of them.

Day 25, midterm, Friday 3.21.2008, no film.

Day 26, Monday 3.24.2008.

Today's lesson is that while Mahavier has not pressured any student to present any problem, he has always allowed students with less presentations to go first. Neither Rasmussen and Phelan had success early on in the course, but neither gave up and now both are having success. Today, Phelan presents Problem 58 PROBLEM 58
Assume that \(\begin{equation}I_1,I_2,I_3,...\end{equation}\) is a sequence of intervals such that for each positive integer \(\begin{equation}n\end{equation}\), \(\begin{equation}I_{n+1} \subseteq I_n\end{equation}\) and the length of \(\begin{equation}I_n\end{equation}\) is less than \(\begin{equation}1/n\end{equation}\). Show that there are not two points common to all the intervals of the sequence \(\begin{equation}I_1,I_2,I_3,...\end{equation}\).
successfully.

Just as a mathematician who breaks new ground can typically make a nice string of theorems based on the one key idea, Valero mastered the concepts of greatest lower bound and least upper bound early on and she has consistently solved problems associated with these concepts. Today she presents Problem 60 PROBLEM 60
Show that if \(\begin{equation}M\end{equation}\) is a closed and bounded point set, then \(\begin{equation}M\end{equation}\) has a rightmost point.
. Mahavier uses this problem as a springboard to ask whether the class can give an example, other than an interval, of a closed and bounded set where every point is a limit point. He allows lots of time for the class to think on this and says that he will ask the question again at the next class meeting.

Day 27, Wednesday 3.26.2008.

The last transition has occurred. We are seeing success from the majority of students and the problems are taking more time to present as they become more difficult. Problem 54 has been outstanding for weeks and multiple students have attempted it. This illustrates the often misunderstood notion of competition in a Moore Method course. Almost surely, several students in this particular course really want to resolve this outstanding problem. And likely some want to get it before anyone else does. This type of friendly competition can be healthy to motivate students. Finally, today Mr. Herring, who has presented less than many other students, successfully presents Problem 54 PROBLEM 54
Assume that \(\begin{equation}p\end{equation}\) is a limit point of the point set \(\begin{equation}M\end{equation}\). Show that there is a sequence of points of \(\begin{equation}M\end{equation}\) all different and each different from \(\begin{equation}p\end{equation}\) which converges to \(\begin{equation}p\end{equation}\).
. Mahavier is clearly very pleased with this success and complements the class on how far they have come since the beginning. Mahavier has not pressured any student, has allowed time for students to mature and learn from the process, and this patience has been rewarded as students who did not present a lot in the beginning are having success on difficult problems.

Day 28, Friday 3.28.2008.

Winterhalter gives an inductive argument that that the sum of first n odd integers is \(\begin{equation}n^2\end{equation}\) and Herring attempts Problem 61 PROBLEM 61
Show that if \(\begin{equation}p_1,p_2,p_3,...\end{equation}\) is a point sequence that converges to the point \(\begin{equation}c\end{equation}\) and \(\begin{equation}q_1,q_2,q_3,...\end{equation}\) is a subsequence of
\(\begin{equation}p_1,p_2,p_3,...\end{equation}\), then \(\begin{equation}q_1,q_2,q_3,...\end{equation}\)
converges to \(\begin{equation}c\end{equation}\).
.

Day 29, Monday 3.31.2008.

Mahavier uses questions to introduce students to mathematical research, where one is often working on a problem that s/he believes to be true or false, but does not know. Both DeMitchell and Valero attempt to resolve Question 4 QUESTION 4
Assume that \(\begin{equation}H\end{equation}\) and \(\begin{equation}K\end{equation}\) are closed point sets such that \(\begin{equation}H\cup K=[0,1]\end{equation}\). Must \(\begin{equation}H\end{equation}\) and \(\begin{equation}K\end{equation}\) must have a point in common?
and DeMitchell tries to give a counter-example and then Valero attempts to give a proof. Meanwhile, Herring continually comments that be believes that the unresolved Question 2 QUESTION 2
If \(\begin{equation}M_1,M_2,M_3,\dots\end{equation}\) is a sequence of point sets such that for each positive integer \(\begin{equation}n\end{equation}\), \(\begin{equation}M_{n+1} \subseteq M_n\end{equation}\), must there be a point that is common to all the sets of the sequence \(\begin{equation}M_1,M_2,M_3,\ldots\end{equation}\)?
will resolve this problem. After Valero's attempt to prove it, Herring attempts a counter example. This clip really shows the collaborative, cooperative attitude of the class as they struggle toward understanding the question.

The next pair of clips demonstrate nicely how just-in-time lecturing can produce rapid progress.

Students are struggling with Rolles Theorem, Problem 62 PROBLEM 62
Assume that \(\begin{equation}f\end{equation}\) is a continuous graph with domain the interval \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}f(a)>0\end{equation}\) and \(\begin{equation}f(b)<0\end{equation}\), then there is a number \(\begin{equation}c \in (a,b)\end{equation}\) such that \(\begin{equation}f(c)=0\end{equation}\).
. Mahavier says that he doesn't often prove things, but points out that the class is not taking full advantage of the Completeness Axiom. He then guides an interactive discussion that leads to an alternative proof of Problem 41 PROBLEM 41
If \(\begin{equation}M\end{equation}\) is an infinite and bounded point set, then \(\begin{equation}M\end{equation}\) has a limit point.
that every infinite bounded set has a limit point. His proof makes careful use of the Completeness Axiom and this understanding immediately produces nice proofs of Rolle's Theorem and the Intermediate Value Theorem.

The problems are becoming more difficult, even if some easier ones are interspersed. Students are often at the board for much longer as the student tries to get a correct proof up and others try to understand. Mahavier has led them from the most elementary problems to problems that challenge all of them and take considerable time to resolve, present and sometimes present a second time.

Day 30, Wednesday 4.2.2008.

Zahn gives a very nice short proof of Problem 62 PROBLEM 62
Assume that \(\begin{equation}f\end{equation}\) is a continuous graph with domain the interval \(\begin{equation}[a,b]\end{equation}\) and \(\begin{equation}f(a)>0\end{equation}\) and \(\begin{equation}f(b)<0\end{equation}\), then there is a number \(\begin{equation}c \in (a,b)\end{equation}\) such that \(\begin{equation}f(c)=0\end{equation}\).
, Rolle's Theorem.

Mahavier spends a full 6 minutes questioning students about what they are working on. He appears completely unhurried, ready to listen to any ideas they have and willing to let any student who wants to put something up go to the board whether they feel they have a proof or not. This is key, that one is welcome to go to the board and present an idea, simply to see what it is they understand and what it is they do not. Several problems are discussed and Mahavier is attempting to get to the board any student who has not presented recently and to see what they are working on.

Herring finally goes to the board with his idea for Problem 64 PROBLEM 64
Assume that \(\begin{equation}f\end{equation}\) is a graph, and \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is a sequence of points in the domain of \(\begin{equation}f\end{equation}\) converging to to the point \(\begin{equation}c\end{equation}\) in the domain of \(\begin{equation}f\end{equation}\) and \(\begin{equation}f(x_1),f(x_2),f(x_3),\ldots\end{equation}\) converges to the point \(\begin{equation}y\end{equation}\) and \(\begin{equation}f(c)\ne y\end{equation}\). Show that \(\begin{equation}f\end{equation}\) is not continuous at the point \(\begin{equation}(x,f(x))\end{equation}\).
and gives a very nice proof, even as he patches portions at the board.

Valero then says she does not have Problem 65 PROBLEM 65
Assume that \(\begin{equation}f\end{equation}\) is a graph, and \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is a sequence of points in the domain of \(\begin{equation}f\end{equation}\) converging to to the point \(\begin{equation}c\end{equation}\) in the domain of \(\begin{equation}f\end{equation}\), and for each positive integer \(\begin{equation}n\end{equation}\), \(\begin{equation}f(x_n)=n\end{equation}\). Show that \(\begin{equation}f\end{equation}\) is not continuous at \(\begin{equation}c\end{equation}\).
, Mahavier encourages her to present what she has at the board and she proceeds to the board and presents a very nice proof for Problem 65 PROBLEM 65
Assume that \(\begin{equation}f\end{equation}\) is a graph, and \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is a sequence of points in the domain of \(\begin{equation}f\end{equation}\) converging to to the point \(\begin{equation}c\end{equation}\) in the domain of \(\begin{equation}f\end{equation}\), and for each positive integer \(\begin{equation}n\end{equation}\), \(\begin{equation}f(x_n)=n\end{equation}\). Show that \(\begin{equation}f\end{equation}\) is not continuous at \(\begin{equation}c\end{equation}\).
.

Day 31, Friday 4.4.2008.

To me, one of the most surprising results of teaching via the Moore Method has been that apparently weak students often thrive under the method. In Mahavier's diary he notes early on that DeMitchell needs some help. We don't have a record of what transpired. Did he give her early guidance in his office? Did he help her more with early theorems? But we do have the result. In this clip, DeMitchell gives a very nice presentation of Problem 63 PROBLEM 63
Assume that \(\begin{equation}f\end{equation}\) is a continuous graph whose domain includes an 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 \(\begin{equation}L\end{equation}\). Show that 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 \(\begin{equation}L\end{equation}\).
, the Intermediate Value Theorem.

Success such as we saw in the previous clip depends on unyielding patience, as sometimes the method can be painful. In this clip Winterhalter presents Problem 67 PROBLEM 67
Assume \(\begin{equation}f\end{equation}\) is a graph, and \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is a sequence of points in the domain of \(\begin{equation}f\end{equation}\) converging to to the point \(\begin{equation}c\end{equation}\) in the domain of \(\begin{equation}f\end{equation}\), and the range of the sequence \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is not bounded. Show that \(\begin{equation}f\end{equation}\) is not continuous at \(\begin{equation}c\end{equation}\).
.

Day 32, Monday 4.7.2008.

Yesterday, Winterhalter struggled at the board for 21 minutes on Problem 67. Today, Winterhalter presents a flawless proof for Problem 67 PROBLEM 67
Assume \(\begin{equation}f\end{equation}\) is a graph, and \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is a sequence of points in the domain of \(\begin{equation}f\end{equation}\) converging to to the point \(\begin{equation}c\end{equation}\) in the domain of \(\begin{equation}f\end{equation}\), and the range of the sequence \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is not bounded. Show that \(\begin{equation}f\end{equation}\) is not continuous at \(\begin{equation}c\end{equation}\).
.

Then DeMitchell presents Problem 69 PROBLEM 69
Show that if the set \(\begin{equation}M\end{equation}\) is not closed then there is a sequence of points of \(\begin{equation}M\end{equation}\) that converges to a point \(\begin{equation}p\end{equation}\) that is not in \(\begin{equation}M\end{equation}\).
efficiently.

A student wants to know the answer to the question Mahavier raised previously, "Does there exists a sequence of points, \(\begin{equation}p_1, p_2, p_3, \dots\end{equation}\) so that the range of the sequence is the interval [0,1]?" Mahavier responds (Editor's note: I was surprised that he provided the answer. I would have told the student that I was glad he was frustrated by the problem as this meant I was doing my job by providing him with a problem worthy of his intellect. I would have told him that I hope it keeps him awake at night until he gets it!) Mahavier takes another path, telling the students that there is no such sequence and stating it carefully as a theorem. He then encourages the class to aim for the Fundamental Theorem of Calculus because they are making such good progress and gives a full lecture on integration. We see now that once he has established the students' abilities to create mathematics, he is quite willing to lecture and give examples to better educate them. The key to the Moore Method is that until he has developed their ability to do mathematics, he won't show them mathematics.

Day 33, Wednesday 4.9.2008.

When a student is struggling, Mahavier will provide guidance by either helping the student clarify the problem or by stating a related problem that will lead to success as he does here. Herring attempts Problem 68 PROBLEM 68
Assume that \(\begin{equation}f\end{equation}\) is a simple graph whose domain is the interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\). Show that the range of \(\begin{equation}f\end{equation}\) is bounded.
and Mahavier carefully states a related problem. While Herring does not resolve this problem, all of Day 35 goes to Winterhalter working this problem at the board with contributions from Herring.

Zahn presents Problem 73 PROBLEM 73
Assume \(\begin{equation}f\end{equation}\) is a simple graph, and \(\begin{equation}x\end{equation}\) is in the domain of \(\begin{equation}f\end{equation}\), and the derivative of \(\begin{equation}f\end{equation}\) at \(\begin{equation}x\end{equation}\) is \(\begin{equation}m\end{equation}\) and \(\begin{equation}k\end{equation}\) is a number different from \(\begin{equation}m\end{equation}\). Show that \(\begin{equation}k\end{equation}\) is not the derivative of \(\begin{equation}f\end{equation}\) at \(\begin{equation}x\end{equation}\). That is: show that a function \(\begin{equation}f\end{equation}\) can't have two derivatives at a number \(\begin{equation}x\end{equation}\).
, derivatives are unique.

Rasmussen presents Problem 75 PROBLEM 75
Suppose that \(\begin{equation}\epsilon\end{equation}\) is a postive number. Show that the segment \(\begin{equation}(x-\epsilon,x+\epsilon)\end{equation}\) is the set of all points \(\begin{equation}t\end{equation}\) such that \(\begin{equation}|t-x|<\epsilon\end{equation}\).
.

This mini-lecture is the only example in the course where he touches on multiple topics in one lecture:

1. He reinforces that the standard way of showing two sets, \(\begin{equation}H\end{equation}\) and \(\begin{equation}K\end{equation}\) are equal is by showing that every element of \(\begin{equation}H\end{equation}\) is in \(\begin{equation}K\end{equation}\) and vice versa.

2. He passes out more sheets and gives a definition for upper and lower Riemann sums.

3. He introduces equivalence relations using the example where the set is all points in the plane and \(\begin{equation}(x,y)\end{equation}\) is related to \(\begin{equation}(a,b)\end{equation}\) if \(\begin{equation}|x|+|y|=|a|+|b|\end{equation}\). He shows that this is an equivalence relation and then asks what the set of all points equivalent to a given point \(\begin{equation}(x,y)\end{equation}\) would be.

Day 34, Friday 4.11.2008.

This clip demonstrates what to do when students don't have anything to present. Mahavier is surprised that no one has Problem 68 PROBLEM 68
Assume that \(\begin{equation}f\end{equation}\) is a simple graph whose domain is the interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\). Show that the range of \(\begin{equation}f\end{equation}\) is bounded.
, so uses his standard tactic of stating a new problem (lemma) which he believes will help.

When no student claims Problem 72 PROBLEM 72
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 \(\begin{equation}f'(0)=0\end{equation}\).
, Mahavier encourages DeMitchell, who declines, and then Valero, who works on Problem 72 PROBLEM 72
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 \(\begin{equation}f'(0)=0\end{equation}\).
at the board to lead the class to a better understanding of the problem. Mahavier concludes by making his second reference to Moore when he uses one of my favorite responses that Moore used. When a student asks if s/he needs to prove something, Moore would respond, "Only if you can't."

Day 35, Monday 4.14.2008.

The entire class went to Winterhalter attempting Problem 68 PROBELM 68
Assume that \(\begin{equation}f\end{equation}\) is a simple graph whose domain is the interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\). Show that the range of \(\begin{equation}f\end{equation}\) is bounded.
at the board while Mahavier and Herring make contributions.

Day 36, Wednesday 4.16.2008.

Mahavier is willing to spend as much time on a problem as needed to address student questions, even when this requires presenting a problem a second time. At the beginning of this class, Mahavier asks if there are questions about Problem 68 PROBLEM 68
Assume that \(\begin{equation}f\end{equation}\) is a simple graph whose domain is the interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\). Show that the range of \(\begin{equation}f\end{equation}\) is bounded.
that Winterhalter showed yesterday. DeMitchell asks to see it again and Winterhalter re-presents Problem 68 PROBLEM 68
Assume that \(\begin{equation}f\end{equation}\) is a simple graph whose domain is the interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f\end{equation}\) is continuous at each number in \(\begin{equation}[a,b]\end{equation}\). Show that the range of \(\begin{equation}f\end{equation}\) is bounded.
. Mahavier uses the second presentation to comment on the writing of mathematics and to help clarify the points that students have difficulty with.

Mahavier asks about Problem 70 PROBLEM 70
Assume \(\begin{equation}f\end{equation}\) is a continuous graph and \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is a sequence of points in the domain of \(\begin{equation}f\end{equation}\) whose range has \(\begin{equation}p\end{equation}\) as a limit point. Assume that \(\begin{equation}p\end{equation}\) is in the domain of \(\begin{equation}f\end{equation}\). Assume that the sequence \(\begin{equation}f(x_1),f(x_2)f(x_3),...\end{equation}\) converges to the number \(\begin{equation}c\end{equation}\). Show that \(\begin{equation}f(p)=c\end{equation}\).
and Norman has it, but only if a certain previous problem about sequences has a slightly different conclusion. The lemma he needs is indeed true and he goes to the board, is questioned by Mahavier, but once Mahavier realizes he has it then Mahavier helps with the lemma that is needed and Norman successfully presents Problem 70 PROBLEM 70
Assume \(\begin{equation}f\end{equation}\) is a continuous graph and \(\begin{equation}x_1,x_2,x_3,...\end{equation}\) is a sequence of points in the domain of \(\begin{equation}f\end{equation}\) whose range has \(\begin{equation}p\end{equation}\) as a limit point. Assume that \(\begin{equation}p\end{equation}\) is in the domain of \(\begin{equation}f\end{equation}\). Assume that the sequence \(\begin{equation}f(x_1),f(x_2)f(x_3),...\end{equation}\) converges to the number \(\begin{equation}c\end{equation}\). Show that \(\begin{equation}f(p)=c\end{equation}\).
.

Friday 4.18.2008, Good Friday, no class.

Day 37, missing tape, Monday 4.21.2008

We know from the following tape that Norman presents Problem 72 PROBLEM 72
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 \(\begin{equation}f'(0)=0\end{equation}\).
.

Day 37, Wednesday 4.23.2008.

Zahn completes half of Problem 76 PROBLEM 76
Show that the function \(\begin{equation}f\end{equation}\) is continuous at the number \(\begin{equation}c\end{equation}\) if and only if
1. \(\begin{equation}c\end{equation}\) is in the domain of \(\begin{equation}f\end{equation}\) and
2. if \(\begin{equation}\epsilon\end{equation}\) is a positive number, there is a positive number \(\begin{equation}\delta\end{equation}\) such that if \(\begin{equation}x\end{equation}\) is a number in the domain of \(\begin{equation}f\end{equation}\) and \(\begin{equation}|x-c| < \delta\end{equation}\), then \(\begin{equation}|f(x)-f(c)| < \epsilon\end{equation}\).
which is to show the equivalence between Mahavier's (Moore's) geometric definition of continuity with the usual epsilon-delta definition of continuity.

Here Mahavier makes his only off-topic comments of the semester. He points out something that is almost universally true, in my experience, of the Moore Method. He states his regret that the department has cut this course from a two semester course to a one semester course because, in his words, "at the end of the first semester, we are on a roll. We could cover enormous amounts of material in one more semester." Of course, if multiple people teach via the method then the time to adapt to the new style of teaching is mitigated and the students know how to prove theorems, so courses can cover significantly more material.

Day 38, Friday, 4.25.2008.

Rasmussen presents Problem 82 PROBLEM 82
Show that if \(\begin{equation}f\end{equation}\) is is a function whose domain is the interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}f(m)\end{equation}\) and \(\begin{equation}f(M)\end{equation}\) are the smallest and largest values of \(\begin{equation}f(x)\end{equation}\) for \(\begin{equation}x \in [a,b]\end{equation}\) and \(\begin{equation}P=\{t_0,t_1,...,t_n\}\end{equation}\) is any partition of \(\begin{equation}[a,b]\end{equation}\), then \(\begin{equation}U_P f \le f(M) (b-a)\end{equation}\) and \(\begin{equation}L_P f \ge f(m) (b-a)\end{equation}\).
.

Zahn completes the second half of Problem 76 PROBLEM 76
Show that the function \(\begin{equation}f\end{equation}\) is continuous at the number \(\begin{equation}c\end{equation}\) if and only if
1. \(\begin{equation}c\end{equation}\) is in the domain of \(\begin{equation}f\end{equation}\) and
2. if \(\begin{equation}\epsilon\end{equation}\) is a positive number, there is a positive number \(\begin{equation}\delta\end{equation}\) such that if \(\begin{equation}x\end{equation}\) is a number in the domain of \(\begin{equation}f\end{equation}\) and \(\begin{equation}|x-c| < \delta\end{equation}\), then \(\begin{equation}|f(x)-f(c)| < \epsilon\end{equation}\).
which is to show the equivalence between Mahavier's (Moore's) geometric definition of continuity and the usual epsilon-delta definition of continuity.

Mahavier very briefly discusses the way calculus books often use limits of sequences as well for continuity, pointing out that they have proved half of this.

Valero attempts Problem 84 PROBLEM 84
If \(\begin{equation}f\end{equation}\) is a continuous 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 \(\begin{equation}_U \int_a^b f>0\end{equation}\).
and understands it, but does not complete the proof. A brief, but productive clarification of what she has done and what must be done follows.

Rasmussen presents Problem 83 PROBLEM 83
If \(\begin{equation}f\end{equation}\) is a continuous function with domain the interval \(\begin{equation}[a,b]\end{equation}\), and \(\begin{equation}P\end{equation}\) is a partition of \(\begin{equation}[a,b]\end{equation}\), then \(\begin{equation}L_ P (f) \le U_P (f)\end{equation}\).
finishing just as class ends.

Day 39, no video, Monday 4.28.2008.

Day 40, no video, Final Exam, 5.1.2008.