# College Math Teaching

## August 28, 2017

### Integration by parts: why the choice of “v” from “dv” might matter…

We all know the integration by parts formula: $\int u dv = uv - \int v du$ though, of course, there is some choice in what $v$ is; any anti-derivative will do. Well, sort of.

I thought about this as I’ve been roped into teaching an actuarial mathematics class (and no, I have zero training in this area…grrr…)

So here is the set up: let $F_x(t) = P(0 \leq T_x \leq t)$ where $T_x$ is the random variable that denotes the number of years longer a person aged $x$ will live. Of course, $F_x$ is a probability distribution function with density function $f$ and if we assume that $F$ is smooth and $T_x$ has a finite expected value we can do the following: $E(T_x) = \int^{\infty}_0 t f_x(t) dt$ and, in principle this integral can be done by parts….but…if we use $u = t, dv = f_x(t), du = dt, v = F_x$ we have:

\

$t(F_x(t))|^{\infty}_0 -\int^{\infty}_0 F_x(t) dt$ which is a big problem on many levels. For one, $lim_{t \rightarrow \infty}F_x(t) = 1$ and so the new integral does not converge..and the first term doesn’t either.

But if, for $v = -(1-F_x(t))$ we note that $(1-F_x(t)) = S_x(t)$ is the survival function whose limit does go to zero, and there is usually the assumption that $tS_x(t) \rightarrow 0$ as $t \rightarrow \infty$

So we now have: $-(S_x(t) t)|^{\infty}_0 + \int^{\infty}_0 S_x(t) dt = \int^{\infty}_0 S_x(t) dt = E(T_x)$ which is one of the more important formulas.

## August 1, 2017

### Numerical solutions to differential equations: I wish that I had heard this talk first

The MAA Mathfest in Chicago was a success for me. I talked about some other talks I went to; my favorite was probably the one given by Douglas Arnold. I wish I had had this talk prior to teaching numerical analysis for the fist time.

Confession: my research specialty is knot theory (a subset of 3-manifold topology); all of my graduate program classes have been in pure mathematics. I last took numerical analysis as an undergraduate in 1980 and as a “part time, not taking things seriously” masters student in 1981 (at UTSA of all places).

In each course…I. Made. A. “C”.

Needless to say, I didn’t learn a damned thing, even though both professors gave decent courses. The fault was mine.

But…I was what my department had, and away I went to teach the course. The first couple of times, I studied hard and stayed maybe 2 weeks ahead of the class.
Nevertheless, I found the material fascinating.

When it came to understanding how to find a numerical approximation to an ordinary differential equation (say, first order), you have: $y' = f(t,y)$ with some initial value for both $y'(0), y(0)$. All of the techniques use some sort of “linearization of the function” technique to: given a step size, approximate the value of the function at the end of the next step. One chooses a step size, and some sort of schemes to approximate an “average slope” (e. g. Runga-Kutta is one of the best known).

This is a lot like numerical integration, but in integration, one knows $y'(t)$ for all values; here you have to infer $y'(t)$ from previous approximations of %latex y(t) \$. And there are things like error (often calculated by using some sort of approximation to $y(t)$ such as, say, the Taylor polynomial, and error terms which are based on things like the second derivative.

And yes, I faithfully taught all that. But what was unknown to me is WHY one might choose one method over another..and much of this is based on the type of problem that one is attempting to solve.

And this is the idea: take something like the Euler method, where one estimates $y(t+h) \approx y(t) + y'(t)h$. You repeat this process a bunch of times thereby obtaining a sequence of approximations for $y(t)$. Hopefully, you get something close to the “true solution” (unknown to you) (and yes, the Euler method is fine for existence theorems and for teaching, but it is too crude for most applications).

But the Euler method DOES yield a piecewise linear approximation to SOME $f(t)$ which might be close to $y(t)$ (a good approximation) or possibly far away from it (a bad approximation). And this $f(t)$ that you actually get from the Euler (or other method) is important.

It turns out that some implicit methods (using an approximation to obtain $y(t+h)$ and then using THAT to refine your approximation can lead to a more stable system of $f(t)$ (the solution that you actually obtain…not the one that you are seeking to obtain) in that this system of “actual functions” might not have a source or a sink…and therefore never spiral out of control. But this comes from the mathematics of the type of equations that you are seeking to obtain an approximation for. This type of example was presented in the talk that I went to.

In other words, we need a large toolbox of approximations to use because some methods work better with certain types of problems.

I wish that I had known that before…but I know it now. 🙂

### Big lesson that many overlook: math is hard

Filed under: advanced mathematics, conference, editorial, mathematician, mathematics education — Tags: — collegemathteaching @ 11:43 am

First of all, it has been a very long time since I’ve posted something here. There are many reasons that I allowed myself to get distracted. I can say that I’ll try to post more but do not know if I will get it done; I am finishing up a paper and teaching a course that I created (at the request of the Business College), and we have a record enrollment..many of the new students are very unprepared.

Back to the main topic of the post.

I just got back from MAA Mathfest and I admit that is one of my favorite mathematics conferences. Sure, the contributed paper sessions give you a tiny amount of time to present, but the main talks (and many of the simple talks) are geared toward those of us who teach mathematics for a living and do some research on the side; there are some mainstream “basic” subjects that I have not seen in 30 years!

That doesn’t mean that they don’t get excellent people for the main speaker; they do. This time, the main speaker was Dusa McDuff: someone who was a member of the National Academy of Sciences. (a very elite level!)

Her talk was on the basics of symplectec geometry (introductory paper can be found here) and the subject is, well, HARD. But she did an excellent job of giving the flavor of it.

I also enjoyed Erica Flapan’s talk on graph theory and chemistry. One of my papers (done with a friend) referenced her work.

I’ll talk about Douglas Arnold’s talk on “when computational math meets geometry”; let’s just say that I wish I had seen this lecture prior to teaching the “numerical solutions for differential equations” section of numerical analysis.

Well, it looks as if I have digressed yet again.

There were many talks, and some were related to the movie Hidden Figures. And the cheery “I did it and so can you” talks were extremely well attended…applause, celebration, etc.

The talks on sympletec geometry: not so well attended toward the end. Again, that stuff is hard.

And that is one thing I think that we miss when we encourage prospective math students: we neglect to tell them that research level mathematics is difficult stuff and, while some have much more talent for it than others, everyone has to think hard, has to work hard, and almost all of us will fail, quite a bit.

I remember trying to spend over a decade trying to prove something, only to fail and to see a better mathematician get the result. One other time I spent 2 years trying to “prove” something…and I couldn’t “seal the deal”. Good thing too, as what I was trying to prove was false..and happily I was able to publish the counterexample.

## June 7, 2016

### Pop-math: getting it wrong but being close enough to give the public a feel for it

Space filling curves: for now, we’ll just work on continuous functions $f: [0,1] \rightarrow [0,1] \times [0,1] \subset R^2$.

A curve is typically defined as a continuous function $f: [0,1] \rightarrow M$ where $M$ is, say, a manifold (a 2’nd countable metric space which has neighborhoods either locally homeomorphic to $R^k$ or $R^{k-1})$. Note: though we often think of smooth or piecewise linear curves, we don’t have to do so. Also, we can allow for self-intersections.

However, if we don’t put restrictions such as these, weird things can happen. It can be shown (and the video suggests a construction, which is correct) that there exists a continuous, ONTO function $f: [0,1] \rightarrow [0,1] \times [0,1]$; such a gadget is called a space filling curve.

It follows from elementary topology that such an $f$ cannot be one to one, because if it were, because the domain is compact, $f$ would have to be a homeomorphism. But the respective spaces are not homeomorphic. For example: the closed interval is disconnected by the removal of any non-end point, whereas the closed square has no such separating point.

Therefore, if $f$ is a space filling curve, the inverse image of a points is actually an infinite number of points; the inverse (as a function) cannot be defined.

And THAT is where this article and video goes off of the rails, though, practically speaking, one can approximate the space filling curve as close as one pleases by an embedded curve (one that IS one to one) and therefore snake the curve through any desired number of points (pixels?).

So, enjoy the video which I got from here (and yes, the text of this post has the aforementioned error)

## February 5, 2016

### More fun with selective sums of divergent series

Just a reminder: if $\sum_{k=1}^{\infty} a_k$ is a series and $c_1, c_2, ...c_n ,,$ is some sequence consisting of 0’s and 1’s then a selective sum of the series is $\sum_{k=1}^{\infty} c_k a_k$. The selective sum concept is discussed in the MAA book Real Infinite Series (MAA Textbooks) by Bonar and Khoury (2006) and I was introduced to the concept by Ferdinands’s article Selective Sums of an Infinite Series in the June 2015 edition of Mathematics Magazine (Vol. 88, 179-185).

There is much of interest there, especially if one considers convergent series or alternating series.

This post will be about divergent series of positive terms for which $lim_{n \rightarrow \infty} a_n = 0$ and $a_{n+1} < a_n$ for all $n$.

The first fun result is this one: any selected $x > 0$ is a selective sum of such a series. The proof of this isn’t that bad. Since $lim_{n \rightarrow \infty} a_n = 0$ we can find a smallest $n$ such that $a_n \leq x$. Clearly if $a_n = x$ we are done: our selective sum has $c_n = 1$ and the rest of the $c_k = 0$.

If not, set $n_1 = n$ and note that because the series diverges, there is a largest $m_1$ so that $\sum_{k=n_1}^{m_1} a_k \leq x$. Now if $\sum_{k=n_1}^{m_1} a_k = x$ we are done, else let $\epsilon_1 = x - \sum_{k=n_1}^{m_1} a_k$ and note $\epsilon_1 < a_{m_1+1}$. Now because the $a_k$ tend to zero, there is some first $n_2$ so that $a_{n_2} \leq \epsilon_1$. If this is equality then the required sum is $a_{n_2} + \sum_{k=n_1}^{m_1} a_k$, else we can find the largest $m_2$ so that $\sum_{k=n_1}^{m_1} a_k + \sum_{k=n_2}^{m_2} a_k \leq x$

This procedure can be continued indefinitely. So if we label $\sum_{k=n_j}^{m_{j}} a_k = s_j$ we see that $s_1 + s_2 + ...s_{n} = t_{n}$ form an increasing, bounded sequence which converges to the least upper bound of its range, and it isn’t hard to see that the least upper bound is $x$ because $x-t_{n} =\epsilon_n < a_{m_n+1}$

So now that we can obtain any positive real number as the selective sum of such a series, what can we say about the set of all selective sums for which almost all of the $c_k = 0$ (that is, all but a finite number of the $c_k$ are zero).

Answer: the set of all such selective sums are dense in the real line, and this isn’t that hard to see, given our above construction. Let $(a,b)$ be any open interval in the real line and let $a < x < b$. Then one can find some $N$ such that for all $n > N$ we have $x - a_n > a$. Now consider our construction and choose $m$ large enough such that $x - t_m > x - a_n > a$. Then the $t_m$ represents the finite selected sum that lies in the interval $(a,b)$.

We can be even more specific if we now look at a specific series, such as the harmonic series $\sum_{k=1}^{\infty} \frac{1}{k}$. We know that the set of finite selected sums forms a dense subset of the real line. But it turns out that the set of select sums is the rationals. I’ll give a slightly different proof than one finds in Bonar and Khoury.

First we prove that every rational in $(0,1]$ is a finite select sum. Clearly 1 is a finite select sum. Otherwise: Given $\frac{p}{q}$ we can find the minimum $n$ so that $\frac{1}{n} \leq \frac{p}{q} < \frac{1}{n-1}$. If $\frac{p}{q} = \frac{1}{n}$ we are done. Otherwise: the strict inequality shows that $pn-p < q$ which means $pn-q < p$. Then note $\frac{p}{q} - \frac{1}{n} = \frac{pn-q}{qn}$ and this fraction has a strictly smaller numerator than $p$. So we can repeat our process with this new rational number. And this process must eventually terminate because the numerators generated from this process form a strictly decreasing sequence of positive integers. The process can only terminate when the new faction has a numerator of 1. Hence the original fraction is some sum of fractions with numerator 1.

Now if the rational number $r$ in question is greater than one, one finds $n_1$ so that $\sum^{n_1}_{k=1} \frac{1}{k} \leq r$ but $\sum^{n_1+1}_{k=1} \frac{1}{k} > r$. Then write $r-\sum^{n_1+1}_{k=1} \frac{1}{k}$ and note that its magnitude is less than $\frac{1}{n_1+1}$. We then use the procedure for numbers in $(0,1)$ noting that our starting point excludes the previously used terms of the harmonic series.

There is more we can do, but I’ll stop here for now.

## January 20, 2016

### Congratulations to the Central Missouri State Mathematics Department

Filed under: advanced mathematics, editorial, number theory — Tags: — collegemathteaching @ 10:43 pm

The largest known prime has been discovered by mathematicians at Central Missouri State University.

For what it is worth, it is: $2^{74,207,281} -1$.

Now if you want to be depressed, go to the Smithsonian Facebook page and read the comment. The Dunning-Kruger effect is real. Let’s just say that in our era, our phones are smarter than our people. 🙂

## July 28, 2015

### J. H. Conway, Terry Tao and avoiding work

Filed under: advanced mathematics, algebra, media — Tags: , , , , , — collegemathteaching @ 7:48 pm

The mainstream media recently had some excellent articles on two mathematical giants:

John Conway and Terrance Tao. I’ve never met Terry Tao though I do read (or try to follow) his blog.

I did meet John Conway when he visited the University of Texas. He is a friend of my dissertation advisor and gave some talks on knot diagram colorings.

I had a private conversation with him at a party, and he gave me some ideas which resulted in three papers for me! Here is one of them.

Yes, I am avoiding studying a book on the theory of interest; I am teaching that course this fall and need to get ahead of the game.

Unfortunately, when I don’t teach, my use of time becomes undisciplined.

## June 25, 2015

### Workshop in Geometric Topology: TCU 2015 morning session 1

Filed under: advanced mathematics, conference, editorial, topology — Tags: — collegemathteaching @ 3:59 pm

I’ll be blunt: I’ve been teaching at a 11-12 hour load (mostly 11; one time I had a 9 hour load; 3 courses) since fall, 1991. Though I’ve published, most of what I’ve done has been extremely “bare handed”; it is tough to learn the most advanced techniques (which is a full time job in and of itself)

So, at math conferences, I get to see how much further behind I’ve fallen.

But these things help in the following way:

1. They are an excellent change of pace from the usual routine of teaching calculus.
2. I do learn things, even if it is “looking up” a definition or two; for example I looked up the definition of “pure braid group” in between the 20 minute talks.
3. I have to review my own stuff to see if I am indeed making progress; I don’t want to say something idiotic in front of some very smart, informed people.

But yes, the talks have been given by smart, (mostly) young, energetic people who have been studying the topic that they are talking about very intensely for a long time; frequently it is tough to hang in to the second half of the 20 minute talks. But I can see WHAT is being studied, what tools are being used and, as I said before, find stuff to look up.

The final talk: didn’t understand much beyond the general gist but it was well organized, well presented..exactly what you get when you have a brilliant energetic young researcher working full time in mathematical research.

On one hand, I envy his talent. On the other hand, I am glad that we have some smart humans among us; they benefit all of us.

The trip here The plane was about 2.5 hours late getting in, then there was a long ride to the car rental place and a 35 minute drive to campus, then finding my way around in the dark. So no morning run; I might do a gentle “after the talks” focused walk (5K-ish?).

I talk at 9 am tomorrow and I want to make it worth their while.

## June 19, 2015

### Scientific American article about finite simple groups

Filed under: advanced mathematics, algebra, mathematician — Tags: , — collegemathteaching @ 2:42 pm

For those of you who are a bit rusty: a finite group is a group that has a finite number of elements. A simple group is one that has no proper non-trivial normal subgroups (that is, only the identity and the whole group are normal subgroups).

It is a theorem that if $G$ is a finite simple group then $G$ falls into one of the following categories:

1. Cyclic (of prime order, of course)
2. Alternating (and not isomorphic to $A_4$ of course)
3. A member of a subclass of Lie Groups
4. One of 26 other groups that don’t fall into 1, 2 or 3.

Scientific American has a nice article about this theorem and the effort to get it written down and understood; the problem is that the proof of such a theorem is far from simple; it spans literally hundreds of research articles and would take thousands of pages to be complete. And, those who have an understanding of this result are aging and won’t be with us forever.

Here is a link to the preview of the article; if you don’t subscribe to SA it is probably in your library.

## April 10, 2015

### Cantor sets and countable products of discrete spaces (0, 1)^Z

Filed under: advanced mathematics, analysis, point set topology, sequences — Tags: , , , — collegemathteaching @ 11:09 am

This might seem like a strange topic but right now our topology class is studying compact metric spaces. One of the more famous of these is the “Cantor set” or “Cantor space”. I discussed the basics of these here.

Now if you know the relationship between a countable product of two point discrete spaces (in the product topology) and Cantor spaces/Cantor Sets, this post is probably too elementary for you.

Construction: start with a two point set $D = \{0, 2 \}$ and give it the discrete topology. The reason for choosing 0 and 2 to represent the elements will become clear later. Of course, $D_2$ is a compact metric space (with the discrete metric: $d(x,y) = 1$ if $x \neq y$.

Now consider the infinite product of such spaces with the product topology: $C = \Pi^{\infty}_{i=1} D_i$ where each $D_i$ is homeomorphic to $D$. It follows from the Tychonoff Theorem that $C$ is compact, though we can prove this directly: Let $\cup_{\alpha \in I} U_{\alpha}$ be any open cover for $C$. Then choose an arbitrary $U$ from this open cover; because we are using the product topology $U = O_1 \times O_2 \times ....O_k \times (\Pi_{i=k+1}^{\infty} D_i )$ where each $O_i$ is a one or two point set. This means that the cardinality of $C - U$ is at most $2^k -1$ which requires at most $2^k -1$ elements of the open cover to cover.

Now let’s examine some properties.

Clearly the space is Hausdorff ($T_2$ ) and uncountable.

1. Every point of $C$ is a limit point of $C$. To see this: denote $x \in C$ by the sequence $\{x_i \}$ where $x_i \in \{0, 2 \}$. Then any open set containing $\{ x_i \}$ is $O_1 \times O_2 \times...O_k \times \Pi^{\infty}_{i=k+1} D_i$ and contains ALL points $y_i$ where $y_i = x_i$ for $i = \{1, 2, ...k \}$. So all points of $C$ are accumulation points of $C$; in fact they are condensation points (or perfect limit points ).

(refresher: accumulation points are those for which every open neighborhood contains an infinite number of points of the set in question; condensation points contain an uncountable number of points, and perfect limit points are those for which every open neighborhood contains as many points as the set in question has (same cardinality).

2. $C$ is totally disconnected (the components are one point sets). Here is how we will show this: given $x, y \in C, x \neq y,$ there exists disjoint open sets $U_x, U_y, x \in U_x, y \in U_y, U_x \cup U_y = C$. Proof of claim: if $x \neq y$ there exists a first coordinate $k$ for which $x_k \neq y_k$ (that is, a first $k$ for which the canonical projection maps disagree ($\pi_k(x) \neq pi_k(y)$ ). Then
$U_x = D_1 \times D_2 \times ....\times D_{k-1} \times x_k \times \Pi^{\infty}_{i=k+1} D_i$,

$U_y = D_1 \times D_2 \times.....\times D_{k-1} \times y_k \times \Pi^{\infty}_{i = k+1} D_i$

are the required disjoint open sets whose union is all of $C$.

3. $C$ is countable, as basis elements for open sets consist of finite sequences of 0’s and 2’s followed by an infinite product of $D_i$.

4. $C$ is metrizable as well; $d(x,y) = \sum^{\infty}_{i=1} \frac{|x_i - y_i|}{3^i}$. Note that is metric is well defined. Suppose $x \neq y$. Then there is a first $k, x_k \neq y_k$. Then note

$d(x,y) = \frac{|x_k - y_k|}{3^k} + \sum^{\infty}_{i = k+1} \frac{|x_i - y_i|}{3^i} \rightarrow |x_k -y_k| =2 = \sum^{\infty}_{i=1} \frac{|x_{i+k} -y_{i+k}|}{3^i} \leq \frac{1}{3} \frac{1}{1 -\frac{2}{3}} =1$

which is impossible.

5. By construction $C$ is uncountable, though this follows from the fact that $C$ is compact, Haudorff and dense in itself.

6. $C \times C$ is homeomorphic to $C$. The homeomorphism is given by $f( \{x_i \}, \{y_i \}) = \{ x_1, y_1, x_2, y_2,... \} \in C$. It follows that $C$ is homeomorphic to a finite product with itself (product topology). Here we use the fact that if $f: X \rightarrow Y$ is a continuous bijection with $X$ compact and $Y$ Hausdorff then $f$ is a homeomorphism.

Now we can say a bit more: if $C_i$ is a copy of $C$ then $\Pi^{\infty}_{i =1 } C_i$ is homeomorphic to $C$. This will follow from subsequent work, but we can prove this right now, provided we review some basic facts about countable products and counting.

First lets show that there is a bijection between $Z \times Z$ and $Z$. A bijection is suggested by this diagram:

which has the following formula (coming up with it is fun; it uses the fact that $\sum^k _{n=1} n = \frac{k(k+1)}{2}$:

$\phi(k,1) = \frac{(k)(k+1)}{2}$ for $k$ even
$\phi(k,1) = \frac{(k-1)(k)}{2} + 1$ for $k$ odd
$\phi(k-j, j+1) =\phi(k,1) + j$ for $k$ odd, $j \in \{1, 2, ...k-1 \}$
$\phi(k-j, j+1) = \phi(k,1) - j$ for $k$ even, $j \in \{1, 2, ...k-1 \}$

Here is a different bijection; it is a fun exercise to come up with the relevant formulas:

Now lets give the map between $\Pi^{\infty}_{i=1} C_i$ and $C$. Let $\{ y_i \} \in C$ and denote the elements of $\Pi^{\infty}_{i=1} C_i$ by $\{ x^i_j \}$ where $\{ x_1^1, x_2^1, x_3^ 1....\} \in C_1, \{x_1^2, x_2 ^2, x_3^3, ....\} \in C_2, ....\{x_1^k, x_2^k, .....\} \in C_k ...$.

We now describe a map $f: C \rightarrow \Pi^{\infty}_{i=1} C_i$ by

$f(\{y_i \}) = \{ x^i_j \} = \{y_{\phi(i,j)} \}$

Example: $x^1_1 = y_1, x^1_2 = y_2, x^2_1 = y_3, x^3_1 =y_4, x^2_2 = y_5, x^1_3 =y_6,...$

That this is a bijection between compact Hausdorff spaces is immediate. If we show that $f^{-1}$ is continuous, we will have shown that $f$ is a homeomorphism.

But that isn’t hard to do. Let $U \subset C$ be open; $U = U_1 \times U_2 \times U_3.... \times U_{m-1} \times \Pi^{\infty}_{k=m} C_k$. Then there is some $k_m$ for which $\phi(k_m, 1) \geq M$. Then if $f^i_j$ denotes the $i,j$ component of $f$ we wee that for all $i+j \geq k_m+1, f^i_j(U) = C$ (these are entries on or below the diagonal containing $(k,1)$ depending on whether $k_m$ is even or odd.

So $f(U)$ is of the form $V_1 \times V_2 \times ....V_{k_m} \times \Pi^{\infty}_{i = k_m +1} C_i$ where each $V_j$ is open in $C_j$. This is an open set in the product topology of $\Pi^{\infty}_{i=1} C_i$ so this shows that $f^{-1}$ is continuous. Therefore $f^{-1}$ is a homeomorphism, therefore so is $f$.

Ok, what does this have to do with Cantor Sets and Cantor Spaces?

If you know what the “middle thirds” Cantor Set is I urge you stop reading now and prove that that Cantor set is indeed homeomorphic to $C$ as we have described it. I’ll give this quote from Willard, page 121 (Hardback edition), section 17.9 in Chapter 6:

The proof is left as an exercise. You should do it if you think you can’t, since it will teach you a lot about product spaces.

What I will do I’ll give a geometric description of a Cantor set and show that this description, which easily includes the “deleted interval” Cantor sets that are used in analysis courses, is homeomorphic to $C$.

Set up
I’ll call this set $F$ and describe it as follows:

$F \subset R^n$ (for those interested in the topology of manifolds this poses no restrictions since any manifold embeds in $R^n$ for sufficiently high $n$).

Reminder: the diameter of a set $F \subset R^n$ will be $lub \{ d(x,y) | x, y \in F \}$
Let $\epsilon_1, \epsilon_2, \epsilon_3 .....\epsilon_k ...$ be a strictly decreasing sequence of positive real numbers such that $\epsilon_k \rightarrow 0$.

Let $F^0$ be some closed n-ball in $R^n$ (that is, $F^)$ is a subset homeomorphic to a closed n-ball; we will use that convention throughout)

Let $F^1_{(0) }, F^1_{(2)}$ be two disjoint closed n-balls in the interior of $F^0$, each of which has diameter less than $\epsilon_1$.

$F^1 = F^1_{(0) } \cup F^1_{(2)}$

Let $F^2_{(0, 0)}, F^2_{(0,2)}$ be disjoint closed n-balls in the interior $F^1_{(0) }$ and $F^2_{(2,0)}, F^2_{(2,2)}$ be disjoint closed n-balls in the interior of $F^1_{(2) }$, each of which (all 4 balls) have diameter less that $\epsilon_2$. Let $F^2 = F^2_{(0, 0)}\cup F^2_{(0,2)} \cup F^2_{(2, 0)} \cup F^2_{(2,2)}$

To describe the construction inductively we will use a bit of notation: $a_i \in \{0, 2 \}$ for all $i \in \{1, 2, ...\}$ and $\{a_i \}$ will represent an infinite sequence of such $a_i$.
Now if $F^k$ has been defined, we let $F^{k+1}_{(a_1, a_2, ...a_{k}, 0)}$ and $F^{k+1}_{(a_1, a_2,....,a_{k}, 2)}$ be disjoint closed n-balls of diameter less than $\epsilon_{k+1}$ which lie in the interior of $F^k_{(a_1, a_2,....a_k) }$. Note that $F^{k+1}$ consists of $2^{k+1}$ disjoint closed n-balls.

Now let $F = \cap^{\infty}_{i=1} F^i$. Since these are compact sets with the finite intersection property ($\cap^{m}_{i=1}F^i =F^i \neq \emptyset$ for all $m$ ), $F$ is non empty and compact. Now for any choice of sequence $\{a_i \}$ we have $F_{ \{a_i \} } =\cap^{\infty}_{i=1} F^i_{(a_1, ...a_i)}$ is nonempty by the finite intersection property. On the other hand, if $x, y \in F, x \neq y$ then $d(x,y) = \delta > 0$ so choose $\epsilon_m$ such that $\epsilon_m < \delta$. Then $x, y$ lie in different components of $F^m$ since the diameters of these components are less than $\epsilon_m$.

Then we can say that the $F_{ \{a_i} \}$ uniquely define the points of $F$. We can call such points $x_{ \{a_i \} }$

Note: in the subspace topology, the $F^k_{(a_1, a_2, ...a_k)}$ are open sets, as well as being closed.

Finding a homeomorphism from $F$ to $C$.
Let $f: F \rightarrow C$ be defined by $f( x_{ \{a_i \} } ) = \{a_i \}$. This is a bijection. To show continuity: consider the open set $U = y_1 \times y_2 ....\times y_m \times \Pi^{\infty}_{i=m} D_i$. Under $f$ this pulls back to the open set (in the subspace topology) $F^{m+1}_{(y1, y2, ...y_m, 0 ) } \cup F^{m+1}_{(y1, y2, ...y_m, 2)}$ hence $f$ is continuous. Because $F$ is compact and $C$ is Hausdorff, $f$ is a homeomorphism.

This ends part I.

We have shown that the Cantor sets defined geometrically and defined via “deleted intervals” are homeomorphic to $C$. What we have not shown is the following:

Let $X$ be a compact Hausdorff space which is dense in itself (every point is a limit point) and totally disconnected (components are one point sets). Then $X$ is homeomorphic to $C$. That will be part II.

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