# College Math Teaching

## March 16, 2015

### Compact Spaces and Tychonoff’s Theorem II

Filed under: advanced mathematics, topology — Tags: , , — collegemathteaching @ 6:10 pm

Ok, now lets prove the following: If $X, Y$ are compact spaces, then $X \times Y$ is compact (in the usual product topology). Note: this effectively proves that the finite product of compact spaces is compact. One might call this a “junior” Tychonoff Theorem.

Proof. We will prove this theorem a couple of times; the first proof is the more elementary but less elegant proof. It can NOT be easily extended to show that the arbitrary product of compact spaces is compact (which is the full Tychonoff Theorem).

We will show that an open covering of $X \times Y$ by basis elements of the form $U \times V$, $U$ open in $X$ and $V$ open in $Y$ has a finite subcover.

So let $\mathscr{U}$ be an open cover of $X \times Y$. Now fix $x_{\beta} \in X$ and consider the subset $x_{\beta} \times Y$. This subset is homeomorphic to $Y$ and is therefore compact; therefore there is a finite subcollection of $\mathscr{U}$ which overs $x_{\beta} \times Y$, say $\cup^{\beta, k}_{i=1} U_{\beta, i} \times V_{\beta, i}$ Note that each $U_{\beta, i}$ is an open set in $X$ which contains $x_{\beta}$ and there are only a finite number of these. Hence $\cap^{\beta k}_{i=1} U_{\beta i} = U_{\beta}$ is also an open set which contains $x_{\beta}$. Also know that $U_{\beta} \times Y \subset \cup^{\beta, k}_{i=1} U_{\beta, i} \times V_{\beta, i}$

We can do this for each $x_{\beta} \in X$ and so obtain an open cover of $X$ by $\cup_{x_{\beta} \in X} U_{\beta}$ and because $X$ is compact, a finite subcollection of these covers $X$. Call these $U_1, U_2, U_3....U_m$. For each one of these, we have $U_j \times Y \subset \cup^{j, k}_{i=1} U_{j, i} \times V_{j, i}$.

So, our finite subcover of $X \times Y$ is $\cup^m_{j=1}\cup^{j, k}_{i=1} U_{j, i} \times V_{j, i}$.

Now while this proof is elementary, it doesn’t extend to the arbitrary infinite product case.

So, to set up such an extension, we’ll give some “equivalent” definitions of compactness. Note: at some point, we’ll use some elementary cardinal arithmetic.

## April 27, 2013

### Unsolicited advise to young professors at heavy teaching load universities: Go to Research Conferences anyway!

This is coming to you from Ames, Iowa at the Spring American Mathematical Society Meeting. I am here to attend the sessions on the Topology of 3-dimensional manifolds.

Note: I try to go to conferences regularly; I have averaged about 1 conference a year. Sometimes, the conference is a MAA Mathfest conference. These ARE fun and refreshing. But sometimes (this year), I go to a research oriented conference.

I’ll speak for myself only.

Sometimes, these can be intimidating. Though many of the attendees are nice, cordial and polite, the fact is that many (ok, almost all of them) are either the best graduate students or among the finest researchers in the world. The big names who have proved the big theorems are here. They earn their living by doing cutting edge research and by guiding graduate students through their research; they are not spending hours and hours convincing students that $\sqrt{x^2 + y^2} \ne x + y$.

So, the talks can be tough. Sure, they do a good job, but remember that most of the audience is immersed in this stuff; they don’t have to review things like “normal surface theory” or “Haken manifold”.

Therefore, it is VERY easy to start lamenting (internally) “oh no, I am by far the dumbest one here”. That, in my case, IS true, but it is unimportant.
What I found is that, if I pay attention to what I can absorb, I can pick up a technique here and there, which I can then later use in my own research. In fact, just today, I picked up something that might help me with a problem that I am pondering.

Also, the atmosphere can be invigorating!

I happen to enjoy the conferences that are held on university campuses. There is nothing that gets my intellectual mood pumped up more than to hang around the campus of a division I research university. For me, there is nothing like it.

This conference
A few general remarks:
1. I didn’t realize how pretty Iowa State University is. I’d rank it along with the University of Tennessee as among the prettiest campuses that I’ve ever seen.

2. As far as the talks: one “big picture” technique that I’ve seen used again and again is the technique of: take an abstract set of objects (say, the Seifert Surfaces of a knot; say of minimal genus. Then to each, say, ambient isotopy class of Seifert Surface, assign a vertex of a graph or simplicial complex. Then group the vertices together either by a segment (in some settings) or a simplex (if, in one setting, the Seifert Surfaces admit disjoint representatives). Then one studies the complex or the graph.

In one of the talks (talking about essential closed surfaces in the complement of a knot), one assigned such things to the vertex of a graph (dendron actually) and set up an algorithm to search along such a graph; it turns out that is one starts near the top of this dendron, one gains the opportunity to prune lower branches of the group by doing the calculation near the top.

Sidenote
The weather couldn’t be better; I found time over lunch to do a 5.7 mile run near my hotel. The run was almost all on bike paths (albeit a “harder” surface than I’d like).

## April 1, 2013

### Fun for my Facebook Friends

Filed under: advanced mathematics, knot theory, topology — Tags: , — collegemathteaching @ 9:59 pm

Fun question one: can anyone see the relation between the following three figures? Note: I made a (sort of subtle) mistake in one of them….the one where the graph lines are showing)

Fun question two (a bit harder):

What is the relation between these figures?

And for the win: what is going on here? (this is ambiguous)

Ok, I’ll help you with the last one: imagine this process (one solid torus (think: doughnut or bagel) inside a larger one, and repeat this process (think: those Russian dolls that are nested). If you then take the infinite intersection, you get a simple closed curve (not obvious) that is so badly embedded, it fails to pierce a disk at any of its points (and certainly fails to have a tangent vector anywhere).

## March 25, 2013

### Mary Ellen Rudin Tribute (1924-2013): Shelling a triangulated disk or ball.

Filed under: advanced mathematics, editorial, famous mathematicians, topology — Tags: , — collegemathteaching @ 1:29 am

Back in January 1991, I attended the American Mathematical Society meeting in San Francisco. I was there mostly to absorb the talks, and to help myself with the job search. It was there where I made my first contact with my current university.

One evening, I attended some point set talks. Since this was not my area, I kept my mouth shut; I didn’t ask many questions. In one talk, someone raised a question about a possible “bisection” type process in a certain case. I thought “that question makes no sense because there is no metric” but figured that I had missed something.

Then in the other corner of the room, a white haired old lady said something to the effect “that question makes no sense because there is no mention of a metric on this space.” π

Then I realized who this was: Mary Ellen Rudin.

Because she was known for topology and because she was a student at the University of Texas (Ph. D. in 1949), I had heard stories about her and her interactions with R. H. Bing. It turns out that her (late) husband was also famous, although in analysis (yes, he is THAT Rudin; the one associated with the analysis books that we studied for our Ph. D. comprehensive examinations.)

I also remember reading her paper on shellings; it was the first one that I tackled as a graduate student. The paper itself is terse (and very brief) but is not overly technical; it just requires time and effort to understand. You don’t have to spend half a decade learning gauge theory. π Here is a workthrough of her paper.

So, in her honor, I’ll talk about this topic. I’ll keep it as basic as possible and try not to lie too much. π
(“lie” as in, leaving out some details that a rigorous presentation would cover).

Shellings: what are they?
First of all, let’s discuss what topology is about. Two objects $X, Y$ are said to be topologically equivalent if there exists a function $f:X \rightarrow Y$ where $f$ is a bijection (one to one and onto) and bicontinuous; that is $f$ is continuous and its inverse function $f^{-1}$ is also continuous. Note: I am using “continuous” in a topological sense, but for the spaces we are talking about, the beginner can imagine that $X, Y$ are subsets of $R^n$ for some $n \ge 1$ and that “continuous” is the “calculus” definition of continuous. This is NOT always true in general, but it works for the kind of objects we discuss here. If two objects are topologically equivalent, we say that they are “homeomorphic” and call $f$ the “homeomorphism”.

Now geometric topology restricts itself to the study of certain nice objects, and often we require that $f$ be differentiable (“differentiable topology”) or take polygonal objects to polygonal objects (“piecewise linear topology”, denoted by p. l.). We will work with polygonal maps, but one can make the translation to differentiable maps (“smooth category”).

Now consider a disk in the plane; the calculus student might think of the unit disk: $\{(x,y) | x^2 + y^2 \le 1 \}$.
In the p. l. category these objects are represented by triangles. It is an exercise to see that a traditional “round” disk is homeomorphic to a triangle (with interior).

Now consider a polygonal disk which is the finite union of triangles put together “in a nice way” (e. g., two triangles touch along a common face or exactly at a vertex, every interior point has an open set in the disk that surrounds it and the boundary consists of a finite number of line segments; faces of the triangles.). This crude illustration shows a “triangulated p. l. disk”

(figures: click to see larger)

One technical problem is this: it is clear that the whole disk is topologically equivalent to any one of its triangles. However it is often useful to see WHAT homeomorphism gives us the equivalence we want and to be able to describe this homeomorphism in a finite number of steps. Homeomorphisms which can be described “triangle by triangle” are especially useful. Now look at the labeled disk on the left. It is clear that the whole disk is homeomorphic to the disk that has all of the triangles EXCEPT for the first one. True, the process of simply removing triangle 1 is not a homeomorphism, but there is a homeomorphism that starts with the whole disk and, in effects, “pushes” triangle 1 into the larger disk

In spirit, this is a bit like the one dimensional homeomorphism $f: [0,1] \times [0,1] \rightarrow [0,1] \times [0,\frac{1}{2}]$ given by $f(x,y) =(x, \frac{1}{2}y)$ which compresses the unit square into a rectangle of half its original height.

So, in the disk on the left hand side of the first figure, we can find a series of homeomorphisms which, in step by step fashion, takes the whole disk to the triangle labeled 10. $f_1$ has the effect of taking the original disk and mapping it to the disk with triangle 1 missing. $f_2$ then starts with this modified disk and has the effect of removing triangle 2, and so on. One can check out that $f_1 \circ f_2 \circ ....\circ f_9$ is a homeomorphism from the original disk to the triangle labeled 10.

So, we an think about these maps as a labeling of this triangulated p. l. disk and such a labeling is called a “shelling” of the p. l. disk with the triangle 10 saved for last. More precisely: a shelling is a labeling of a triangulated p. l. disk with $N$ triangles $\Delta_i$ so that, for ALL $1 \le k \le N, \cup^{N}_{i=k} \Delta_i$ is a p. l. disk.

Note: not just any labeling produces a shelling! For example, if you look at the first figure and look at the disk on the right, the removal of the triangle labeled “1” leaves us with an annulus (ring with a hole) and NOT a disk; hence there is no homeomorphism that does that!

So here is a question: does every 2 dimensional p. l. disk have a shelling? The answer: “yes” and not only that: we can choose, in advance, the ending triangle. There is always a path to the last triangle (though we’ve seen that we do NOT have our choice of starting triangle).

To see a proof, see Bing’s book The Geometric Topology of 3-manifolds:

This is an outline of the proof. It is fun to try it yourself.

Ok, good enough. But the follow on question:
if we now try to extend this “shellable 3-balls”: if we have a 3-dimensional triangulated (into tetrahedra) ball (homeomorphic to $\{(x,y,z)| x^2+y^2+z^2 \le 1 \}$) can you always shell it?

The answer is NO. That is the subject of Mary Ellen Rudin’s paper that I talked about at the start. However there are easier counterexamples, and I’ll close with my favorite, which is suggested by this figure:

Here is an idea: start with a cube (homeomorphic to a ball) and break this into a bunch of small cubes. Then start at the top of the cube and start removing a cube in the middle of the top face of the cube. Then work your way down, removing small cubes so as to leave a “knotted tunnel” from the top face of the cube to the bottom face. Of course what is left is NOT a ball due to this hole; you have “added genus”. Now we need to make this a ball, so we add the last “removed cube” back (“plugging the hole”); hence the “cube with knotted tunnel” is now a cube with an indentation in the form of an “almost hole”; hence it remains homeomorphic to a ball.

Now divide the remaining small cubes into tetrahedra to obtain a triangulated 3-ball.

It turns out that this ball cannot be shelled. Here is why:

1. If one removes the “plug” by shelling before ALL of the cubes above the plane of the plug cube have been removed (by removing tetrahedra) one changes the genus from zero (a ball) to non-zero (not a ball).

2. If one tries to remove all of the cubes in the plane above plug (by removing the tetrahedra), this has the effect of taking the knotted tunnel to the sides of the cube via a homeomorphism of space; this is impossible if the tunnel truly is knotted! (that is why having an honest to goodness knot is essential).

So the existence of knots in 3-space really complicates things. Hence “knot theory” remains an active field of research in mathematics.

## February 26, 2013

### Undergraduate Topology: what is that?

Filed under: advanced mathematics, editorial, topology — Tags: , — collegemathteaching @ 1:44 am

I remember looking at the course schedule when I was near the end of my first semester, senior year. I noticed that a “topology” course was being offered; I also remember reading a bit about “topology” in the Time-Life book on mathematics. I remembered a donut shaped object (called torus), Klein Bottles, Mobius bands and the like. I wondered if the course would be a sequence of parlor tricks.

Then when I got to the course; well….it wasn’t what I expected.

Ok, what is going on? I heard terms like “open”, “closed”, “basis”, “Hausdorf”, “Regular”, “Normal”, “second countable”, “first countable” etc. It reminded me of “analysis on steroids”. I began to wonder if the Time-Life book was just making it all up.

But it wasn’t; toward the back of our text (Munkres) there was even a drawing of a double torus.

So, what in the world is going on?

Well, I’ll just write informally; if you can catch my omissions you already know this stuff. π

Probably the most basic question topology asks is: “when are two spaces “the same””. So, what do you mean by “the same”?

In topology the answer is almost always: given spaces $X, Y$ is there a continuous bijection (one to one and onto function) $f: X \rightarrow Y$ which as a continuous inverse $f^{-1}$? If there is such a function, the spaces are said to be “topologically equivalent” or “homeomorphic”. You might notice the term “continuous” and wonder what it means in an abstract context like this.

The usual calculus “epsilon-delta” definition works if $X, Y$ are real n-spaces with the usual open intervals/disks/balls, etc.
So here is an elementary example: the unit interval $[0,1]$ is homeomorphic to any other closed interval $[a,b]$. Here is the proof: the map $f$ (called a “homeomorphism”) is given by $f(x) = (b-a)x + a$. Notice that $f$ is a bijection and has a continuous inverse. On the other hand, $[0,1]$ is NOT homeomorphic to $(0,1]$. The standard way to see the latter is to use the tools developed in either analysis courses or a beginning topology course that I am describing; the quick answer is that the closed interval is “compact” whereas the half-open interval is not. Or, another way: $[0,1]$ has all of its limit points; $(0,1]$ is missing a limit point.

You go on to talk about what an open set is. The calculus notion is that an open set is one that is built up as the collection of open intervals (in the real line) or open disks, open balls, etc. In topology, you take a collection of sets $T$ and if this collection meets the following properties: the whole space and the empty sets belong to $T$ and if an arbitrary union of subsets of $T$ belong to $T$ and any FINITE intersection of elements of $T$ belong to $T$, then $T$ is said to be a topology for the space.

Clearly, the standard “open” intervals form a topology for the real line; we do calculus with these. But one can form a topology generated by, say, half open intervals (these things are bizarre) or other stranger collection of sets.

Yes, the complement of an open set is a “closed set”, and yes, in some topological spaces, there are sets that are both open and closed at the same time. (grrr…) Example: given a set $X$, declare EVERY point in $X$ to be an open set.

It is the study of these things that often constitute the first part of a first topology course. And yes, you DO need to know this stuff.

So what about the geometric stuff?
Well, let’s start small. You know that there is a bijection between $[0,1)$ and the unit circle $x^2 + y^2 =1$ (the bijection is $f(x) = e^{i2\pi x}$. You also know that $[0,1]$ and the unit circle are both compact sets (think: “closed and bounded” if you are new). But they are NOT homeomorphic sets. Learning why starts you in the more geometric direction. One easy way: if you look at the intervals, removal of any point except the end points separates the intervals into two pieces, where no one point separates the unit circle into two pieces. That is a more “global” property.

So in this more global view, you’ll learn not only geometric type arguments but also how do use algebra (yes, at advanced levels, math is not as compartmentalized as it appears to many undergraduates). That is part of algebraic topology.

And, of course, to simplify the types of objects studied, one might want to but a differential structure on a space (assign the notion of a derivative and “tangents”) by attaching something called the “tangent bundle” to a space. That is the subject of differential topology. Here the homeomorphisms are often required to be infinitely differentiable as well.

So, yes, there IS a connection between the “rubber sheet” geometry stuff that you read about in the popular media and the abstract sounding stuff that you sometimes get at the start of an undergraduate topology class. It just takes a bit of time and effort to get there.

Now the type of topology that really never gets to the geometric stuff is called “point set” topology; there is something called “general topology” too. (don’t ask). π