# College Math Teaching

## August 11, 2016

### Post Promotion Summer

Filed under: editorial, topology — Tags: — collegemathteaching @ 12:02 am

This is my first “terminal promotion” summer. And while I have something that I have “sort of” written up…I just don’t like the result; it basically fills in some gaps in a survey article. But I think that my thinking about this article has lead me to something that I can add to the paper so that I’ll actually LIKE what I submit.

Then again, my quandary can be summed up in this tweet:

If I wait until I am absolutely in love with my work before I send it out, it will never get sent out.

Hopefully, I’ll have more material to add to this blog this semester.

What I am working on: equivalence classes of simple closed curves; these are one to one, continuous images of the unit circle in 3-space. The objects that I am studying are so pathological that these curves fail to have a tangent at ANY point. One of these beasts can be constructed by taking the intersection of these nested, solid tori.

## 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)

## May 20, 2016

### Student integral tricks…

Ok, classes ended last week and my brain is way out of math shape. Right now I am contemplating how to show that the complements of this object

and of the complement of the object depicted in figure 3, are NOT homeomorphic.

I can do this in this very specific case; I am interested in seeing what happens if the “tangle pattern” is changed. Are the complements of these two related objects *always* topologically different? I am reasonably sure yes, but my brain is rebelling at doing the hard work to nail it down.

Anyhow, finals are graded and I am usually treated to one unusual student trick. Here is one for the semester:

$\int x^2 \sqrt{x+1} dx =$

Now I was hoping that they would say $u = x +1 \rightarrow u-1 = x \rightarrow x^2 = u^2-2u+1$ at which case the integral is translated to: $\int u^{\frac{5}{2}} - 2u^{\frac{3}{2}} + u^{\frac{1}{2}} du$ which is easy to do.

Now those wanting to do it a more difficult (but still sort of standard) way could do two repetitions of integration by parts with the first set up being $x^2 = u, \sqrt{x+1}dx =dv \rightarrow du = 2xdx, v = \frac{2}{3} (x+1)^{\frac{3}{2}}$ and that works just fine.

But I did see this: $x =tan^2(u), dx = 2tan(u)sec^2(u)du, x+1 = tan^2(x)+1 = sec^2(u)$ (ok, there are some domain issues here but never mind that) and we end up with the transformed integral: $2\int tan^5(u)sec^3(u) du$ which can be transformed to $2\int (sec^6(u) - 2 sec^4(u) + sec^2(u)) tan(u)sec(u) du$ by elementary trig identities.

And yes, that leads to an answer of $\frac{2}{7}sec^7(u) +\frac{4}{5}sec^5(u) + \frac{2}{3}sec^3(u) + C$ which, upon using the triangle

Gives you an answer that is exactly in the same form as the desired “rationalization substitution” answer. Yeah, I gave full credit despite the “domain issues” (in the original integral, it is possible for $x \in (-1,0]$ ).

What can I say?

## January 27, 2016

### A popular video and covering spaces…

Filed under: media, popular mathematics, topology — Tags: , , , , — collegemathteaching @ 11:16 pm

Think back to how you introduced the sine and cosine functions on the real line. Ok, you didn’t do it quite this way, but what you did, in effect, is to define $sin(u) = Im(e^{iu})$ and $cos(u) = Re(e^{iu})$ and then use “elementary trigonometry” to relate the “angle” $u$ to the arc length subtended on the circle $|z| = 1$. One notes that the map $\rho: R^1 \rightarrow C^1$ defined by $\rho(u) = e^{iu}$ has period $2\pi$

Note: the direction “to the right” on the real line is taken to be “counterclockwise” on the circle (red arrows).

Skip if you haven’t had a topology class
The top line is known as the “universal covering space” for the circle. The reason for the terminology has to do with topology. Depending on how long ago you had your topology course, you might remember that the fundamental group of the real line is trivial and the associated group of deck transformations is infinite cyclic (generated by the map $d(u) = u + 2\pi$ ). One then shows that the fundamental group of the circle is the quotient of the group of deck transformations with the fundamental group of the real line; hence the fundamental group of the circle is infinite cylic.

Resume if you haven’t had a topology class

Notice the following: if one, say, “takes a walk” along the line in the direction of the red arrow, the action of the “covering mapping” is to take the same walk in the counter clockwise direction of the circle. That is, the covering action does the following: a walk on the line in the direction of points $A_1, B_1, C_1, A_2, B_2, C_2....$ corresponds to a walk on the circle $A, B, C, A, B, C....$. That is, walking from $A_1$ to $A_2$ corresponds to a complete lap of the circle.

(that is, on the real line, $A_{n+1} = A_{n} + 2\pi$)

Now note the following: for BOTH the line and the circle, the direction is well defined. “To the right” on the real line” is “counter clockwise” on the circle.

However: on the real line, it makes perfect sense to say that $A_1$ is “before” $B_1$ which is “before” $C_1$ which is “before” $A_2$ and so on; this is merely:

$A_1 < B_1 < C_1 < A_2 < B_2 ...$. This is order is valid no matter where one starts on the line.

However, this “universal ordering” makes no sense on the circle, UNLESS one specifies a start point. True, one moves from $A$ to $B$ to $C$ and back to $A$ again..but if one started at $B$ and started to walk, it would appear that $A$ came AFTER $B$ and not before.

So what?

This quirky animation from CraveFX starts off innocently enough, a janitorial worker mops up a leaky refrigerator and then picks up a coin on the ground. It’s not until you see what causes the refrigerator to leak and why the coin is on the ground that you realize that you’re watching an intricate moving puzzle piece before your eyes. The characters are stuck in an infinite loop caused by another character in their own infinite loop. It’s chaotic and great and hard to keep up with.

The video is below. Now the question: “what action occurred before what other action”? and the answer is “it depends on when you started watching”. The direction of time corresponds to the red arrows in the above diagrams; THAT is well defined. Why? The reason is the Second Law of Thermodynamics; spills do NOT reverse themselves, hence the direction is set in stone, so to speak. But as far as order, it depends ON WHEN THE VIEWER STARTED WATCHING.

Anyway, this video reminded me of covering spaces.

## October 29, 2015

### A quick break from the routine…

Filed under: editorial, topology — Tags: — collegemathteaching @ 9:23 pm

But this semester, I’ve been up to my eyeballs in this new (to me) course. If I never see actuarial mathematics again, it will be too soon. 🙂

## July 1, 2015

### Embarrassing gaps in my mathematical knowledge

Filed under: mathematician, topology — Tags: , — collegemathteaching @ 1:56 pm

Yes, mathematics is a huge, huge subject and no one knows everything. And, when I was a graduate student, I could only focus on 1-2 advanced courses at a time, and when I was working on my thesis, I almost had a “blinders on” approach to finishing that thing up. I think that I had to do that, given my intellectual limitations.

So, even in “my area”, my knowledge outside of a very narrow area was weak at best.

Add to this: 20+ years of teaching 3 courses per semester; I’ve even forgotten some of what I once knew well, though in return, I’ve picked up elementary knowledge in disciplines that I didn’t know before.

But, I have many gaps in my own “area”. One of these is in the area of hyperbolic geometry and the geometry of knot complements (think of this way: take a smooth simple closed curve in $R^3$, add a point at infinity to get $S^3$ (a compact space), now take a solid torus product neighborhood of the knot (“thicken” the knot up into a sort of “rope”) then remove this “rope” from $S^3$. What you have left over is a “knot complement” manifold.

Now these knot complements fall into one of 3 different types: they are torus knot complements (the knot can live on the “skin” of a torus),

satellite knot complements (the knot can live inside the solid torus that is the product neighborhood of a different, mathematically inequivalent knot,

or the knot complement is “hyperbolic”; it can be given a hyperbolic structure. At least for “most” knots of small “crossing number” (roughly: how many crossings the knot diagram has), are hyperbolic knots.

So it turns out that the complement of such knots can be filled with “horoballs”; roughly speaking, these are the interior of spheres which are “tangent to infinity”; infinity is the “missing stuff” that was removed when the knot was removed from $S^3$. And, I really never understood what was going on at all.

I suppose that one can view the boundary of these balls (called “horospheres”) as one would view, say, the level planes $z = k$ in $R^3$; those planes become spheres when the point at infinity is added. This is a horoball packing of the complement of the figure 8 knot; missing is the horosphere at $z = 1$ which can be thought of as a plane.

But the internet is a wonderful thing, and I found a lecture based on the work of Anastasiia Tsvietkova and Morwen Thistlethwaite (who generated the horoball packing photo above) and I’ll be trying to wrap my head around this.

## 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.

## May 31, 2015

### And a Fields Medalist makes me feel better

Filed under: calculus, editorial, elementary mathematics, popular mathematics, topology — Tags: — collegemathteaching @ 10:30 pm

I have subscribed to Terence Tao’s blog.

$\frac{d^{k+1}}{dx^{k+1}}(1+x^2)^{\frac{k}{2}}$ for $k \in \{1, 2, 3, ... \}$. Try this yourself and surf to his post to see the “slick, 3 line proof”.

But that really isn’t the point of this post.

This is the point: I often delight in finding something “fun” and “new to me” about an established area. I thought “well, that is because I am too dumb to do the really hard stuff.” (Yes, I’ve published, but my results are not Annals of Mathematics caliber stuff. 🙂 )

But I see that even the smartest, most accomplished among us can delight in the fun, simple things.

That makes me feel better.

Side note: I haven’t published much on this blog lately, mostly because I’ve been busy updating this one. It is a blog giving notes for my undergraduate topology class. That class was time consuming, but I had the teaching time of my life. I hope that my students enjoyed it too.

## March 17, 2015

### Compact Spaces and the Tychonoff Theorem IV: conclusion

Filed under: topology — Tags: , , — collegemathteaching @ 9:14 pm

We are finishing up a discussion of the Tychonoff Theorem: an arbitrary product of compact spaces is compact (in the product topology, of course). The genesis of this discussion comes from this David Wright article.

In the first post in this series, we gave an introduction to “compactness”.

In the second post, we gave a proof that the finite product of compact spaces is compact.

In the third post, we gave come equivalent definitions of compactness

In particular, we showed that:

1. A space is compact if and only if the space has the following property: if $A \subset X$ is an infinite union of open sets with no finite subcover, then $A$ is a proper subset of $X$; that is, $X -A \neq \emptyset$ and

2. A space is compact if and only if the space has the following property: every infinite subset $E$ has a perfect limit point. Note: a perfect limit point for a set $E$ is a point $x \in X$ such that, for every open $U, x \in U$, $|U \cap E| = |E|$ (the intersection of every open neighborhood of a perfect limit point with $E$ has the same cardinality as $E$.

Note the following about these two facts: each of these facts promises the existence of a specific point rather than the existence/non-existence of a cover of a particular type. Fact 1 promises the existence of an excluded point, and fact 2 promises the existence of a perfect limit point.

When it comes to a point in an infinite of topological spaces, constructing a point is really like constructing a sequence of points (in the case of countable products) or a net of points (in the case of uncountable products). That is, if one wants to construct a point in an infinite product of spaces, one can assume some well ordering of the index used in the product, then construct the first coordinate of the point from the first factor space, the second coordinate from the second factor space, and so on.

We’ll use fact one: the excluded point property to prove Tychonoff’s Theorem.

Proof. Assume that $X = \Pi_{\alpha \in I} X_{\alpha}$ and that $I$ is well ordered. We start out by showing that the product of two compact spaces is compact, and use recursion to get the general result.

Let $\mathscr{O}$ be an infinite union of open sets in $X_1 \times X_2$ with no finite subcover. First, we show that there is some $a \in X_1$ such that for each open set $U, a \in U$, no finite subcollection of $\mathscr{O}$ covers $U \times X_2$. Now if there is some open $U \subset X_1$ where $U$ is disjoint from every $\pi_1 (O), O \in \mathscr{O}$ we are done with this step. So assume not; assume that every $U$ is a subset of the first factor of some $O \in \mathscr{O}$. If it isn’t the case that there is $x \in X_1$ where $U_x \times X_2$ has no finite subcover of elements of $\mathscr{O}$, for each such $U_x \subset X_1, x \in X_1$ there is a finite number of elements of $\mathscr{O}$ that covers $U_x \times X_2$. Now since $X_1$ is compact, a finite number of $U_x$ covers $X_1$, hence a finite subcover of $\mathscr{O}$ covers ALL of $X_1 \times X_2$. Hence some point $a \in X_1$ exists such that no finite subcover of $\mathscr{O}$ covers $U \times X_2$ for any open $U \subset X_1, a \in U$.

Similarly, we can find $b \in X_2$ so that for all open $V \subset X_2, b \in V$, no finite subcollection of $\mathscr{O}$ covers $U \times V$ where $U$ is a basic open set in $X_1$ that contains $a$. This shows that $(a,b) \notin \cup_{O \in \mathscr{O}}$ because, if it were, this single point would lie in some basic open set $U \times V$ which, by definition, is a finite subcover.

Now given an arbitrary product with a well ordered index set $I$ we can now assume that there is some collection of open sets that lacks a finite subcover and inductively define $a_{\gamma} \in X_{\gamma}$ so that, if $U$ is any basic open set containing $\Pi_{\alpha \leq \gamma} \{a_{\alpha} \} \times \Pi_{\alpha > \gamma} X_{\alpha}$ then no finite subcollection of $\mathscr{O}$ covers $U$. The point $(a_{\gamma})$ thus constructed lies in no $\mathscr{O}$.

Note: if you are wondering why this “works”, note that we assumed NOTHING about the compactness of the remaining product space factors $\Pi_{\alpha > \gamma} X_{\alpha}$.
And remember that we are using the product topology: an open set in this topology has the entire space as factors for all but a finite number of indices. So we only exploit the compactness of the leading factors.

### Compact Spaces and the Tychonoff Theorem III

Filed under: topology — Tags: , , — collegemathteaching @ 2:49 am

We continue on our quest to prove the Tychonoff Theorem: an arbitrary product of compact spaces is compact. We just show that this is true for the FINITE product of compact spaces.

It is our goal to do this by using elementary tools and avoiding things like nets (for example, Willard uses ultranets)

We will basically adding background and commentary to David Wright’s excellent 1994 paper which appeared in the Proceedings of the American Mathematical Society. We will use a bit of cardinal arithmetic and facts about ordinals at times.

Yes, we do need some background, but the background we are providing is necessary for the understanding of any mathematics that uses topology anyway.

Conditions which are equivalent to compactness

1. If $X \subset R^n$ in the usual topology, then $X$ is compact if $X$ is closed and bounded.
Proof: Let $X$ be compact. Then $X$ is closed because the usual topology for $R^n$ is Hausdorff. $X$ is bounded as well, as if it weren’t, the if $x \in X, d(x,0) = M, cover$latex X \$ by $\cup_{x \in X} B_x(\frac{1}{M})$. This open cover has no finite subcover as $M$ is unbounded.

Now let $X$ be closed and bounded. Then $X \subset \Pi^n_{i=1} [a,b]$ for some real $a, b$, which is compact by our “junior” Tychonoff Theorem. So $X$ is a closed subset of a compact set and therefore compact.

2. I’ll call this the excluded point condition: Let $U = \cup_{\alpha \in I} U_{\alpha}$, where each $U_{\alpha}$ is open. We say that $U$ lacks a finite subcover if there is no finite subcollection of the $U_{\alpha}$ that covers U. A topological space $X$ is said to have the excluded point condition if any subset $U$ which has an open cover which lacks a finite subcover must exclude at least one point of $X$; that is, any set which has an open cover with no finite subcover must be a proper subset of $X$.

Example of an open cover which lacks a finite subcover: consider $[0,1]$ as a subset of $R^1$ in the usual topology, and let $U = \cup^{\infty}_{n=3}(\frac{1}{n}, \frac{n-1}{n})$; here $\{0, 1 \}$ are the excluded points from this open cover.

Theorem: a space is compact if and only if it has the excluded point condition for open covers.
Proof. If $X$ is compact then any open cover of $X$ has a finite subcover, hence any subset of $X$ which has an open cover which lacks a finite subcover cannot be all of $X$.
Now assume that $X$ has the excluded point condition. Let $U$ be any open cover of $X$. $U$ cannot lack a finite subcover as any subset which has an open cover lacking a finite subcover must exclude a point of $X$.

3. The finite intersection property condition: let $\mathscr{C}$ be any collection of closed sets. We say that $\mathscr{C}$ has the finite intersection property if the following holds: if the intersection of any finite subcollection of elements of $\mathscr{C}$ is non-empty.

Example: in $R^1$, $\mathscr{C} = \{ [1 - \frac{1}{n}, 1+ \frac{1}{n} ], n \in \{1, 2, ...\} \}$ has the finite intersection property. On the other hand, $\{ [n, n+1], n \in \{..-2, -1, 0, 1, 2, ..\} \}$ does not have this property as there are finite subcollections of this set that have an empty intersection.

Theorem: $X$ is compact if and only if the following holds: if $\mathscr{C}$ is a collection of closed sets with the finite intersection property, then an arbitrary intersection of elements of $\mathscr{C}$ is non-empty.

Proof: Let $X$ be compact. Let $\mathscr{C}$ be an infinte collection of closed sets with the finite intersection property. This means that no finite collection of the complements of these sets can cover $X$ Then $X - \cap_{C \in \mathscr{C}} C = \cup_{C \in \mathscr{C}}(X - C)$ is an open cover of a subset of $X$. Since no finite subcollection of these closed set complements (open sets) can cover all of $X$, then $X - \cup_{C \in \mathscr{C}}(X - C)$ is non-empty and therefore so is $\cap_{C \in \mathscr{C}} C$

Now let the finite intersection property hold for $X$. Let $\mathscr{U}$ be any open cover for $X$. This means that $X-\cup_{U \in \mathscr{U}} U = \cap_{U \in \mathscr{U}} (X -U)$ is empty. Hence the collection of closed sets $\{ (X-U), U \in \mathscr{U} \}$ cannot have the finite intersection property which means that there is some finite subcollection $F \subset \mathscr{U}$ where $\cap_{U \in F} (X-U)$ is empty which means $\cup_{U \in F} U$ covers $X$.

4. Limit point compactness: part I. Theorem: a space $X$ is compact if and only if every infinite subset $E \subset X$ has a limit point.
Note: we can actually prove a bit more; that will be in part II. This is a “junior theorem” which can lead the beginner to understanding the “varsity theorem”.

Proof. Let $X$ be compact and let $E$ be an infinite subset of $X$. Consider: $U_{x \in X} U_x$ where $U_x$ is some open set containing $x$. If $E$ has no limit point we can assume that the $U_x$ are chosen so that each $|E \cap U_x|$ is finite for each $x$. Now a finite subcollection of the $U_x$ covers..say $\cup_{i=1}^k U_{x_i}$ and $E = \cup_{i=1}^k (U_{x_i} \cap E)$. This is impossible as each $|(U_{x_i} \cap E|$ is finite but $E$ is infinite.

Now assume that $X$ is limit point compact in that every infinite subset has a limit point. Let $\mathscr{U}$ be an open cover which has no finite subcover. We assume that this open cover is efficient in that for each $U \in \mathscr{U}, U \not \subset \cup_{V \in \mathscr{U}, V \neq U} V$; that is, any $U$ in the open cover contains at least one point not contained in the union of the other open cover sets. Then the set $x_{U} \in U$ is an infinite set with no limit point.

5. Let $E$ be a set with cardinality $c$. We say that $x$ is a perfect limit point of $E$ if for all open sets $U_x$ containing $x$, $|U_x \cap E| = c$. Example: $[0,1]$ has every point as a perfect limit point (usual topology) as $[0,1]$ has the cardinality of the real numbers and if $U$ is open in $R^1$ and contains any point of $[0,1]$ then $U \cap [0,1]$ has the cardinality of the real numbers.

Now the stronger theorem is this: a topological space $X$ is compact if and only if every infinite subset $E$ has a perfect limit point.

Proof. First, assume that $X$ is compact. Let $E$ be an infinite subset with cardinality $c$. Cover $X$ by open sets $\cup_{x \in X} U_x$ where $x \in U_x$. Suppose that for all $U_x, |U_x \cap E| < c$. Now this open cover of $X$ has a finite subcover $U_1, U_2, ...U_k$ and so we have $E = \cup^k_{i=1} U_i \cap E$ and so $|E| \leq |U_1 \cap E| + |U_2 \cap E| + ...+|U_k \cap E|$ where each $|U_j \cap E| < c$. This is impossible because $c$ is an infinite cardinal (a limit cardinal) and it is impossible to reach a limit cardinal by a finite sum of strictly smaller cardinals.

If you are new to this and are a bit confused, start by assuming that $c$ is, say, the first countably infinite cardinal. ALL lesser cardinals are finite cardinals, and it is impossible for a finite sum of finite cardinals to add up to any infinite cardinal. Then, imagine $c$ being the first uncountable cardinal. One can not reach any uncountable cardinal by the finite sum of countable (or smaller) cardinals (the finite sum of countable cardinals is still countable). That is more or less what is going on here.

Now, suppose that every infinite set has a perfect limit point. Let $\cup_{\alpha \in I} U_{\alpha}$ be an open cover which has no finite subcover. We can assume that $I$ is the index of smallest cardinality for which this is true and that the cover is efficient: $U_{\beta} \not \subset \cup_{\alpha < \beta}U_{\alpha}$ that is, the open subcover is built by adding open sets which contain at least one point not contained by the previously added open sets. Also we put a well ordering on $I$ where the cardinality of $\{ \alpha \in I | \alpha < \beta \} < |I|$. If this confuses you a bit, think of a countable index set where the cardinality of the previous indices are finite, or of an uncountable index set where the smaller cardinals are all countable.

So, for each $\beta$ let $x_{\beta} \in U_{\beta} - \cup_{\alpha < \beta} U_{\alpha}$. Now let $E = \cup_{\alpha \in I} x_{\alpha}$ and note $|E| = |I|$ by design.

Now if $x \in X$, there is some $\alpha < I$ where $x \in U_{\alpha}$ but $|E \cap U_{\alpha}| \leq |I|$ as all $\alpha \in I$ have smaller cardinality than $I$ Therefore $E$ has no perfect limit point.

Again, the person new to topology can run through this with $I$ first being the countable ordinal (and every previous ordinal having finite cardinality) or $I$ being the first uncountable ordinal with every previous ordinal having at most countable cardinality.

We now have the background to give a simple proof of the full strength Tychonoff Theorem, which we will do in the next post.

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