Peking University Stuffs

I encountered this challenge while I was browsing through the AoPS forum. It caught my attention that @watery replied in that thread about why it is hard to prove. This got me thinking, because that’s exactly what the Tube Lemma is used to prove. So I started trying to solve this problem from Peking University (PKU) lol!!

This was the response I gave,

Well here’s much formal version of that, which i actually took more time and effort on writing.

Proof

Let $\mathscr{U}$ be an open cover of $X_1\times X_2$.

Fix $x\in X_1$. The slice $\{x\}\times X_2$ is homeomorphic to $X_2$, hence compact. So there exist

such that

Define the open set

Then $\{x\}\times X_2 \subseteq N_x$, and $N_x$ is open in $X_1\times X_2$.

By the Tube Lemma (with $Y=X_2$), there is an open neighborhood $W_x\subseteq X_1$ of $x$ such that

Now $\{W_x : x\in X_1\}$ is an open cover of $X_1$ because $X_1$ is compact, so we can choose

Each $N_{x_j}$ is a finite union of members of $\mathscr{U}$, so $\bigcup_{j=1}^m N_{x_j}$ is a finite union of members of $\mathscr{U}$. Because of this $\mathscr{U}$ has a finite subcover.

So we just proved that $X_1\times X_2$ is compact! $\square$

Note that this proof is elementary. Since you learn the Tube Lemma early in Topology, the logic doesn’t quite extend to the arbitrary infinite product case. To handle that extension, we’ll introduce ‘equivalent’ definitions of compactness later in the post :)

Note that at some point, we’ll use some elementary cardinal arithmetic.

Introduction to Tychonoff Theorem

Before we get our hands dirty and all those math signs, I will tell you a bit about Tychonoff Theorem, actually lemme make this boring but cool by bringing what Wikipedia told me.

Tychonoff’s theorem states that the product of any collection of compact topological spaces is compact with respect of the product topology. the theorem is named after Andrey Tikhonov (aka. Tychonoff). Who proved it first in 1930 for powers of the closed unit interval and in 1935 states the full theorem along with the remark that its proof was the same as for the special case. The earliest known published proof is contained in a 1935 article by Tychonoff, “Über einen Funktionenraum”

Well this is important because Tychonoff’s theorem is one of the most foundational theorem in topology. And this is also goes for topological spaces that is based on uncertain sets (Sets whose elements have degrees of membership).

That didn’t really showed how important it is right? So lemme give you a scenario.

If we take an infinite product of spaces, the box Topology seems intuitive but it destroy the compactness (This will be important later on).

So what’s the solution? Tychonoff’s theorem. The Tychonoff’s theorem confirms that the Tychonoff topology is the “correct” topology to use since it is the only topology that can render the product of compact spaces compact (you know what i mean) This will validate the definition of the product topology itself.

Aaand also Tychonoff’s theorem is also used to prove the Banach-Alaoglu Theorem.

Bradley University’s “Junior Tychonoff” proof

There’s very standard proof “textbook” style such as Munkres or Kelly which goes for definition of compactness (open covers) directly without considering convergence.

While standard proofs often rely on covers and sub bases, the filter based approach goes a a bit more “elegant” approach (Honestly both of them are cursed still). This was done by treating points and “directions” of convergence algebraically via filters, the proof of the product theorem reduces to a co ordinate wise verification.

Sounds cool right? Let’s do it!

This is assignment from Bradley University I found when I was searching about Tychonoff. And I think it was worth mentioning so here it is.

What Bradley University is doing is basically the tube lemma we did above, but they built the tube by hand using basic opens $U\times V$

Any open cover of $X\times Y$ can be refined by a cover consisting of basic rectangles $U\times V$ with $U\subseteq X$, and $V\subseteq opened because those rectangles form a basis for the product topology.

Proof

Let $\mathscr{B}$ be an open cover of $X\times Y$ by basis rectangles $U\times V$.

Fix $x\in X$. The slice $\{x\}\times Y$ is homeomorphic to $Y$, hence compact. So choose finitely many rectangles

that cover $\{x\}\times Y$.

Since each $U_{x,i}$ is an open neighborhood of $x$, the intersection

is open in $X$ and still contains $x$

Now check the “tube” inclusion

Why?

For instance, take any $(x',y)\in U_x\times Y$ Since $\{x\}\times Y$ is covered by the rectangles, we can choose $i$ such that

For example

Hence

So $\{U_x : x\in X\}$ is an open cover of $X$. Compactness of $X$ gives $x_1,\dots,x_m$ such that

Then

which is a finite union of rectangles from $\mathscr{B}$! $\square$

Why the infinite product case is genuinely different

For a finite product, you can basically “thicken a slice” in one coordinate and boom, you are done.

For an infinite product $\prod_{\alpha\in I} X_\alpha$ this kind of thickening becomes subtler since a basic open set only restricts finitely many coordinates, and compactness interacts with that finiteness in an essential way.

This is also where the box topology vs product topology split shows up, In the box topology, you allow all coordinates to be restricted at once since products of opens with no finiteness condition but in other hand for the product topology, only finitely many coordinates are restricted in a basic neighborhood.

Tychonoff’s theorem talks about that the compactness is possible in the product topology, but in general sense, it fails for the box topology on infinite products.

Filters, ultrafilters, and compactness

To go full Tychonoff, I’ll use the ultrafilter characteristics of compactness.

Now you will ask me what is filters, ultrafilters, and compactness. Great question But i’m kinda lazy for now, so I will maybe make a new blog post about this.

For some disclaimer, I’ll use

• $\mathscr{U}$ for an open cover
• $\mathcal{U}$ for an ultrafilter

Proof

($\Rightarrow$) Assume $X$ is compact and let $\mathcal{U}$ be an ultrafilter on $X$.

If $\mathcal{U}$ converges nowhere, then for each $x\in X$ there is an open neighborhood $U_x\ni x$ with $U_x\notin\mathcal{U}$.
Since $\mathcal{U}$ is an ultrafilter it’s $X\setminus U_x\in\mathcal{U}$ for each $x$.

But $\{U_x\}_{x\in X}$ is an open cover so choose a finite subcover $U_{x_1},\dots,U_{x_n}$.

By closure under finite intersections, the left side would belong to $\mathcal{U}$ This contradicts $\varnothing\notin\mathcal{U}$. Hence $\mathcal{U}$ converges somewhere.

($\Leftarrow$) Assume every ultrafilter converges. Let $\{F_j\}_{j\in J}$ be a family of closed sets with the Finite Intersection Property (FIP). The family of all finite intersections of the $F_j$’s forms a filter base so it generates a filter $\mathcal{F}$. By the Ultrafilter Lemma (a form of Choice) we can extend $\mathcal{F}$ to an ultrafilter $\mathcal{U}\supseteq \mathcal{F}$ By assumption $\mathcal{U}\to x$ for some $x\in X$.

Since $\mathcal{F}\subseteq\mathcal{U}$, every $F_j$ belongs to $\mathcal{U}$. Suppose $x\notin F_j$. Then $X\setminus F_j$ is an open neighborhood of $x$, so convergence implies $X\setminus F_j \in \mathcal{U}$. By finite intersection, we get

which is impossible. Thus $x \in F_j$ for all $j$, meaning $\bigcap_{j\in J} F_j\neq\varnothing$. This proves compactness. $\square$

Filters on products

The next lemma is the whole reason filters make Tychonoff feel “coordinatewise”.

Proof

($\Rightarrow$) The projection $\pi_\alpha$ is continuous. Kinda standard fact but if $\mathcal{F}\to x$ and $f$ is continuous, then $f_*(\mathcal{F})\to f(x)$. Apply it with $f=\pi_\alpha$

($\Leftarrow$) Assume $(\pi_\alpha)_*(\mathcal{F}) \to \pi_\alpha(x)$ for all $\alpha$.
Let $U$ be a basic open neighborhood of $x$ in the product topology.

where each $V_k$ is open in $X_{\alpha_k}$ and $\pi_{\alpha_k}(x)\in V_k$ because of

we have $V_k\in (\pi_{\alpha_k})_*(\mathcal{F})$

By definition of pushforward, this means $\pi_{\alpha_k}^{-1}(V_k)\in \mathcal{F}$. Filters are closed under finite intersections, so $U\in\mathcal{F}$.

Hence $\mathcal{F}\to x$. $\square$

Tychonoff’s Theorem

Proof

Let $\mathcal{U}$ be an ultrafilter on $X$. We will show $\mathcal{U}$ converges.

Fix $\alpha\in I$. Consider the pushforward ultrafilter $(\pi_\alpha)_*(\mathcal{U})$ on $X_\alpha$. Since $X_\alpha$ is compact, Proposition 4.4 implies $(\pi_\alpha)_*(\mathcal{U})$ converges to some point $x_\alpha\in X_\alpha$.

Do this for every $\alpha$, and define $x\in X$ by the coordinates $x=(x_\alpha)_{\alpha\in I}$. Then for each $\alpha$,

By Lemma 5.2, this implies $\mathcal{U}\to x$ in $X$.

So every ultrafilter on $X$ converges; by Proposition 4.4, $X$ is compact. $\square$

It was quite a journey right?? Well for me it was, took me a while writing this. But I’m happy that I was able to finish this off. This was one of the “TODO” on my list lol.

Summary

Honestly, While i was searching through some tube lemma related resources, I was quite surprised on my findings that even in the nLab had typo on Tychonoff, I feel bad for Andrey Tikhonov (Also known as Tychonoff) that had his name typo’ed in a lot of places. So I fixed it while in doing it :)

But while I was writing i was asking “what if… we go for full proof for Tychonoff theorem like Watery was suggesting..”, But I was like nahhhh Imma LET IT ROLL lol

Q: why make life harder? A: I have so much free time and i have no idea on new blog topic.

References

These were the reference I also used besides nLab.

• Munkres, Topology
• Willard, General Topology
• Kelley, General Topology