What is the importance of Zorn's lemma in Tychonoff theorem proof?

I read that Tychonoff's theorem states that given a family of topological spaces $$\{(X_i,\tau_i),i \in I\}$$, the product topology $$(X,\tau)=\prod_{i \in I}(X_i,\tau_i)$$ is compact iff each $$(X_i,\tau_i)$$ is compact.

The proof states that given any family $$\mathcal{F}$$ of closed subsets of $$X$$ with the finite intersection property, there is a maximal family $$\mathcal{H}$$ that contains $$\mathcal{F}$$ and which also has the finite intersection property. The existence of $$\mathcal{H}$$ is shown using Zorn's lemma.

The proof goes on to show that $$\bigcap_{H \in \mathcal{H}}\bar{H} \neq \emptyset \implies \bigcap_{F \in \mathcal{F}}F \neq \emptyset$$, from which the theorem follows. I won't go into the details, but the proof took a 'slice' from each $$H \in \mathcal{H}$$, using a projection function $$p_i$$, so that $$p_i(H) \in X_i$$. Since each $$p_i(H)$$ has the finite intersection property for $$i \in I$$, and $$X_i$$ is compact, we can get the intersection $$x_i$$ of all $$p_i(H)$$, and form $$x=\prod_{i\in I}x_i \in X$$. It follows that with $$x$$, we have $$\bigcap_{H \in \mathcal{H}}\bar{H} \neq \emptyset$$

I think am missing a step, but what is the importance of having to use Zorn's lemma to get a maximal $$\mathcal{H}$$ and show that $$\bigcap_{H \in \mathcal{H}}\bar{H} \neq \emptyset$$. Since $$\mathcal{F}$$ already has the finite intersection property, can we not use any arbitrary $$\mathcal{F}$$?

• The axiom of choice could be directly used to simultaneously choose an element $x_i$ from each $\bigcap_{F\in\mathcal F} p_i(F)$. This step is somehow traded off for the usage of Zorn lemma. – Berci Jun 14 '19 at 11:17
• Are there still people who prove Tychonoff without using ultrafilters ? – Maxime Ramzi Jun 14 '19 at 12:34
• @Max: I mean, $\mathcal{H}$ here is an ultrafilter. This is the ultrafilter proof just without saying the word "ultrafilter". – Eric Wofsey Jun 14 '19 at 16:51
• Right, if you develop the theory of convergence of (ultra)filters first, then that greatly shortens the proof, because a big chunk of the work of the proof is hidden in that theory. – Eric Wofsey Jun 14 '19 at 17:03
• @Max You could also prove Alexander's subbase lemma using Zorn, and then Tychonoff is a relatively easy consequence. Or use nets and universal nets (as I did here), e.g. But Tychonoff implies AC (see here) or Zorn too, equivalently, so we cannot avoid it. Tychonoff himself did his proof using points of complete accumulation (no FIP families or ultradilters at all) and only for $[0,1]^I$ originally. – Henno Brandsma Jun 14 '19 at 21:41

So the step you are missing is that the proof you have, very lightly concludes that $$\bigcap\overline{H}\neq\emptyset$$. Which should be understood as $$x\in\bigcap\overline{H}$$. But why? This is unclear and to prove that you need $$\mathcal{H}$$ to be maximal.

First I will show you an example when it fails if $$\mathcal{H}$$ is not maximal. Consider $$X_1=X_2=\{0,1\}$$ with discrete topology. Then $$X=X_1\times X_2$$ has four elements. Let $$\mathcal{H}=\{K\}$$ consist of a single closed subset $$K=\{(0,0),(1,1)\}$$. Obviously $$\mathcal{H}$$ has FIP (Finite Intersection Property). But $$\mathcal{H}$$ is not maximal since we can extend it with for example $$\{(1,1)\}$$ and maintain FIP.

Now lets construct $$x$$. Apply projection $$\pi_1$$ to $$\mathcal{H}$$ to get $$\pi_1(K)=\{0,1\}$$. So we have two choices for $$x_1$$. Since the choice is arbitrary then pick $$x_1=0$$. Now apply $$\pi_2$$ to $$\mathcal{H}$$ to get $$\pi_2(K)=\{0,1\}$$ as well. And again the choice of $$x_2$$ is arbitrary, so lets take $$x_2=1$$ this time. So we end up with $$x=(0,1)$$ which doesn't belong to $$K$$ and therefore it doesn't belong to $$\bigcap_{H\in\mathcal{H}}\overline{H}$$.

One way to fix that situation is too somehow pick $$x_\alpha$$ in a smart way. But there doesn't seem to be a way to do that in general. It turns out that it is easier to enlarge $$\mathcal{H}$$ as much as we can so that any choice of $$x_\alpha$$ becomes valid.

So as you can see the assumption about $$\mathcal{H}$$ being maximal is crucial. That's why we can't apply the reasoning to $$\mathcal{F}$$, because it might not be big enough. Actually $$\mathcal{F}$$ hardly ever is maximal because it contains closed subsets only.

Anyway the full proof is a bit too long for Math StackExchange so please have a look at details here:

Notice that the statement "$$x\in\overline{H}$$ for each $$H$$" is not trivial at all. And I'm surprised that whoever wrote the proof you have, treated it as obvious.
• Okay thanks, but just a short question. Does $\mathcal{H}$ being maximal imply that it can 'take' in all possible combinations of inverse projections $x_i$ from the $X_i$? (where the inverse projections are used in constructing $x$) – Link L Jun 15 '19 at 13:07