Comments on How can one prove Tychonoff's theorem using ultrafilters?
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How can one prove Tychonoff's theorem using ultrafilters?
Tychonoff's theorem, as a reminder, says that the product of a possibly infinite number of compact spaces is compact (a compact space being a space every open cover of which has a finite subcover; this is sometimes called quasi-compactness, as in the Stacks Project).
Interlude: the typical way of proving Tychonoff's theorem
If one proves the Alexander subbase theorem (that, given a subbase $S$ of a topological space $X$, $X$ is compact whenever all open coverings of $X$ that are subsets of $S$ have finite subcoverings), it is not hard to prove Tychonoff's theorem. To do so, let us set the $X$ and $S$ needed for applying the Alexander subbase theorem: let our topological space $X$ be $\Pi_n X_n$,1 the product space of infinitely many $X_n$, and let our subbase $S$ be the set of inverse images of open sets under the projections $\pi_n: X → X_n$. Then let $\mathcal{O} \subseteq S$ be an open covering of $S$.
It is then clear that every element of $\mathcal{O}$ will be a product of all but one of the $X_n$ with an open set from the remaining $X_n$. Then either some of the elements of $\mathcal{O}$ will completely cover one of the $X_n$ and thus have a finite subcovering that includes all of $X$, or we will be able to find an element of $X_n$ that $\mathcal{O}$ does not cover by taking the product of the closed sets that $\mathcal{O}$ leaves uncovered in each $X_n$, basically as illustrated below:
And to prove the Alexander subbase lemma, in turn, is not hard to do using Zorn's lemma: without further ado, let $\mathcal{O}$ be a maximally coarse open covering of $X$ with no finite subcovering; then $\mathcal{O}$ must be a subset of $S$2, but all open coverings that are subsets of $S$ have finite subcoverings by hypothesis, hence there is no maximally coarse open covering of $X$ with no finite subcovering, hence all open coverings of $X$ have finite subcoverings and $X$ is compact.
What about proving it using the ultrafilter lemma?
I have heard that Tychonoff's theorem is actually a quite straightforward consequence of basic theorems on ultrafilters. Proving it this way would have two advantages: the conceptual advantage of linking it to the ultrafilter-based understanding of compactness (useful elsewhere, e.g. in Stone-Čech compactification) and the axiomatic advantage of using the ultrafilter lemma instead of the more powerful Zorn's lemma. (But I'm much more interested in the conceptual advantage.)
But I have not as yet figured out how to formulate the right ultrafilter-based understanding of compactness for this task. Speaking vaguely, the approach of saying that quasi-compact spaces are those all of whose "ultrafilters" (in some sense) are principal does not seem to be getting me anywhere (I can show that products of non-principal ultrafilters extend to non-principal ultrafilters, but that's just affirming the consequent).
Hence, to restate my question: what am I missing? What is the right ultrafilter-based conception of compactness? And how does it apply to proving Tychonoff's theorem?
1Where the $n$ are elements of a not necessarily countable indexed set $I$.
2 How do we prove this? Suppose, for a contradiction, that $\mathcal{O}\not\subseteq S$. Then there exists $O_1 \not\in S$; since $S$ is a subbase $O_1$ is the intersection of a finite number $n$ of $S_1,…S_n \in S$. Then we have $n$ coarsenings of $\mathcal{O}$ achieved by replacing $O_1$ with $S_1$, $S_2$, … $S_n$. All $n$ coarsenings (by maximality of $\mathcal{O}$) must have finite subcoverings, and we can take the union of these $n$ finite subcoverings and replace the $S_i$ with $O_1$ to get a finite subcovering of $\mathcal{O}$, contradiction. (It's very similar, for some reason, to a technique I used here, in footnote 2…)
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| clemens |
Thread: Works for me See my comment: https://math.codidact.com/comments/thread/11565#comment-28620 |
Mar 1, 2026 at 03:19 |
Presumably, the approach is the one mentioned on Wikipedia, specifically the third proof by Cartan and Bourbaki. It is summarized there, under the assumption, which requires the Ultrafilter Lemma, that a topological space is compact if and only if all ultrafilters on that space converge.
Given that, the proof is simply we given an ultrafilter on the product consider the ultrafilter generated by the image of a projection. That ultrafilter converges to some point in the corresponding component space, because it's compact. The ultrafilter on the product space then converges to tuple of all these points. Thus, every ultrafilter on the product space converges, and the product space is compact.
This is a close paraphrase of what is stated on Wikipedia, but it is a bit misleading. If you skip to the next paragraph and continue on, you'll see a proof that uses the Ultrafilter Lemma but not the Axiom of Choice. Since the Ultrafilter Lemma is weaker than the Axiom of Choice, but Tychonoff's Theorem is equivalent to it, the Axiom of Choice must be coming in somewhere, and here is where it is. The theorem does not give a unique point for each component. We only end up with a non-empty set. We need the Axiom of Choice to make a tuple where each component chooses an element from the corresponding non-empty set.
With that out of the way, the real substance is thus the claim that a space being compact is equivalent to all ultrafilters converging given the Ultrafilter Lemma.
First, to actually define convergence of (ultra)filters, an (ultra)filter converges to a point $x$ if the (ultra)filter contains the neighborhood filter of $x$. The neighborhood filter is the set of all subsets that contain an open neighborhood of $x$. Alternatively, it is the upward closure of all open neighborhoods of $x$. This is easily verified to actually be a filter.
For this, it's best to use the contrapositive characterization of compactness:
A space is compact if and only if every family of closed subsets having the finite intersection property has non-empty intersection.
The finite intersection property simply states that for a collection of sets, any finite subset of them has a non-empty intersection. The important thing is that any non-empty collection of sets with the finite intersection property is a filter subbase from which we can generate a proper filter.
With this, we can prove that compactness implies that every ultrafilter converges. Proof: $\implies$ Given an ultrafilter on a compact space, it satisfies the finite intersection property by definition. In particular, if we restrict to the necessarily proper filter generated by the basis of closed subsets contained in the ultrafilter, those also satisfy the finite intersection property, and, thus, by compactness, their intersection is non-empty. Let $x$ be a point in that intersection and let $U$ be any open neighborhood of $x$. Then, by definition, either $U$ is in the ultrafilter or its complement is in the ultrafilter. But its complement is a closed subset that doesn't contain $x$, so it can't be in the ultrafilter. Therefore, every open neighborhood of $x$ is in the ultrafilter and the ultrafilter converges to $x$. $\square$ (Note, this doesn't exclude it from also converging to a different point, e.g. if the points are topologically indistinguishable.)
Proving the other direction is where the Ultrafilter Lemma comes in. It states that every proper filter on a set is contained in some ultrafilter on that set.
With this, it's easy to show that every ultrafilter converging implies compactness. Proof: $\impliedby$ Given any collection of closed subsets satisfying the finite intersection property, we get an ultrafilter that converges to some point, $x$, i.e. that ultrafilter contains all the subsets in the neighborhood filter of $x$. If some closed subset in the original collection (and thus in the ultrafilter) doesn't contain $x$, then its complement does and is open. It is thus an open neighborhood of $x$ and contained in the ultrafilter. But an ultrafilter can't contain both a set and its complement, so this is a contradiction, so no such closed subset exists. Which is to say, the intersection of all the original closed subsets contains $x$. $\square$
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