Lemma 2.7.28734A1486521

Proof. This is Tychonoff's theorem; we prove it with ultrafilters, using the axiom of choice in the form of Zorn's lemma. A filter on a set $Z$ is a family $\mathcal F$ of subsets of $Z$ such that $Z\in \mathcal F$, $\varnothing \notin \mathcal F$, and $\mathcal F$ is closed under finite intersections and under passing to supersets. An ultrafilter is a filter which is maximal for inclusion. Since the union of a chain of filters containing a given filter is again a filter, Zorn's lemma shows that every filter is contained in an ultrafilter.

A filter $\mathcal U$ on $Z$ is an ultrafilter if and only if, for every $A\subset Z$, either $A\in \mathcal U$ or $Z\setminus A\in \mathcal U$. Indeed, let $\mathcal U$ be an ultrafilter and $A\subset Z$. If $A$ meets every member of $\mathcal U$, the sets containing some $U\cap A$ with $U\in \mathcal U$ form a filter containing $\mathcal U$ and $A$, so $A\in \mathcal U$ by maximality; otherwise some $U\in \mathcal U$ is disjoint from $A$, and $Z\setminus A\supset U$ belongs to $\mathcal U$. Conversely, if $\mathcal U$ has this property, a strictly larger filter would contain some $A\notin \mathcal U$ together with $Z\setminus A$, hence $\varnothing $.

A filter on a topological space $Z$ converges to $z$ if every neighborhood of $z$ belongs to it. We show that $Z$ is compact if and only if every ultrafilter on $Z$ converges. If $Z$ is compact and $\mathcal U$ is an ultrafilter, the closures of the members of $\mathcal U$ have the finite intersection property and hence a common point $z$. If an open neighborhood $N$ of $z$ were not in $\mathcal U$, the closed set $Z\setminus N$ would belong to $\mathcal U$. Since $z$ lies in the closure of every member of $\mathcal U$, this would give $z\in Z\setminus N$, which is absurd. Hence $\mathcal U$ converges to $z$. Conversely, suppose that every ultrafilter converges, and let $\mathcal C$ be a family of closed sets with the finite intersection property. The sets containing a finite intersection of members of $\mathcal C$ form a filter, which lies in an ultrafilter $\mathcal U$ converging to some $z$. If $z\notin C$ for some $C\in \mathcal C$, then both $C$ and its complement, a neighborhood of $z$, would belong to $\mathcal U$. Hence $z$ lies in every member of $\mathcal C$, and $Z$ is compact.

Now let $Z=\prod _\alpha X_\alpha $ with compact factors and projections $p_\alpha $; if some $X_\alpha $ is empty, so is $Z$, and there is nothing to prove. Let $\mathcal U$ be an ultrafilter on $Z$. For each $\alpha $, the family $\{ A\subset X_\alpha :p_\alpha ^{-1}(A)\in \mathcal U\} $ is a filter with the property characterizing ultrafilters, hence an ultrafilter, and since $X_\alpha $ is compact it converges to some $z_\alpha \in X_\alpha $. Put $z=(z_\alpha )$. A basic neighborhood $\bigcap _{l=1}^ np_{\alpha _ l}^{-1}(V_ l)$ of $z$, with $V_ l$ a neighborhood of $z_{\alpha _ l}$, is a finite intersection of members of $\mathcal U$, so every neighborhood of $z$ belongs to $\mathcal U$. Thus every ultrafilter on $Z$ converges, and $Z$ is compact. $\square$