Tychonoff’s Theorem via Filters

To give myself some practice with filters, we describe the proof of Tychonoff’s theorem via filters.

Definition. A filter on a set \(X\) is a nonempty collection \(\mathcal F\) of subsets of \(X\) which does not contain \(\varnothing\), is closed under finite unions, and is closed upwards (i.e., if \(A\subseteq B\) for some \(A\in\mathcal F\), then \(B\in\mathcal F\)). An ultrafilter is a maximal filter.

Example. For a subset \(A\subseteq X\), let \(\mathcal F_A\) be the collection of subsets containing \(A\) Then \(\mathcal F_A\) can easily be checked to be a filter. In fact, if \(A\) is a singleton \(\{x\}\), then \(\mathcal F_A\) is an ultrafilter: if \(\mathcal F_A\) is contained in a filter \(\mathcal F\), then any \(B\in\mathcal F\) must have \(\{x\}\cap B\ne\varnothing\) because \(\varnothing\notin\mathcal F\), so it follows that \(x\in B\) and so \(B\in\mathcal F_x\).

The filters in the previous example are called the principal filters. Here are a couple more constructions which we will use in the sequel.

Example. Given two filters \(\mathcal F_1\) and \(\mathcal F_2\), we let \(\mathcal F_1\cdot\mathcal F_2\) denote the collection \[\{A_1\cap A_2:A_1\in\mathcal F_1,A_2\in\mathcal F_2\}.\] If \(\varnothing\notin\mathcal F_1\cdot\mathcal F_2\), we claim that this is a filter containing \(\mathcal F_1\) and \(\mathcal F_2\). Certainly \(\mathcal F_1\cdot\mathcal F_2\) is closed under finite intersections and is closed upwards. To show that \(\mathcal F_1\subseteq\mathcal F_1\cdot\mathcal F_2\), we note that \(X\in\mathcal F_2\), so any \(A_1\in\mathcal F_1\) has \(A_1\cap X\) in \(\mathcal F_1\cdot\mathcal F_2\).

Example. Let \(\mathcal F\) be a filter on a set \(X\). For any function \(f\colon X\to Y\), then \[f_*(\mathcal F)=\left\{B\subseteq Y:f^{-1}(B)\in\mathcal F\right\}\] continues to be a filter, respectively. For example, \(\varnothing\notin\mathcal F\) implies \(\varnothing\in f_*\mathcal F\); similarly, having \(f^{-1}(B_1)\) and \(f^{-1}(B_2)\) in \(\mathcal F\) implies that their intersection \(f^{-1}(B_1\cap B_2)\) is too. Being closed upwards is similarly easy to check.

Remark. It follows by Zorn’s lemma (or equivalently, transfinite induction) that any filter is contained in an ultrafilter. The main point is to show that the union \(\mathcal F\) of filters \(\{\mathcal F_i\}\) in an ascending chain continues to be a filter. For one, if \(A\) and \(B\) are in \(\mathcal F\), then there is some \(\mathcal F_i\) containing both, so \(A\cap B\) is in \(\mathcal F_i\) and hence in \(\mathcal F\). Similarly, if \(A\subseteq B\) for \(A\in\mathcal F\), then \(A\in\mathcal F_i\) for some \(i\), so \(B\in\mathcal F_i\), so \(B\in\mathcal F\). (And of course \(\varnothing\notin\mathcal F\).)

Lemma 1. We note that a filter \(\mathcal F\) is an ultrafilter if and only if \(A\in\mathcal F\) or \(X\setminus A\in\mathcal F\) for all \(A\subseteq X\).

Proof. In one direction, suppose that \(\mathcal F\) satisfies the conclusion, and we want to show that \(\mathcal F\) is an ultrafilter. Well, suppose \(\mathcal F\subseteq\mathcal F’\) for any filter \(\mathcal F’\). Then for any \(A\notin\mathcal F\), we see that \(X\setminus A\in\mathcal F\), so \(X\setminus A\in\mathcal F’\), so \(A\notin\mathcal F’\) because \(\varnothing\notin\mathcal F’\). Taking the contraposition reveals \(\mathcal F=\mathcal F’\).

In the other direction, suppose that \(\mathcal F\) is an ultrafilter not containing \(A\), and we will show that \(X\setminus A\in\mathcal F\). Well, we note that \(\mathcal F\cdot\mathcal F_A\) cannot be a filter containing \(\mathcal F\). By a previous example (and the fact that \(A\notin\mathcal F\), it follows that \(\varnothing\notin\mathcal F\cdot\mathcal F_A\). Thus, \[\varnothing=B\cap C\] for some \(B\in\mathcal F\) and \(C\supseteq A\). In particular, \(B\subseteq X\setminus A\), so \(X\setminus A\in\mathcal F\) follows. \(\blacksquare\)

Remark. Given a function \(f\colon X\to Y\), we can use the above lemma to check that \(f_*\mathcal F\) is an ultrafilter when \(\mathcal F\) is. Indeed, for a subset \(B\subseteq Y\), we see that one of \(f^{-1}(B)\) or \(X\setminus f^{-1}(B)=f^{-1}(Y\setminus B)\) is in \(\mathcal F\).

Let’s begin doing some topology.

Definition. Fix a topological space \(X\). A filter \(\mathcal F\) converges to a point \(x\) if and only if \(\mathcal F\) contains all open neighborhoods of \(x\). We may write \(\mathcal F\to x\).

Remark. The collection of subsets containing an open neighborhood of \(x\) is a filter. Of course, this collection is nonempty and closed upwards, and it is closed under finite intersections because open sets are too.

Approximately speaking, we can think of filters as somehow being a fuzzy point, where the idea is that principal filters correspond to actual points. As with nets, we can characterize continuity with filter convergence.

Proposition 2. A function \(f\colon X\to Y\) of topological spaces is continuous if and only if any filter \(\mathcal F\) on \(X\) converging to \(x\) has \(f_*\mathcal F\to f(x)\).

Proof. We show the two directions independently.

• Suppose that \(f\) is continuous, and suppose that \(\mathcal F\to x\). Then we need to show that \(f_*\mathcal F\to f(x)\). Well, for any open neighborhood \(V\) of \(f(x)\), we see that \(f^{-1}(V)\) is an open neighborhood of \(x\), so \(V\in f_*\mathcal F\).

• Suppose that \(f\) preserves converging filters. Given some \(x\in X\), we will directly show that \(f\) is continuous at \(x\). Well, for any open neighborhood \(V\) of \(f(x)\), we need to find some open neighborhood inside \(f^{-1}(V)\). For this, we let \(\mathcal F\) be the filter generated by the open neighborhoods of \(x\). Then \(\mathcal F\to x\) by construction, so \(f_*\mathcal F\to f(x)\), so \(V\in f_*\mathcal F\), so \(f^{-1}(V)\in\mathcal F\), so \(f^{-1}(V)\) contains an open neighborhood of \(x\). \(\blacksquare\)

For our application to Tychonoff’s theorem, we need to understand how products and compactness behave with filters. Because it is mildly easier, let’s start with products; we will handle compactness after.

Lemma 3. Fix a collection \(\{X_i\}\) of topological spaces. Then a filter \(\mathcal F\) of \(\prod_iX_i\) converges to some \((x_i)\) if and only if \(\operatorname{pr}_{i*}\mathcal F\to x_i\) for each \(i\).

Proof. For brevity, let \(X\) denote the product space. The forward direction follows from Proposition 2 because the projections \(\operatorname{pr}_i\colon X\to X_i\) are continuous maps.

It remains to show the reverse direction. To show that \(\mathcal F\to(x_i)\), it is enough to check that \(\mathcal F\) contains a basis of open neighborhoods of \((x_i)\). In other words, we would like to show that \[\prod_iU_i\stackrel?\in\mathcal F,\] where each \(U_i\) is an open neighborhood of \(x_i\), and all but finitely many \(i\) have \(U_i=X_i\).

Well, \(\operatorname{pr}_{i*}\mathcal F\to x_i\) implies that \(U_i\in\operatorname{pr}_{i*}\mathcal F\), so \[U_i\times\prod_{j\ne i}X_j\in\mathcal F.\] It follows that \[\bigcap_{i\in I}\left(U_i\times\prod_{j\ne i}X_j\right)\in\mathcal F\] for any finite subset \(I\) of indices. In particular, the desired claim follows by taking \(I\) to be the finite collection of indices with \(U_i\ne X_i\). \(\blacksquare\)

Lemma 4. A topological space \(X\) is compact if and only if any ultrafilter converges to some point.

Proof. We show the implications separately.

• Suppose that \(X\) is compact, and suppose for the sake of contradiction that we have an ultrafilter \(\mathcal F\) not converging to any point. Then each \(x\in X\) has an open neighborhood \(U_x\) not in \(\mathcal F\). By compactness of \(x\), the open cover \(\{U_x\}_{x\in X}\) of \(X\) admits a finite subcover \(\{U_i\}_i\) of open subsets not in \(\mathcal F\).

  Now, because \(\mathcal F\) is an ultrafilter, we see that \(X\setminus U_i\in\mathcal F\) for each \(i\) by Lemma 1, but this implies that \(\varnothing\in\mathcal F\) by taking the intersection. This is our contradiction!

• Suppose that all ultrafilters converge to some point. Then we will show that \(X\) is compact by showing that any collection \(\mathcal A\) of closed sets with the finite intersection property has nonempty intersection. To see this is enough, note that if \(X\) did admit an open cover \(\mathcal U\) with no finite subcover, then taking \(\mathcal A=\{X\setminus U:U\in\mathcal U\}\) would provide a collection of closed sets with the finite intersection property but having empty intersection!

  We will use the hypothesis to find a point in \(\mathcal A\). Because \(\mathcal A\) has the finite intersection property, the collection of subsets containing an element in \(\mathcal A\) is a filter, which we can then extend to an ultrafilter \(\mathcal F\), which then converges to some point \(x\in X\).

  We claim that \(x\in\bigcap_{A\in\mathcal A}A\). Well, for any \(A\in\mathcal A\), we would like to show that \(x\in A\); because \(A\) is closed, it is enough to show that \(x\in\overline A\). Thus, choose an open neighborhood \(U\) of \(x\), and we would like to show that \(U\cap A\ne\varnothing\). For this, we note \(U\in\mathcal F\) (because \(\mathcal F\to x\)) and \(A\in\mathcal F\) (by construction of \(\mathcal F\)), so \(U\cap A\ne\varnothing\) follows. \(\blacksquare\)

We are now ready for our theorem.

Theorem (Tychonoff). Fix a collection \(\{X_i\}\) of topological spaces. Then the product space \(\prod_iX_i\) is still compact.

Proof. By Lemma 4, it is enough to show that any ultrafilter \(\mathcal F\) converges to some point \((x_i)\). Well, for each \(i\), the compactness of \(X_i\) implies that \(\operatorname{pr}_{i*}\mathcal F\) converges to some point \(x_i\in X_i\) by Lemma 4, so it follows that \(\mathcal F\to(x_i)\) by Lemma 3. \(\blacksquare\)