Tychonoff’s Theorem By Filters
Tychonoff’s Theorem By Filters
Statement
Tychonoff’s theorem says that an arbitrary product of compact topological spaces is compact. More precisely, if \((X_i)_{i \in I}\) is a family of compact spaces, then the product space \[ X=\prod_{i\in I} X_i \] with the product topology is compact.
The proof below uses filters. The point of the filter proof is that compactness becomes a clean convergence statement, and the product topology is exactly designed so that convergence in the product is the same as coordinatewise convergence.
Filters
Let \(X\) be a set. A filter \(\mathcal F\) on \(X\) is a nonempty collection of subsets of \(X\) satisfying the following three conditions.
- \(\varnothing \notin \mathcal F\).
- If \(A,B\in\mathcal F\), then \(A\cap B\in\mathcal F\).
- If \(A\in\mathcal F\) and \(A\subset B\subset X\), then \(B\in\mathcal F\).
We think of the sets in \(\mathcal F\) as the “large” subsets of \(X\). The first condition says that the empty set is not large. The second says that the intersection of two large sets is large. The third says that any superset of a large set is large.
A basic example comes from a point \(x\in X\): the neighborhood filter at \(x\) is \[ \mathcal N(x)=\{A\subset X: A \text{ contains an open neighborhood of } x\}. \] A filter \(\mathcal F\) on a topological space \(X\) converges to \(x\), written \(\mathcal F\to x\), if \[ \mathcal N(x)\subset \mathcal F. \] In words: every neighborhood of \(x\) is large for the filter.
Finer Filters And Ultrafilters
Let \(\mathcal F\) and \(\mathcal G\) be filters on the same set \(X\). We say that \(\mathcal G\) is finer than \(\mathcal F\) if \[ \mathcal F\subset \mathcal G. \] A finer filter has more large sets, so it imposes more conditions.
An ultrafilter is a filter that is maximal among filters on \(X\): it is not properly contained in any strictly finer filter. Equivalently, a filter \(\mathcal U\) on \(X\) is an ultrafilter if and only if, for every subset \(A\subset X\), \[ A\in\mathcal U \quad\text{or}\quad X\setminus A\in\mathcal U. \] Exactly one of these two alternatives holds, since a filter cannot contain both \(A\) and \(X\setminus A\).
Existence Of Ultrafilters
Every filter is contained in an ultrafilter.
This result uses Zorn’s lemma. Let \(\mathcal F\) be a filter on \(X\). Consider the partially ordered set of all filters on \(X\) containing \(\mathcal F\), ordered by inclusion. Every chain has an upper bound, namely the union of all filters in the chain. The union is again a filter: finite intersections already occur inside one member of the chain, since the chain is linearly ordered by inclusion. Hence Zorn’s lemma gives a maximal filter containing \(\mathcal F\), which is an ultrafilter.
This is the only serious set-theoretic ingredient in the proof.
Image Filters
Let \(f:X\to Y\) be a function, and let \(\mathcal F\) be a filter on \(X\). The image filter \(f\mathcal F\) on \(Y\) is the filter generated by the sets \(f(A)\) with \(A\in\mathcal F\).
Equivalently, \[ B\in f\mathcal F \] if and only if there exists \(A\in\mathcal F\) such that \[ f(A)\subset B. \]
If \(\mathcal F\) is an ultrafilter, then \(f\mathcal F\) is an ultrafilter on \(Y\), provided \(f(X)=Y\) is not needed; more generally, it is an ultrafilter on the image support determined by \(f\). For our application, the coordinate projections from a nonempty product are surjective, so no subtlety arises.
The key convergence fact is simple: if \(\mathcal F\to x\) in \(X\) and \(f:X\to Y\) is continuous, then \[ f\mathcal F\to f(x). \] Indeed, if \(V\) is a neighborhood of \(f(x)\), then \(f^{-1}(V)\) is a neighborhood of \(x\), hence \(f^{-1}(V)\in\mathcal F\), and therefore \(V\in f\mathcal F\).
Compactness Via Ultrafilters
A topological space \(X\) is compact if and only if every ultrafilter on \(X\) converges to at least one point of \(X\).
We prove both directions.
First suppose \(X\) is compact, and let \(\mathcal U\) be an ultrafilter on \(X\). Suppose, for contradiction, that \(\mathcal U\) does not converge to any point. Then for each \(x\in X\), there is an open neighborhood \(V_x\) of \(x\) such that \[ V_x\notin\mathcal U. \] Since \(\mathcal U\) is an ultrafilter, \(X\setminus V_x\in\mathcal U\) for every \(x\). The family \((V_x)_{x\in X}\) is an open cover of \(X\). By compactness, there are \(x_1,\ldots,x_n\) such that \[ X=V_{x_1}\cup\cdots\cup V_{x_n}. \] Therefore \[ \varnothing=(X\setminus V_{x_1})\cap\cdots\cap(X\setminus V_{x_n}). \] But each \(X\setminus V_{x_k}\) belongs to \(\mathcal U\), and filters are closed under finite intersections. Hence \(\varnothing\in\mathcal U\), a contradiction. Therefore \(\mathcal U\) converges.
Conversely, suppose every ultrafilter on \(X\) converges. We show that \(X\) is compact. Let \((G_\alpha)_{\alpha\in A}\) be an open cover of \(X\). Suppose it has no finite subcover. Then the closed sets \[ F_\alpha=X\setminus G_\alpha \] have the finite intersection property: every finite intersection is nonempty. The family of finite intersections of the \(F_\alpha\) generates a filter \(\mathcal F\). By the ultrafilter existence theorem, extend \(\mathcal F\) to an ultrafilter \(\mathcal U\).
By assumption, \(\mathcal U\to x\) for some \(x\in X\). Since the \(G_\alpha\) cover \(X\), choose \(\alpha\) with \(x\in G_\alpha\). Then \(G_\alpha\) is a neighborhood of \(x\), so \(G_\alpha\in\mathcal U\). But \(F_\alpha=X\setminus G_\alpha\) belongs to the original filter base, hence \(F_\alpha\in\mathcal U\). Therefore \(\varnothing=G_\alpha\cap F_\alpha\in\mathcal U\), impossible. Hence every open cover has a finite subcover, and \(X\) is compact.
Convergence In Products
Let \[ X=\prod_{i\in I}X_i, \] and let \(\pi_i:X\to X_i\) be the \(i\)th coordinate projection.
A filter \(\mathcal F\) on \(X\) converges to \(x=(x_i)_{i\in I}\) if and only if, for every \(i\in I\), \[ \pi_i\mathcal F\to x_i. \] The forward implication follows because each projection \(\pi_i\) is continuous.
For the reverse implication, suppose \(\pi_i\mathcal F\to x_i\) for every \(i\). We must show that every neighborhood of \(x\) belongs to \(\mathcal F\). By definition of the product topology, every neighborhood of \(x\) contains a basic neighborhood of the form \[ \bigcap_{k=1}^n \pi_{i_k}^{-1}(V_{i_k}), \] where \(V_{i_k}\) is a neighborhood of \(x_{i_k}\) in \(X_{i_k}\). Since \(\pi_{i_k}\mathcal F\to x_{i_k}\), each \(V_{i_k}\) belongs to \(\pi_{i_k}\mathcal F\). Hence each inverse image \(\pi_{i_k}^{-1}(V_{i_k})\) belongs to \(\mathcal F\). By closure under finite intersections, \[ \bigcap_{k=1}^n \pi_{i_k}^{-1}(V_{i_k})\in\mathcal F. \] Since \(\mathcal F\) is upward closed, every neighborhood of \(x\) belongs to \(\mathcal F\). Therefore \(\mathcal F\to x\).
This lemma is the heart of the proof. Product convergence is exactly coordinatewise convergence.
Proof Of Tychonoff’s Theorem
Let \((X_i)_{i\in I}\) be compact spaces, and let \[ X=\prod_{i\in I}X_i \] with the product topology. We prove that \(X\) is compact by showing that every ultrafilter on \(X\) converges.
Let \(\mathcal U\) be an ultrafilter on \(X\). For each \(i\in I\), form the image ultrafilter \[ \mathcal U_i=\pi_i\mathcal U \] on \(X_i\).
Since \(X_i\) is compact, the ultrafilter characterization of compactness implies that \(\mathcal U_i\) converges to some point \(x_i\in X_i\). Now define \[ x=(x_i)_{i\in I}\in\prod_{i\in I}X_i. \] For every \(i\), we have \[ \pi_i\mathcal U=\mathcal U_i\to x_i. \] By the product convergence lemma, \(\mathcal U\to x\) in \(X\).
Thus every ultrafilter on \(X\) converges. By the ultrafilter characterization of compactness, \(X\) is compact.
This proves Tychonoff’s theorem.
Why The Proof Works
The proof is short because filters isolate the two essential facts.
First, compactness is exactly the statement that every ultrafilter has a limit. An ultrafilter is a generalized way of saying “eventually,” without requiring a sequence or a countable index set.
Second, the product topology is exactly the topology for which convergence is coordinatewise convergence. A neighborhood in the product only restricts finitely many coordinates. A filter therefore converges in the product precisely when its projections converge in each coordinate.
Tychonoff’s theorem follows by combining these two observations: project an ultrafilter to each compact coordinate space, take a coordinatewise limit, and then use the product topology to assemble those coordinatewise limits into a genuine product limit.