From Compact Factors to a Compact Product
Compactness is defined by open covers: every open cover must have a finite subcover. In a product, however, open sets can involve many possible coordinates, and an open-cover proof can obscure how compactness of the individual factors combines. Ultrafilters provide a way to organize the argument coordinate by coordinate. A single ultrafilter on the product induces one on each factor; compactness gives a limit in every factor, and the product topology turns those coordinate limits into a limit in the product.
We use the product topology throughout. A basic open set in \(\prod_{i\in J}X_i\) restricts coordinates in only a finite subset of the index set \(J\); coordinates outside that finite subset are unrestricted. The index set \(J\) may be empty.
We use two earlier results: the ultrafilter extension principle, which extends any proper filter to an ultrafilter, and the characterization of compactness by ultrafilters. For clarity, the latter says that a space is compact if and only if every ultrafilter on its underlying set converges to at least one point. Here is the argument that connects this criterion to the open-cover definition.
Proof. Suppose first that \(X\) is compact, and let \(\mathcal{U}\) be an ultrafilter on \(X\). The family of closed sets \(\{\overline{A}:A\in\mathcal{U}\}\) has the finite intersection property: for \(A_1,\ldots,A_n\in\mathcal{U}\), their intersection \(A_1\cap\cdots\cap A_n\) belongs to \(\mathcal{U}\), so it is nonempty and is contained in \(\overline{A_1}\cap\cdots\cap\overline{A_n}\). Compactness implies that all these closures have a common point \(x\). To see this implication directly, if the closed sets had empty intersection, their open complements would cover \(X\); a finite subcover would give a finite intersection of the closed sets equal to \(\varnothing\), contradicting the finite intersection property.
Let \(V\) be any open neighborhood of \(x\). If \(V\notin\mathcal{U}\), the ultrafilter dichotomy gives \(X\setminus V\in\mathcal{U}\). Since \(X\setminus V\) is closed, its closure is itself. The choice of \(x\) would then imply \(x\in\overline{X\setminus V}=X\setminus V\), contradicting \(x\in V\). Thus \(V\in\mathcal{U}\), and \(\mathcal{U}\) converges to \(x\).
Conversely, suppose every ultrafilter on \(X\) converges, and let \(\{G_\alpha:\alpha\in A\}\) be an open cover with no finite subcover. The closed sets \(X\setminus G_\alpha\) have the finite intersection property: a finite intersection being empty would mean the corresponding finite collection of open sets covers \(X\). They therefore generate a proper filter, which extends to an ultrafilter \(\mathcal{U}\) by the ultrafilter extension principle. By assumption, \(\mathcal{U}\) converges to some \(x\in X\). Choose a cover member \(G_\alpha\) containing \(x\). Convergence gives \(G_\alpha\in\mathcal{U}\), while the construction gives \(X\setminus G_\alpha\in\mathcal{U}\). Their intersection is empty, contradicting that \(\mathcal{U}\) is proper. Hence every open cover has a finite subcover, and \(X\) is compact. \(\square\)
The extension principle used here relies on the axiom of choice, in its usual ultrafilter form. The product theorem below also uses choice to select a limit in each factor when limits are not unique.
Ultrafilters Induced by Coordinate Maps
In the product proof, \(f\) will be a coordinate projection \(\pi_i:\prod_{j\in J}X_j\to X_i\). The image collection really is an ultrafilter: preimages preserve inclusions, finite intersections, and complements; also \(f^{-1}(Y)=X\) and \(f^{-1}(\varnothing)=\varnothing\). Thus the filter rules and the ultrafilter dichotomy for \(\mathcal{U}\) give the corresponding rules for \(f_*\mathcal{U}\).
Proof. Suppose \(\mathcal{U}\) converges to \(x\). If \(V_i\) is an open neighborhood of \(x_i\) in \(X_i\), then \(\pi_i^{-1}(V_i)\) is an open neighborhood of \(x\) in the product. Hence \(\pi_i^{-1}(V_i)\in\mathcal{U}\), which means \(V_i\in(\pi_i)_*\mathcal{U}\). Therefore each image ultrafilter converges to its coordinate.
Conversely, suppose every \((\pi_i)_*\mathcal{U}\) converges to \(x_i\). Any open neighborhood \(W\) of \(x\) contains a basic open neighborhood \(C=\bigcap_{i\in F}\pi_i^{-1}(V_i)\), where \(F\subseteq J\) is finite and each \(V_i\) is open with \(x_i\in V_i\). Coordinate convergence gives \(V_i\in(\pi_i)_*\mathcal{U}\), so \(\pi_i^{-1}(V_i)\in\mathcal{U}\) for each \(i\in F\). Closure under finite intersections gives \(C\in\mathcal{U}\), and closure under supersets gives \(W\in\mathcal{U}\). If \(F\) is empty, \(C=X\in\mathcal{U}\), so the same conclusion holds. Thus \(\mathcal{U}\) converges to \(x\). \(\square\)
Tychonoff’s Theorem
Proof. If some \(X_i\) is empty, then the product is empty. The empty space is compact because every open cover of it has the empty finite subcover. If \(J\) is empty, the product is a singleton, which is compact: any open cover has a member containing its one point. It remains to consider a nonempty index set with all factors nonempty.
Write \(X=\prod_{i\in J}X_i\). Let \(\mathcal{U}\) be any ultrafilter on \(X\). For each \(i\in J\), the collection \((\pi_i)_*\mathcal{U}\) is an ultrafilter on \(X_i\). Since \(X_i\) is compact, the ultrafilter characterization gives a point \(x_i\in X_i\) to which \((\pi_i)_*\mathcal{U}\) converges. Choose such a point for each \(i\), using the axiom of choice if necessary, and put \(x=(x_i)_{i\in J}\in X\). The coordinate criterion shows that \(\mathcal{U}\) converges to \(x\). We have proved that every ultrafilter on \(X\) converges; the ultrafilter characterization now implies that \(X\) is compact. \(\square\)
The proof uses the product topology at exactly the point where a neighborhood is reduced to a basic open set involving only finitely many coordinates. It does not require the factors to be Hausdorff, and it does not require the limits in the factors to be unique. The argument is stated in the usual setting with the axiom of choice: the ultrafilter extension principle and the selection of coordinate limits are choice-dependent steps.
Three Ways to See the Coordinate Argument
Worked Example: A Nonprincipal Ultrafilter and a Limit in the Interval
Let \(\mathcal{U}\) be a nonprincipal ultrafilter on \(\mathbb{N}=\{1,2,\ldots\}\). Such an ultrafilter contains every cofinite set: it contains no finite set, so for every finite \(F\subseteq\mathbb{N}\), the ultrafilter dichotomy forces \(\mathbb{N}\setminus F\in\mathcal{U}\). Define \(f:\mathbb{N}\to[0,1]\) by \(f(n)=n/(n+1)\), and give \([0,1]\) its usual subspace topology from \(\mathbb{R}\).
For every \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(1/(N+1)<\varepsilon\). If \(n\geq N\), then \(0\leq 1-f(n)=1/(n+1)\leq1/(N+1)<\varepsilon\). Thus \(\{n:f(n)\in(1-\varepsilon,1]\}\) contains the cofinite set \(\{n:n\geq N\}\), so it belongs to \(\mathcal{U}\). Every relative open neighborhood of \(1\) contains \((1-\varepsilon,1]\) for some \(\varepsilon>0\); consequently \(f_*\mathcal{U}\) converges to \(1\).
Check: for \(\varepsilon=1/10\), take \(N=10\). At \(n=10\), \(f(10)=10/11\) and \(1-f(10)=1/11<1/10\), as required. The general estimate above applies to every \(n\geq10\), not just this check. This illustrates how a compact factor supplies a limit for an ultrafilter transported into it.
Worked Example: A Principal Ultrafilter on a Two-Factor Product
Let \(X=[0,1]\times\{0,1\}\), with the product topology, where \([0,1]\) has its usual topology and \(\{0,1\}\) has the discrete topology. Both factors are compact: \([0,1]\) is compact by Heine–Borel, and every open cover of a finite set has a finite subcover. Let \(p=(1/3,1)\) and let \(\mathcal{U}=\{A\subseteq X:p\in A\}\), the principal ultrafilter at \(p\).
For the first projection, a set \(B\subseteq[0,1]\) belongs to \((\pi_1)_*\mathcal{U}\) exactly when \(p\in\pi_1^{-1}(B)\), which is exactly when \(1/3\in B\). Hence \((\pi_1)_*\mathcal{U}\) is principal at \(1/3\). The same calculation gives that \((\pi_2)_*\mathcal{U}\) is principal at \(1\). Each principal ultrafilter converges to its defining point, since every neighborhood of that point contains the point. The coordinate criterion therefore gives \(\mathcal{U}\to(1/3,1)\) in the product.
Check: a basic neighborhood has the form \(V\times W\), where \(1/3\in V\subseteq[0,1]\) is open and \(1\in W\subseteq\{0,1\}\) is open. It contains \((1/3,1)\), so it belongs to \(\mathcal{U}\), in agreement with the product convergence conclusion.
Worked Example: Coordinate Limits of Unit Sequences
For each \(n\in\mathbb{N}\), define \(x^{(n)}\in\{0,1\}^{\mathbb{N}}\) by \(x^{(n)}_k=1\) when \(n=k\), and \(x^{(n)}_k=0\) when \(n\ne k\). Each factor \(\{0,1\}\) has the discrete topology and is compact. Fix a nonprincipal ultrafilter \(\mathcal{U}\) on \(\mathbb{N}\), and let \(f(n)=x^{(n)}\). Give the product \(\{0,1\}^{\mathbb{N}}\) the product topology.
Fix a coordinate \(k\). The inverse image under \(n\mapsto x^{(n)}_k\) of \(\{0\}\) is \(\mathbb{N}\setminus\{k\}\), which is cofinite and therefore belongs to \(\mathcal{U}\). The inverse image of \(\{1\}\) is \(\{k\}\), which does not belong to the nonprincipal ultrafilter. Thus the coordinate image ultrafilter is principal at \(0\), and it converges to \(0\). This holds for every \(k\), so the coordinate criterion gives \(f_*\mathcal{U}\to(0,0,\ldots)\).
Check: a basic neighborhood of the zero sequence restricts a finite set \(F\subseteq\mathbb{N}\) of coordinates to \(\{0\}\). Its inverse image under \(f\) is \(\mathbb{N}\setminus F\), a cofinite set in \(\mathcal{U}\). If \(F=\{2,5\}\), for example, the inverse image is \(\mathbb{N}\setminus\{2,5\}\); it excludes exactly the two indices whose sequences have a \(1\) in a restricted coordinate. This verifies the finite-coordinate step directly.
A Common Pitfall: Finite Choices Are Not a Finite Product Argument
A tempting open-cover argument says: choose a finite subcover in each factor, then combine those finite subcovers to get a finite subcover of the product. The last step is not justified for infinitely many factors. For example, suppose each factor contributes two selected sets. For the first \(m\) factors there are \(2^m\) ways to choose one set in each factor; for \(m=3\), this gives \(2\cdot2\cdot2=8\) choices. With infinitely many factors, these finite-stage counts do not produce a finite collection of choices covering the product. The ultrafilter proof avoids this gap: it does not combine factor covers, but instead obtains one limit in each coordinate and checks convergence using basic neighborhoods that restrict only finitely many coordinates.
| Stage | What is used | Why it matters |
|---|---|---|
| Start with an ultrafilter on the product | Coordinate projections | Each projection induces an ultrafilter on its factor |
| Use compactness of each factor | Ultrafilter characterization | Each coordinate ultrafilter has a limit |
| Assemble the coordinate limits | Choice when needed | The selected coordinates define a point of the product |
| Check convergence in the product | Finite-coordinate basic neighborhoods | The original ultrafilter converges to that point |
| Conclude compactness | Ultrafilter characterization again | Every open cover has a finite subcover |
Check Your Understanding
Answer each question by identifying the filter or neighborhood condition being used.
- For a map \(f:X\to Y\) and an ultrafilter \(\mathcal{U}\) on \(X\), express the condition \(B\in f_*\mathcal{U}\) using a preimage.
- Why does convergence of an ultrafilter on a product imply convergence of every coordinate image ultrafilter?
- In the converse direction of the coordinate criterion, why must a basic product neighborhood restrict only finitely many coordinates?
- Which step in the product proof uses compactness of \(X_i\), and which step may use the axiom of choice?
- What happens to the compactness conclusion if one factor is empty? What is the product when the index set is empty?