Proof of Tychonoff's Theorem via Ultrafilters

We present the elegant proof of Tychonoff's theorem using the theory of ultrafilters. This approach makes the role of the axiom of choice transparent and provides a clean conceptual framework.

Statement

Theorem8.13Tychonoff's Theorem

Let $\{X_\alpha\}_{\alpha \in A}$ be a family of compact topological spaces. Then $P = \prod_{\alpha \in A} X_\alpha$ with the product topology is compact.

Proof via Ultrafilters

Proof

We use the characterization: a space is compact if and only if every ultrafilter converges (Theorem 8.4).

Let $\mathcal{U}$ be an ultrafilter on $P$ . We must show $\mathcal{U}$ converges to some point of $P$ .

Step 1: Push forward to each factor.

For each $\alpha \in A$ , the projection $\pi_\alpha: P \to X_\alpha$ is continuous. The pushforward $(\pi_\alpha)_*(\mathcal{U})$ is an ultrafilter on $X_\alpha$ (Theorem 8.5): $(\pi_\alpha)_*(\mathcal{U}) = \{B \subseteq X_\alpha : \pi_\alpha^{-1}(B) \in \mathcal{U}\}.$

Step 2: Convergence in each factor.

Since $X_\alpha$ is compact, every ultrafilter on $X_\alpha$ converges. So $(\pi_\alpha)_*(\mathcal{U})$ converges to some point $x_\alpha \in X_\alpha$ .

This means: for every open set $V_\alpha \subseteq X_\alpha$ containing $x_\alpha$ , we have $V_\alpha \in (\pi_\alpha)_*(\mathcal{U})$ , i.e., $\pi_\alpha^{-1}(V_\alpha) \in \mathcal{U}$ .

Step 3: The product point.

Define $\mathbf{x} = (x_\alpha)_{\alpha \in A} \in P$ . We claim $\mathcal{U} \to \mathbf{x}$ .

Step 4: Verification.

We must show that every open neighborhood of $\mathbf{x}$ in $P$ belongs to $\mathcal{U}$ . Since the product topology is generated by the subbasis $\{\pi_\alpha^{-1}(V_\alpha)\}$ , every open neighborhood of $\mathbf{x}$ contains a basic open set of the form: $W = \pi_{\alpha_1}^{-1}(V_{\alpha_1}) \cap \cdots \cap \pi_{\alpha_n}^{-1}(V_{\alpha_n})$ where each $V_{\alpha_i}$ is an open neighborhood of $x_{\alpha_i}$ in $X_{\alpha_i}$ .

By Step 2, $\pi_{\alpha_i}^{-1}(V_{\alpha_i}) \in \mathcal{U}$ for each $i = 1, \ldots, n$ .

Since $\mathcal{U}$ is a filter (closed under finite intersections): $W = \bigcap_{i=1}^n \pi_{\alpha_i}^{-1}(V_{\alpha_i}) \in \mathcal{U}.$

Since $W \subseteq U$ (where $U$ is the original open neighborhood of $\mathbf{x}$ ) and $\mathcal{U}$ is upward closed, $U \in \mathcal{U}$ .

Therefore $\mathcal{U}$ converges to $\mathbf{x}$ , and $P$ is compact.

■

Analysis of the Proof

RemarkWhere the Axiom of Choice is Used

The axiom of choice enters in two places:

Existence of ultrafilters (Step 1): The ultrafilter lemma (every filter extends to an ultrafilter) is proved via Zorn's lemma.
Choice of limit points (Step 2): For each $\alpha$ , we choose a point $x_\alpha$ to which $(\pi_\alpha)_*(\mathcal{U})$ converges. In a compact space, the set of limit points of an ultrafilter is nonempty (by the convergence characterization of compactness), but selecting one for each $\alpha$ simultaneously requires choice.

The first use is unavoidable (Tychonoff is equivalent to the axiom of choice). The second can be eliminated if each $X_\alpha$ is Hausdorff (then limits are unique).

Comparison with the Alexander Subbasis Proof

RemarkUltrafilter Proof vs. Alexander Proof

Alexander subbasis approach:

Uses Zorn's lemma on the collection of open covers without finite subcovers.
The maximal element gives a "worst-case" cover.
Subbasis structure is exploited to derive a contradiction.

Ultrafilter approach:

Uses Zorn's lemma to extend filters to ultrafilters.
Compactness of factors gives convergence of projected ultrafilters.
Product structure directly assembles the limit point.

Both proofs require the axiom of choice. The ultrafilter proof is arguably more transparent: it reduces Tychonoff to two clean steps (project and lift), while the Alexander proof is more combinatorial.

The Converse Direction

Theorem8.14Converse of Tychonoff

If $\prod_\alpha X_\alpha$ is compact (with the product topology) and each $X_\alpha$ is nonempty, then each $X_\alpha$ is compact.

Proof

Each projection $\pi_\alpha: \prod_\alpha X_\alpha \to X_\alpha$ is continuous and surjective (surjectivity uses the fact that each factor is nonempty -- the axiom of choice guarantees elements in the other factors). The continuous image of a compact space is compact. Therefore each $X_\alpha$ is compact.

■

ExampleThe Hilbert Cube Revisited

The Hilbert cube $H = [0, 1]^{\mathbb{N}}$ is compact by Tychonoff's theorem. It can be metrized by: $d(\mathbf{x}, \mathbf{y}) = \sum_{n=1}^{\infty} \frac{|x_n - y_n|}{2^n}.$

$H$ is a universal space for separable metrizable spaces: every separable metrizable space embeds in $H$ (Urysohn metrization theorem). The compactness of $H$ (via Tychonoff) is crucial for this universality.

ExampleProfinite Completions

Let $G$ be a group and $\{N_i\}$ the family of its normal subgroups of finite index. The profinite completion $\hat{G} = \varprojlim_{i} G/N_i$ is a closed subspace of the product $\prod_i G/N_i$ . Since each $G/N_i$ is finite (hence compact with the discrete topology), the product is compact by Tychonoff, and $\hat{G}$ is compact (closed in compact). This construction is fundamental in algebraic number theory (e.g., $\hat{\mathbb{Z}} = \prod_p \mathbb{Z}_p$ ).