Tychonoff's Theorem

Tychonoff's theorem is one of the most important and powerful theorems in general topology. It states that an arbitrary product of compact spaces is compact in the product topology. The theorem is equivalent to the axiom of choice.

Statement

Theorem8.8Tychonoff's Theorem

Let $\{X_\alpha\}_{\alpha \in A}$ be a family of topological spaces (indexed by an arbitrary set $A$ ). Then the product $\prod_{\alpha \in A} X_\alpha$ with the product topology is compact if and only if each $X_\alpha$ is compact.

The "only if" direction is easy (projections are continuous and surjective, and continuous images of compact spaces are compact). The deep content is the "if" direction: compact factors imply a compact product, even for uncountable index sets.

Significance

RemarkWhy Tychonoff's Theorem Matters

Tychonoff's theorem is fundamental in:

Functional analysis: The Banach--Alaoglu theorem (the unit ball in a dual space is weak-* compact) is a consequence of Tychonoff.
Algebraic geometry: The Zariski topology on $\operatorname{Spec}(R)$ uses compactness of products of finite spaces.
Logic: The compactness theorem for propositional logic follows from Tychonoff applied to $\{0, 1\}^I$ .
Measure theory: Kolmogorov's extension theorem uses compact products.
Stone-Cech compactification: $\beta X$ is defined as a closed subspace of $[0,1]^{\mathcal{F}}$ , which is compact by Tychonoff.

Equivalence with the Axiom of Choice

Theorem8.9Tychonoff $\Leftrightarrow$ AC

Tychonoff's theorem is equivalent to the axiom of choice (over ZF set theory).

Proof

(AC $\Rightarrow$ Tychonoff): This is the standard proof (see Chapter 8, Proof section).

(Tychonoff $\Rightarrow$ AC): Due to Kelley (1950). Given a family of nonempty sets $\{S_\alpha\}_{\alpha \in A}$ , we must show $\prod_\alpha S_\alpha \neq \emptyset$ .

For each $\alpha$ , let $X_\alpha = S_\alpha \cup \{*_\alpha\}$ with the topology $\tau_\alpha = \{\emptyset, \{*_\alpha\}, X_\alpha\}$ (the particular point topology with $*_\alpha$ as the open point). Each $X_\alpha$ is compact: in any open cover, the set $X_\alpha$ itself is a member (or a finite union covers because $X_\alpha$ has only three open sets).

By Tychonoff, $\prod_\alpha X_\alpha$ is compact. The sets $C_\alpha = \prod_\beta Y_\beta$ where $Y_\beta = X_\beta$ for $\beta \neq \alpha$ and $Y_\alpha = S_\alpha = X_\alpha \setminus \{*_\alpha\}$ are closed (since $\{*_\alpha\}$ is open, $S_\alpha$ is closed). The family $\{C_\alpha\}$ has the FIP (any finite intersection is nonempty since each $S_\alpha$ is nonempty). By compactness, $\bigcap_\alpha C_\alpha = \prod_\alpha S_\alpha \neq \emptyset$ .

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Key Consequences

Theorem8.10Banach--Alaoglu Theorem

Let $V$ be a normed vector space and $V^*$ its dual. The closed unit ball $B^* = \{f \in V^* : \|f\| \leq 1\}$ is compact in the weak-* topology.

Proof

For each $v \in V$ , the evaluation $\operatorname{ev}_v: B^* \to \mathbb{R}$ given by $f \mapsto f(v)$ satisfies $|f(v)| \leq \|v\|$ . So $B^*$ embeds into $\prod_{v \in V} [-\|v\|, \|v\|]$ , which is compact by Tychonoff. The image of $B^*$ is closed (intersection of closed hyperplanes), hence compact.

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ExampleApplications of Tychonoff

Cantor set: $\{0, 1\}^{\mathbb{N}} \cong C$ (the Cantor set) is compact.
Hilbert cube: $[0, 1]^{\mathbb{N}}$ is compact, a universal separable metrizable compact space.
Profinite groups: $\varprojlim G_i$ is compact (as a closed subspace of $\prod G_i$ with each $G_i$ finite and discrete, hence compact).
$p$ -adic integers: $\mathbb{Z}_p = \varprojlim \mathbb{Z}/p^n\mathbb{Z}$ is compact (inverse limit of finite groups).
Compactness theorem in logic: The set of models of a propositional theory is a closed subspace of $\{T, F\}^{\text{variables}}$ , which is compact by Tychonoff.

RemarkProduct Topology is Essential

Tychonoff's theorem fails for the box topology. The product $[0, 1]^{\mathbb{N}}$ with the box topology is not compact: the cover $\{\prod_n U_n^{(k)}\}$ with $U_n^{(k)} = ((k-1)/n, (k+1)/n) \cap [0,1]$ has no finite subcover.

This is one of the primary reasons the product topology is preferred over the box topology for infinite products.