Tychonoff's theorem

Definition. Given a family $\{X_i\}$ of topological spaces, the product topology on $X = \prod X_i$ is the coarsest topology in which each projection $\pi_i : X \to X_i$ is continuous. The open sets of the form $\pi_i^{-1}(V)$ where $V$ is open in $X_i$ form a subbasis for the product topology.

Tychonoff’s theorem. A product of compact spaces is compact.

Of course, taken with the product topology. Another natural topology for $X_i$ is to consider the sets of the form $\prod V_i$ where each $V_i$ is open in $X_i$. This forms a basis for the box topology, yet a product of compact spaces may not be compact under the box topology. If the product is finite, the box and product topology coincide, since

$$\prod_{i=1}^n V_i = \bigcap_{i=1}^n \pi_i^{-1}(V_i)$$

But for the infinite product $[0, 1]^\omega$, we can define the nbds $U_n = \prod V_{in}$ where $V_{in} = (0, 1/3)$ if $i \ne n$ and $V_{in} = X_i$ if $i = n$. Clearly $\{U_n\}$ must cover $[0, 1]^{\omega}$, but the point $x = (x_n)$ defined as $x_n = 2/3$ cannot be covered by any strict subcover of $U_n$. The box topology, in a sense, has too many open sets.

To prove the theorem, we will prove every FIP (finite-intersection-property) family of closed sets has a nonempty intersection, which is equivalent to being compact. If $\mathcal{A}$ is FIP, we use Zorn’s lemma to construct a maximal $\mathcal{D} \supseteq \mathcal{A}$ which is FIP, and prove the $\bigcap_{D \in \mathcal{D}} \cl{D}$ is nonempty by proving it contains a point $x$ with coordinates $x_i \in \bigcap_{D \in \mathcal{D}} \pi_i^{-1}(\cl{D})$, using the fact that this is a FIP family of closed sets and that $X_i$ is compact.

Lemma 1. For any family $\mathcal{A}$ totally-ordered by inclusion, if $\mathcal{F}$ is a collection of FIP families $F$ such that $\mathcal{A} \subseteq F$, then $\bigcup \mathcal{F}$ is also a FIP family containing $\mathcal{A}$.

Proof. Clearly $\mathcal{A} \subseteq \bigcup \mathcal{F}$. Let $S = \{x_1, \dots, x_n\}$ be a finite subset of $\bigcup \mathcal{F}$. Then choose for each $i$, an $X_i$ such that $x_i \in X_i$. Since $\mathcal{F}$ is totally ordered, let $X_i$ be the maximum element of $\{X_1, \dots, X_n\}$ such that $S \subseteq X_i$. Since $X_i$ is FIP, $\bigcap S$ is nonempty, proving $\bigcup \mathcal{F}$ is FIP.

Lemma 2. Let $\mathcal{D}$ be a maximal FIP-family of subsets of $X$ that contains $\mathcal{A}$.

(a) For any finite subset $\{D_1, \dots, D_n\} \subseteq \mathcal{D}$, $\bigcap_{i=1}^n D_i \in \mathcal{D}$.

(b) If $A \subseteq X$ is a set which intersects every element of $\mathcal{D}$, then $A \in \mathcal{D}$.

Proof. (a) Let $D = \bigcap_{i=1}^n D_i$. We just have to show $\mathcal{D}' = \mathcal{D} \cup \{D\}$ is FIP, and so by maximality of $\mathcal{D}$, it must be that $D \in \mathcal{D}$. Clearly any finite subset of $\mathcal{D}'$ can be written as a finite subset of $\mathcal{D}$ including or not $D$. So it suffices to show that for any $\{D'_1, \dots, D'_m\} \subseteq \mathcal{D}$, that

$$D \cap \bigcap_{i=1}^m D'_i \ne \emptyset$$

This is clearly true, as it is a finite intersection of elements in $\mathcal{D}$.

(b) Again, we just have to show $\mathcal{D}' = \mathcal{D} \cup \{A\}$ is FIP, and again, we just have to prove $A \cap \bigcap_{i=1}^n D_i$ is nonempty for any finite selection $\{D_1, \dots, D_n\} \subseteq \mathcal{D}$. But by (a), $D = \bigcap_{i=1}^n D_i \in \mathcal{D}$, and by hypothesis, $A \cap D \ne \emptyset$. $\square$

Proposition 1. The axiom of choice implies Tychonoff’s theorem.

Proof. Let $\{X_i\}_{i \in I}$ be a family of compact sets and $X = \prod_{i \in I} X_i$. Let $\mathcal{A}$ be a FIP family in $X$ and we wish to show $\bigcap_{A \in \mathcal{A}} \cl{A} \ne \emptyset$. By Lemma 1, using Zorn’s lemma, there must exist a maximal family $\mathcal{D}$ in $X$ which is FIP and $\mathcal{A} \subseteq \mathcal{D}$. This means $\bigcap_{D \in \mathcal{D}} \cl{D} \subseteq \bigcap_{A \in \mathcal{A}} \cl{A}$, so it suffices to show the former is nonempty.

We show $\mathcal{D}_i = \{ \pi_i(D) \ : \ D \in \mathcal{D} \}$ is FIP. This follows from basic set operation properties.

$$\bigcap_{j=1}^n D_j \ne \emptyset \implies \bigcap_{j=1}^n \pi_i(D_j) \supseteq \pi_i\left(\bigcap_{j=1}^n D_j\right) \ne \emptyset$$

For each $i \in I$, since $X_i$ is compact, we can choose (using the axiom of choice again!) an element

$$x_i \in \bigcap_{D \in \mathcal{D}} \cl{\pi_i(D)}$$

and so $x = (x_i)_{i \in I} \in X$. Now we prove $x \in \bigcap_{D \in \mathcal{D}} \cl{D}$. For each $i$, take any open set $V \ni x_i$ of $X_i$. Since $x_i \in \cl{\pi_i(D)}$ for all $D$, $V \cap \pi_i(D) \ne \emptyset$ and so there must exist some $y \in D$ with $y_i \in V$. This proves $\pi_i^{-1}(V)$ intersects every $D$, and by lemma 2(a), every subbasis element $\pi_i^{-1}(V)$ which contains $x$ must be an element of $\mathcal{D}$. Every basis element is a finite intersection of subbasis elements, which means every basis element containing $x$ is an element of $\mathcal{D}$. That is, for any basis element $B \ni x$, we have that $B \cap D \ne \emptyset$ for all $D \in \mathcal{D}$, since $\{B, D\} \subseteq \mathcal{D}$ is finite. This proves $x \in \cl{D}$ for all $D \in \mathcal{D}$. $\square$

Proposition 2. Tychonoff’s theorem implies the axiom of choice.

Proof. We wish to prove a product of nonempty sets is nonempty. Let $\{X_i\}_{i \in I}$ be any family of nonempty sets. The standard trick is to select an element $\lambda \notin \bigcup_{i \in I} X_i$ so that if we define $Y_i = X_i \cup \{\lambda\}$, we have $X_i \ne Y_i$. Define $Y = \prod_{i \in I} Y_i$. We define the topology $\tau_i$ on $Y_i$ as

$$\tau_i = \{ \emptyset, X_i, \{\lambda\}, Y_i \}$$

This way, we have two properties.

1. Each $X_i$ is closed in $Y_i$ (since its complement is $\{\lambda\}$). Since each projection $\pi_i : Y \to Y_i$ is continuous, we have that $\pi_i^{-1}(X_i)$ is closed in $Y$.

2. Each $Y_i$ is compact. Clearly, as there are only finitely many open sets.

By Tychonoff’s theorem, $Y$ is compact. Furthermore, define the family of closed sets

$$\mathcal{F} = \{ \pi_i^{-1}(X_i) \ : \ i \in I \}$$

If $J \subseteq I$ is finite, we can choose an element $x_j \in X_j$ for each $j \in J$ and define the point $y = (y_i)$ where

$$y_i = \begin{cases} x_i, & \ i \in J \\ \lambda, & \ i \notin J \end{cases}$$

Then $y \in Y$ (this is why $\lambda$ is introduced in the proof at all, as otherwise how would we construct such an element of $Y$ without the axiom of choice?), and furthermore, $y \in \bigcap_{i \in J} \pi_i^{-1}(X_i)$. This proves $\mathcal{F}$ is a FIP family of closed sets, and so by the compactness of $Y$,

$$\bigcap_{i \in I} \pi_i^{-1}(X_i) = \prod_{i \in I} X_i \ne \emptyset \quad \square$$

References

Munkres, Topology: 37 - The Tychonoff Theorem

J. L. Kelley (1950) The Tychonoff Theorem Implies the Axiom of Choice