Tychonoff's Theorem

Theorem
Let $\left \langle {X_i}\right \rangle_{i \in I}$ be a family of non-empty topological spaces, where $I$ is an arbitrary index set.

Let $\displaystyle X = \prod_{i \in I} X_i$ be the corresponding product space.

Then $X$ is compact if and only if each $X_i$ is.

Proof

 * First assume that $X$ is compact.

Since the projections $\operatorname{pr}_i : X \to X_i$ are continuous, it follows that the $X_i$ are compact.


 * Assume now that each $X_i$ is compact.

By the equivalent definitions of compact sets it is enough to show that every ultrafilter on $X$ converges.

Thus let $\mathcal F$ be an ultrafilter on $X$.

For each $i \in I$, the image filter $\operatorname{pr}_i \left({\mathcal F}\right)$ then is an ultrafilter on $X_i$.

Since each $X_i$ is compact by assumption, each $\operatorname{pr}_i \left({\mathcal F}\right)$ therefore converges to a $x_i \in X_i$.

This implies that $\mathcal F$ converges to $x := \left({x_i}\right)_{i \in I}$.

Preliminaries
From the definition of the Tychonoff topology, a basic open set of the natural basis of $X$ is a set of the form:
 * $\prod_{i \in I} U_i$

where:
 * each $U_i$ is a nonempty open subset of $X_i$

and:
 * $U_i = X_i$ for all but finitely many $i \in I$.

The word 'collection will be used to mean set of sets.

Statement
Let $\left({I, <}\right)$ be a well-ordered set.

(The Well-Ordering Theorem, implied by the Axiom of Choice, states that every set is well-orderable.)

Let $\left \langle {X_i} \right \rangle_{i \in I}$ be a family of compact topological spaces.

Denote $\displaystyle X = \prod_{i \in I} X_i$.

Let $F$ be the set-theoretic tree $\left({\bigcup_{i \in I} \prod_{j < i} X_j}\right) \cup X$ of functions defined on initial intervals of $I$ ordered by inclusion.

Consider the set of all subtrees $T \subset F$ with the following property:
 * For every $i \in I$ and every $\displaystyle f \in T \cap \prod_{j<i} X_j$, the set $\left\{{g \left({i}\right): g \in T, f \subsetneq g}\right\}$ is closed in $X_i$.

Suppose that every such subtree $T$ has a branch.

(By the Hausdorff Maximal Principle, every poset has a maximal chain, so every tree has a branch.)

Then $\displaystyle \prod_{i \in I} X_i$ is compact.

Proof
To prove that every open cover of $X$ has a finite subcover, it is enough to prove that every open cover by basic open set of the natural basis has a finite subcover.

Let $\mathcal O$ be a collection of basic open subsets of $X$ such that no finite subcollection of $\mathcal O$ covers $X$.

It is enough to prove that $\mathcal O$ does not cover $X$.

Let $T_{\mathcal O}$ be the set of all elements $f \in F$ such that the set $\left\{{x \in X : f \subset x}\right\}$ is not covered by any finite subcollection of $\mathcal O$.

Then $T_{\mathcal O}$ is a subtree of $F$.

For every $i\in I$ and $f\in T_{\mathcal O}\cap\prod_{j<i}X_j$, denote


 * $C_{\mathcal O}(f) = \{\,g(i): f\subsetneqq g\in T_{\mathcal O}\,\}$

Step 1. For every $i\in I$ and every $f\in T_{\mathcal O}\cap\prod_{j<i}X_j$, $C_{\mathcal O}(f)$ is closed in $X_i$ and nonempty.

To prove this, let $\mathcal W$ be the collection of all open subsets $U$ of $X_i$ such that there exist a finite subcollection $\mathcal P\subset\mathcal O$ such that:
 * $\{\,x\in X\mid f\subset x,\ x(i)\in U\ \,\}\subset\bigcup\mathcal P$

Then $X_i\setminus C_{\mathcal O}(f) = \bigcup\mathcal W$ and hence $C_{\mathcal O}(f)$ is closed.

It also follows that $C_{\mathcal O}(f)$ is nonempty, because otherwise, by compactness of $X_i$, $\mathcal W$ would have a finite subcover for $X_i$, which would yield a finite collection of $\mathcal O$ that covers $\{\,x\in X\mid f\subset x\ \,\}$ in contradiction with the fact that $f\in T_{\mathcal O}$.

(Here the compactness of $X_i$ is used similarly to the usual proof that the product of two compact spaces is compact.)

It is left to show that indeed $X_i\setminus C_{\mathcal O}(f) = \bigcup\mathcal W$.

Consider an arbitrary $a\in X_i\setminus C_{\mathcal O}(f)$ and define $g\in\prod_{j\le i}X_j$ by: $f\subset g$ and $g(i)=a$.

Then $g\notin T_{\mathcal O}$, and therefore there is a finite collection $\mathcal P\subset\mathcal O$ such that:


 * $\left\{{x \in X: f \subset x, x(i) = a}\right\} = \left\{{x \in X: g \subset x}\right\} \subset \bigcup \mathcal P $

and
 * $\left\{{x \in X: f \subset x, x(i) = a}\right\} \cap V \ne \varnothing$ for every $V \in\mathcal P$.

Let $U=\bigcap\{\,\operatorname{pr}_i(V) \mid V\in\mathcal P\,\}$.

Then $U$ is an open subset of $X_i$, $a\in U$, and


 * $\{\,x\in X\mid f\subset x,\ x(i)\in U\ \,\}\subset\bigcup\mathcal P$

Therefore $U\cap C_{\mathcal O}(f)=\varnothing$ and $a\in U\in\mathcal W$.

Step 2.

Every branch of $T_{\mathcal O}$ has the greatest element.

To prove this, suppose that $B$ is a branch of $T_{\mathcal O}$ without the greatest element.

Let $f=\bigcup B$.

Let $i$ be the least element of $I$ that is not in the domain of any element of $B$; then $f\in\prod_{j<i}X_j$.

Since $B$ has no greatest element, $f\notin B$, and since $B$ is a maximal chain in $T_{\mathcal O}$, $f\notin T_{\mathcal O}$.

Let $\mathcal P\subset\mathcal O$ be a finite collection such that:
 * $\left\{{x \in X: f \subset x}\right\} \subset \bigcup \mathcal P$

Let $m$ be the greatest element of the finite set:
 * $\left\{{j \in I: j < i \text{ and } \exists V \in \mathcal P \operatorname{pr}_j \left({V}\right) \ne X_j}\right\}$

Let $g$ be any element of $B$ that is defined on $m$.

Consider an arbitrary $x\in X$ such that $g\subset x$.

Let $y\in X$ be defined by $f\subset y$ and $y(j)=x(j)$ for every $j\ge i$, and choose $V\in\mathcal P$ such that $y\in V$.

Then $x\in V$ (because $V$ does not "take into account" the values of $x(j)$ for $m<j<i$).

Thus:
 * $\left\{{x \in X: g \subset x}\right\} \subset \bigcup \mathcal P$

in contradiction of the fact that $g \in T_{\mathcal O}$.

Step 3.

Every maximal element of $T_{\mathcal O}$ is an element of $X$: an element $f\in T_{\mathcal O}\setminus X$ cannot be maximal in $T_{\mathcal O}$ because $C_{\mathcal O}(f)\ne\varnothing$.

Now it can be shown that there is $f\in X$ such that $f\notin\bigcup\mathcal O$.

Indeed, according to the hypotheses, $T_{\mathcal O}$ has a branch $B$.

Let $f$ be the greatest element of $B$.

Then $f$ is a maximal element of $T_{\mathcal O}$.

Therefore $f\in X$.

Therefore the set $\left\{{f}\right\} = \left\{{x \in X: f \subset x}\right\}$ is not covered by any finite subcollection of $\mathcal O$.

Hence $f \notin \bigcup \mathcal O$.

Corollary 1
The Cartesian product of a finite family of compact topological spaces is compact.

Corollary 2
Let $I$ be a well-orderable set and $(X_i)_{i\in I}$ a family of compact topological spaces.

Suppose that the Cartesian product of all nonempty closed subsets of all $X_i$ is nonempty:
 * $\displaystyle \prod \left\{{C: C \text { is nonempty and closed in } X_i \text { for some } i \in I}\right\} \ne \varnothing$

Then $\displaystyle \prod_{i\in I} X_i$ is compact.

Proof
Let $<$ be a well-order relation on $I$.

Denote $X = \prod_{i\in I}X_i$.

Let $F$ be the tree $\displaystyle \left({\bigcup_{i \in I} \prod_{j < i} X_j}\right) \cup X$ ordered by inclusion.

To apply the theorem, it is enough to verify that if $T$ is a subtree of $F$ such that for every $i\in I$ and every $\displaystyle f \in T \cap \prod_{j<i}X_j$, the set $\left\{{g(i): g \in T,\ f \subsetneq g}\right\}$ is closed in $X_i$, then $T$ has a branch.

Let:
 * $\displaystyle e \in \prod \left\{{C: C \text{ is nonempty and closed in } X_i \text { for some } i \in I}\right\}$

be a choice function.

Let $B_e$ be the minimal tree among all subtrees $S$ of $T$ with the property that for every $i\in I$ and every $f\in S\cap\prod_{j<i}X_j$:


 * $e\left({\left\{{g(i): g \in T,\ f \subsetneq g}\right\}}\right) \in \left\{{g (i): g \in S,\ f \subsetneq g}\right\}$

unless
 * $\left\{{g(i): g \in T,\ f \subsetneq g}\right\} = \varnothing$

Such subtrees of $T$ exist because $T$ itself is such, and the minimal such subtree is the intersection of all such subtrees.

It can be shown that $B_e$ is a branch by assuming that it is not, considering the minimal $i\in I$ where it "branches", and arriving at a contradiction with its minimality.

Applications
The proof that $[0,1]^{\mathbb Z}$ is compact does not require the Axiom of Choice, because the product of all nonempty closed subsets of $[0,1]$ contains, for example, the greatest lower bound function $\inf$ (restricted to the collection of closed subsets of $[0,1]$).