Tychonoff's Theorem/General Case

Theorem

Let $I$ be an indexing set.

Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of non-empty topological spaces.

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

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

Proof 1

First assume that $X$ is compact.

From Projection from Product Topology is Continuous, the projections $\pr_i : X \to X_i$ are continuous.

It follows from Continuous Image of Compact Space is Compact that the $X_i$ are compact.

Assume now that each $X_i$ is compact.

By Equivalence of Definitions of Compact Topological Space it is enough to show that every ultrafilter on $X$ converges.

Thus let $\FF$ be an ultrafilter on $X$.

From Image of Ultrafilter is Ultrafilter, for each $i \in I$, the image filter $\map {\pr_i} \FF$ is an ultrafilter on $X_i$.

Each $X_i$ is compact by assumption.

So by Equivalence of Definitions of Compact Topological Space, each $\map {\pr_i} \FF$ converges.

By Filter on Product Space Converges iff Projections Converge, $\FF$ converges.

$\blacksquare$

Proof 2

First assume that $X$ is compact.

From Projection from Product Topology is Continuous, the projections $\pr_i : X \to X_i$ are continuous.

It follows from Continuous Image of Compact Space is Compact that the $X_i$ are compact.

Assume now that each $X_i$ is compact.

By Compact Space satisfies Finite Intersection Axiom, it suffices to show that:

for each family $\CC$ of closed subsets of $X$ with:
$\ds \bigcap_{C \mathop \in \CC} C = \O$
there exists a finite subset $\SS \subseteq \CC$ such that:
$\ds \bigcap_{G \mathop \in \SS} G = \O$

We show the contrapositive.

That is, let $\CC$ be a family of closed subsets of $X$ such that:

$\ds \bigcap_{G \mathop \in \SS} G \ne \O$

for all finite subsets $\SS \subseteq \CC$.

We say that a family $\DD$ of subsets of $X$ has the "finite intersection property" if:

$\ds \bigcap_{S \mathop \in \SS} S \ne \O$

for all finite subsets $\DD \subseteq \CC$.

We aim to show that:

$\ds \bigcap_{C \mathop \in \CC} C \ne \O$

Let:

$S = \set {\CC \subseteq \AA : \AA \text { has the finite intersection property} }$

Lemma 1

$S$ has a $\subseteq$-maximal element $\AA$.

$\Box$

It will now suffice to show that:

$\ds \bigcap_{A \mathop \in \AA} \map \cl A \ne \O$

where $\map \cl A$ denotes the closure of $A$.

Lemma 2

Let $A, B \in \AA$.

Then $A \cap B \in \AA$.

$\Box$

Lemma 3

Let $A \subseteq X$ be such that:

$A \cap B \ne \O$

for all $B \in \AA$.

Then $A \in \AA$.

$\Box$

We now show that:

$\set {\map \cl { {\pr_i} \sqbrk A} : A \in \AA}$

has the finite intersection property for each $i \in I$, with $\map \cl {\map {\pr_i} A}$ being the closure taken in $X_i$.

Fix $i \in I$.

Pick a finite set:

$\set { {\pr_i} \sqbrk {A_1}, \ldots, {\pr_i} \sqbrk {A_n} } \subseteq \set { {\pr_i} \sqbrk A : A \in \AA}$

with $A_j \in \AA$ for each $1 \le j \le n$.

Since $\AA$ has the finite intersection property, we have:

$\ds \bigcap_{j \mathop = 1}^n A_j \ne \O$

so that:

$\ds {\pr_i} \sqbrk {\bigcap_{j \mathop = 1}^n A_j} \ne \O$

From Image of Intersection under Mapping: General Result, we have:

$\ds \O \ne {\pr_i} \sqbrk {\bigcap_{j \mathop = 1}^n A_j} \subseteq \bigcap_{j \mathop = 1}^n {\pr_i} \sqbrk {A_j}$

and in particular:

$\ds \bigcap_{j \mathop = 1}^n {\pr_i} \sqbrk {A_j} \ne \O$

From Closure of Intersection is Subset of Intersection of Closures, we then have:

$\ds \O \ne \map \cl {\bigcap_{j \mathop = 1}^n {\pr_i} \sqbrk {A_j} } \subseteq \bigcap_{j \mathop = 1}^n \map \cl { {\pr_i} \sqbrk {A_j} }$

in particular:

$\ds \bigcap_{j \mathop = 1}^n \map \cl { {\pr_i} \sqbrk {A_j} } \ne \O$

So:

$\set {\map \cl { {\pr_i} \sqbrk A} : A \in \AA}$ has the finite intersection property.

Since $X_i$ is compact by hypothesis, we have:

$\ds \bigcap_{A \mathop \in \AA} \map \cl { {\pr_i} \sqbrk A} \ne \O$

for each $i \in I$, using Compact Space satisfies Finite Intersection Axiom.

Now pick:

$\ds x_i \in \bigcap_{A \mathop \in \AA} \map \cl { {\pr_i} \sqbrk A}$

for each $i \in I$.

Let $x = \family {x_i}_{i \mathop \in I}$.

We will show that:

$\ds x \in \bigcap_{A \mathop \in \AA} \map \cl A$

We will then have:

$\ds \bigcap_{A \mathop \in \AA} \map \cl A \ne \O$

and in particular:

$\ds \bigcap_{C \mathop \in \CC} C \ne \O$

In order to show:

$x \in \map \cl A$ for each $A \in \AA$

by Condition for Point being in Closure, we can show that for each open neighborhood $V$ of $x$ in $X$, we have:

$V \cap A \ne \O$

Let $V$ be an open neighborhood of $x$ in $X$.

First, suppose that there exists $\delta_1, \ldots, \delta_n \in I$ and open neighborhoods $U_i \in \tau_{\delta_i}$ of $x_i$ such that:

$\ds V = \bigcap_{i \mathop = 1}^n \pr_i^{-1} \sqbrk {U_i}$

Recall that $x$ was constructed so that we have, for each $1 \le i \le n$:

$\ds x_{\delta_i} \in \bigcap_{A \mathop \in \AA} \map \cl {\pr_{\delta_i} \sqbrk A}$

so that, by Condition for Point being in Closure:

$U_i \cap \pr_{\delta_i} \sqbrk A \ne \O$

for each $1 \le i \le n$.

Then there exists $v \in U_i \cap \pr_{\delta_i} \sqbrk A$.

That is, there exists $u \in A$ such that $\map {\pr_{\delta_i} } u = v \in U_i$.

So we have:

$u \in A \cap \pr_{\delta_i}^{-1} \sqbrk {U_i}$

so that:

$A \cap \pr_{\delta_i}^{-1} \sqbrk {U_i} \ne \O$

for each $A \in \AA$ and $1 \le i \le n$.

So by Lemma 3, we have $\pr_{\delta_i}^{-1} \sqbrk {U_i} \in \AA$ for each $1 \le i \le n$.

Then, by Lemma 2, we have:

$\ds V = \bigcap_{i \mathop = 1}^n \pr_{\delta_i}^{-1} \sqbrk {U_i} \in \AA$

So $V$ has the finite intersection property and:

$V \cap A \ne \O$ for each $A \in \AA$.

Now take a general open neighborhood of $x$ in $X$.

Then there exists an indexing set $J$, $n_j \in \N$, $i_{j, 1}, \ldots, i_{j, n_j} \in I$ and $U_k \in \tau_{i_{j, k} }$ for each $1 \le k \le n$ such that:

$\ds V = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \pr_{i_{j, k} }^{-1} \sqbrk {U_k}$

Then, for each $A \in \AA$, we have:

$\ds V \cap A = \bigcup_{j \mathop \in J} \paren {A \cap \bigcap_{k \mathop = 1}^{n_j} \pr_{i_{j, k} }^{-1} \sqbrk {U_k} }$

By what we have already shown, we have that:

$\ds A \cap \bigcap_{k \mathop = 1}^{n_j} \pr_{i_{j, k} }^{-1} \sqbrk {U_k} \ne \O$

for each $1 \le k \le n$, and so $V \cap A \ne \O$.

So we have $x \in \map \cl A$ for each $A \in \AA$, and hence:

$\ds x \in \bigcap_{A \mathop \in \AA} \map \cl A$

Now note that since $\CC \subseteq \AA$, we have:

$\ds \bigcap_{A \mathop \in \AA} \map \cl A \subseteq \bigcap_{C \mathop \in \CC} \map \cl C$

From Set is Closed iff Equals Topological Closure, we have:

$\map \cl C = C$ for each $C \in \CC$

and hence:

$\ds \bigcap_{A \mathop \in \AA} \map \cl A \subseteq \bigcap_{C \mathop \in \CC} C$

We hence obtain:

$\ds x \in \bigcap_{C \mathop \in \CC} C$

so that:

$\ds \bigcap_{C \mathop \in \CC} C \ne \O$

and the proof is complete.

$\blacksquare$

Also see

Tychonoff's Theorem for Hausdorff Spaces, a weaker result requiring only BPI instead of the full AoC.

Tychonoff's Theorem Without Choice, a weaker result that holds in pure ZF theory.