Tychonoff's Theorem/General Case/Proof 2

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

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 Intersection Distributes over Union.

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.

$\blacksquare$

Axiom of Choice

This theorem depends on the Axiom of Choice, by way of Zorn's Lemma.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.