Tychonoff's Theorem/General Case/Proof 2

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:


 * $\ds \bigcap_{C \in \CC} C = \O$


 * there exists a finite subset $\SS \subseteq \CC$ such that:


 * $\ds \bigcap_{G \in \SS} G = \O$

We show the contrapositive.

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


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

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

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


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

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

We aim to show that:


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

Let:


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

Lemma 1
It will now suffice to show that:


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

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

Lemma 3
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:


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

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

Now pick:


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

for each $i \in I$.

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

We will show that:


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

We will then have:


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

and in particular:


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

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 $1 \le i \le n$:


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

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 $J$, $i_{j, 1}, \ldots, i_{j, n} \in I$ and $U_k \in \tau_{i_{j, k} }$ for each $1 \le k \le n$ such that:


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

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


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

by Intersection Distributes over Union.

By what we have already shown, we have that:


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

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:


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

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


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

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


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

and hence:


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

We hence obtain:


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

so that:


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

and we are done.