Topological Product of Compact Spaces

== Theorem ==
Let $T_1$ and $T_2$ be topological spaces.
Let $T_1 \times T_2$ be the product space of $T_1$ and $T_2$.
Then $T_1 \times T_2$ is compact both $T_1$ and $T_2$ are compact.
=== General Result for Finite Product ===
== Proof ==
=== Necessary Condition ===
Let $T_1 \times T_2$ be compact.
From Projection from Product Topology is Continuous, both $T_1$ and $T_2$ are the continuous images of $T_1 \times T_2$ under the projections $\pr_1$ and $\pr_2$ respectively.
So both $T_1$ and $T_2$ are compact by Continuous Image of Compact Space is Compact.
=== Sufficient Condition ===
Now let $T_1$ and $T_2$ be compact.
Let $\WW$ be an open cover for $T_1 \times T_2$.
Let $A \subseteq T_1$ be defined as $\text {good}$ (for $\WW$) if $A \times T_2$ is covered by a finite subset of $\WW$.
We want to prove that $T_1$ is $\text {good}$.
We will do this in stages, because it's complicated.
==== Step 1 ====
Suppose $A_1, A_2, \ldots, A_r$ are all $\text{good}$.
Then so is $\ds A = \bigcup_{i \mathop = 1}^r A_i$.
For any given $i=1, 2, \ldots, r$ we have that $A_i \times T_2$ is covered by a finite subset of $\WW$, say $\WW_i \subseteq \WW$.
Hence $\ds A \times T_2 = \bigcup_{i \mathop = 1}^r \paren{A_i \times T_2}$ is covered by the finite subset $\ds \bigcup_{i \mathop = 1}^r \WW_i$ of $\WW$.
==== Step 2 ====
Now we show that $T_1$ is locally $\text{good}$, in the sense that $\forall x \in T_1$, there is an open set $\map U x$ such that $x \in \map U x$ and $\map U x$ is $\text{good}$.
Consider a fixed $x \in T_1$.
For each $y \in T_2$, we have that $\tuple {x, y} \in \map W y$ for some $\map W y \in \WW$, since $\WW$ covers $T_1$ and $T_2$.
By the definition of the product topology, $\exists \map U y, \map V y$ open in $T_1$ and $T_2$ respectively such that $\tuple {x, y} \in \map U y \times \map V y \subseteq \map W y$.
The set $\set {\map V y : y \in T_2}$ is an open cover for $T_2$.
As $T_2$ is compact, there is a finite subcover of $\map V y$, say $\set {\map V {y_1}, \map V {y_2}, \ldots, \map V {y_r} }$.
Let $\map U x = \map U {y_1} \cap \map U {y_2} \cap \cdots \cap \map U {y_r}$.
For each $i = 1, 2, \ldots, r$, we have $\map U x \times \map V {y_i} \subseteq \map U {y_i} \times \map V {y_i} \subseteq \map W {y_i}$.
Hence $\ds \map U x \times T_2 = \map U x \times \bigcup_{i \mathop = 1}^r \map V {y_i} \subseteq \map W {y_i}$.
So $\map U x$ is $\text {good}$.
And we have that $x \in \map U x$ and $\map U x$ is open in $T_1$ as required.
==== Step 3 ====
Now we pass from local to global.
For each $x \in T_1$, let $\map U x$ be a $\text{good}$ open set containing $x$, by step 2.
Then $\set {\map U x: x \in T_1}$ is an open cover for $T_1$.
Because $T_1$ is compact, $\map U x$ has a finite subcover, say $\set {\map U {x_1}, \map U {x_2}, \ldots, \map U {x_m} }$.
Since each $\map U {x_i}$ is $\text {good}$, then so is $\ds \bigcup_{i \mathop = 1}^m \map U {x_i}$ by step 1.
But $\ds \bigcup_{i \mathop = 1}^m \map U {x_i} = T_1$, so $T_1$ is $\text {good}$, as required.
== Also see ==
* Tychonoff's Theorem, where this result is extended to the product space of any infinite number of topological spaces.
== Sources ==
* : $5$: Compact spaces: $5.6$: Compactness and Constructions: Proposition $5.6.2$
Category:Topology