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.

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.