Topological Product of Compact Spaces

Theorem
Let $$T_1$$ and $$T_2$$ be topological spaces.

Let $$T_1 \times T_2$$ be the topological product of $$T_1$$ and $$T_2$$.

Then $$T_1 \times T_2$$ is compact iff both $$T_1$$ and $$T_2$$ are compact.

General Result
It follows that if $$T_1, T_2, \ldots, T_n$$ are topological spaces, then $$\prod_{i=1}^n T_i$$ is compact if all of $$T_1, T_2, \ldots, T_n$$ are.

Proof of Main Result

 * 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 a Compact Space is Compact.


 * Now let $$T_1$$ and $$T_2$$ be compact.

Let $$\mathcal{W}$$ be an open cover for $$T_1 \times T_2$$.

Let $$A \subseteq T_1$$ be defined as $$\mbox{good}$$ (for $$\mathcal{W}$$) if $$A \times T_2$$ is covered by a finite subset of $$\mathcal{W}$$.

We want to prove that $$T_1$$ is $$\mbox{good}$$.

We will do this in stages, because it's complicated.

Step 1
Suppose $$A_1, A_2, \ldots, A_r$$ are all $$\mbox{good}$$.

Then so is $$A = \bigcup_{i=1}^r A_i$$.

For any given $$i=1, 2, \ldots, r$$ we have that $$A_i \times T_2$$ is covered by a finite subset of $$\mathcal{W}$$, say $$\mathcal{W}_i \subseteq \mathcal{W}$$.

Hence $$A \times T_2 = \bigcup_{i=1}^r \left({A_i \times T_2}\right)$$ is covered by the finite subset $$\bigcup_{i=1}^r \mathcal{W}_i$$ of $$\mathcal{W}$$.

Step 2
Now we show that $$T_1$$ is locally $$\mbox{good}$$, in the sense that $$\forall x \in T_1$$, there is an open set $$U \left({x}\right)$$ such that $$x \in U \left({x}\right)$$ and $$U \left({x}\right)$$ is $$\mbox{good}$$.

Consider a fixed $$x \in T_1$$.

For each $$y \in T_2$$, we have that $$\left({x, y}\right) \in W \left({y}\right)$$ for some $$W \left({y}\right) \in \mathcal{W}$$, since $$\mathcal{W}$$ covers $$T_1$$ and $$T_2$$.

By the definition of the product topology, $$\exists U \left({y}\right), V \left({y}\right)$$ open in $$T_1$$ and $$T_2$$ respectively such that $$\left({x, y}\right) \in U \left({y}\right) \times V \left({y}\right) \subseteq W \left({y}\right)$$.

The set $$\left\{{V \left({y}\right): y \in T_2}\right\}$$ is an open cover for $$T_2$$.

As $$T_2$$ is compact, there is a finite subcover of $$V \left({y}\right)$$, say $$\left\{{V \left({y_1}\right), V \left({y_2}\right), \ldots, V \left({y_r}\right)}\right\}$$.

Let $$U \left({x}\right) = U \left({y_1}\right) \cap U \left({y_2}\right) \cap \cdots \cap U \left({y_r}\right)$$.

For each $$i = 1, 2, \ldots, r$$, we have $$U \left({x}\right) \times V \left({y_i}\right) \subseteq U \left({y_i}\right) \times V \left({y_i}\right) \subseteq W \left({y_i}\right)$$.

Hence $$U \left({x}\right) \times T_2 = U \left({x}\right) \times \bigcup_{i=1}^r V \left({y_i}\right) \subseteq W \left({y_i}\right)$$.

So $$U \left({x}\right)$$ is $$\mbox{good}$$.

And we have that $$x \in U \left({x}\right)$$ and $$U \left({x}\right)$$ is open in $$T_1$$ as required.

Step 3
Now we pass from local to global.

For each $$x \in T_1$$, let $$U \left({x}\right)$$ be a $$\mbox{good}$$ open set containing $$x$$, by step 2.

Then $$\left\{{U \left({x}\right): x \in T_1}\right\}$$ is an open cover for $$T_1$$.

Because $$T_1$$ is compact, $$U \left({x}\right)$$ has a finite subcover, say $$\left\{{U \left({x_1}\right), U \left({x_2}\right), \ldots, U \left({x_m}\right)}\right\}$$.

Since each $$U \left({x_i}\right)$$ is $$\mbox{good}$$, then so is $$\bigcup_{i=1}^m U \left({x_i}\right)$$ by step 1.

But $$\bigcup_{i=1}^m U \left({x_i}\right) = T_1$$, so $$T_1$$ is $$\mbox{good}$$, as required.

Proof of General Result
This can be shown by induction.

Note
This result can also be extended to the topological product of any infinite number of topological spaces, in which form it is known as Tychonoff's Theorem.