Topological Product of Compact Spaces

From ProofWiki
Jump to navigation Jump to search

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 if and only if both $T_1$ and $T_2$ are compact.


General Result for Finite Product

Let $T_1, T_2, \ldots, T_n$ be topological spaces.

Let $\displaystyle \prod_{i \mathop = 1}^n T_i$ be the topological product of $T_1, T_2, \ldots, T_n$.


Then $\displaystyle \prod_{i \mathop = 1}^n T_i$ is compact if and only if all of $T_1,T_2, \ldots,T_n$ are compact.


Proof

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 $\operatorname{pr}_1$ and $\operatorname{pr}_2$ respectively.

So both $T_1$ and $T_2$ are compact by Continuous Image of 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 $\text{good}$ (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 $\text{good}$.

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 $\displaystyle A = \bigcup_{i \mathop = 1}^r A_i$.

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

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


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 $U \left({x}\right)$ such that $x \in U \left({x}\right)$ and $U \left({x}\right)$ is $\text{good}$.


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

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

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

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

$\blacksquare$


Also see


Sources