Cartesian product of compact sets is compact

Prove that if two sets $A$ and $B$ are compact then so is their Cartesian product $A \times B = \{(a,b): a \in A, b\in B\}$.

The hint is to use Bolzano Weiertrass theorem and an argument of sequence to proof the statement.

Solution 1:

There is indeed a topological proof using the open covers definition of compactness.

Let $A$ and $B$ be compact sets and $\{O_\lambda\}_{\lambda\in\Lambda}$ be an open cover of $A \times B$. For each $(a,b) \in A \times B$, we can choose some $\lambda = \lambda(a,b)$ such that $(a,b) \in O_{\lambda(a,b)}$. By construction, $O_{\lambda(a,b)}$ is open, hence the point $(a,b)$ is contained in some open box $X \subset O_{\lambda(a,b)}$ where $X = U_{(a,b)} \times V_{(a,b)}$, where $U_{(a,b)} \subset A$ and $V_{(a,b)} \subset B$.

Suppose we fix $a$ and vary $b$. Then for every point $(a,b)$ we find that the point is contained in an open box in the product $A \times B$, and that box is then itself the product of a subset of $A$ with a subset of $B$. Proceeding in this manner, we observe that the collection of sets $\{V_{(a,b)}\}_{b\in B}$ is an open cover of $B$. Since by assumption $B$ is compact, we can find a finite cover $\{V_{(a,b_j(a))}\}$ of $B$ that consists of finitely many open sets containing points $\{(a,b_j(a))\}$.

Now let $U_a = \bigcap_j U_{(a,b_j(a))}$. Since $U_a$ is the intersection of finitely many open sets, it is itself open. Since $A$ is compact, there are finitely many $a_i$ such that $\{U_{a_i}\}$ forms an open cover of $A$. Then it follows that the collection of sets $\{O_{(a_i,b_j(a_i))}\}$ (for all combinations of $i,j$) is a finite cover of $A \times B$, hence $A \times B$ is compact.

Solution 2:

A set $S$ is compact if from any sequence of elements in $S$ you can extract a sub-sequence with a limit in $S$.

If we are given a sequence $(u_n)$ of $A \times B$, then you can write $u_n=(a_n,b_n)$. Since $A$ is compact, you can find a sub-sequence $(a_{f(n)})$ with a limit in $A$. Then, since B is also compact, you can extract a sub-sequence $(b_{f(g(n))})$ of $(b_{f(n)})$ with a limit in B. Thus, the sub-sequence $(u_{f(g(n))})$ of $(u_n)$ has its limit in $A \times B$. This proves that $A \times B$ is compact.

I hope you can understand my explanation, I know my english is approximative...