Cartesian product of compact sets is compact

Question

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.

Answer 1

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 that you can understand my explanation, as I am still learning English.