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

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.

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

Answer 3

The following proof is adapted from Munkres' Topology (page 167), and it works for topological spaces, too. We shall use the following claim, which is important in its own right:

Lemma (The tube lemma). Suppose $X$ and $Y$ are topological spaces, $Y$ is compact, and $x_0\in X$. If $N$ is an open set of $X\times Y$ such that $\{x_0\}\times Y\subseteq N$, then there is an open neighbourhood $W$ of $x_0$ in $X$ such that $\{x_0\}\times Y\subseteq W\times Y\subseteq N$. (The set $W\times Y$ is often called a tube about $x_0\times Y$.)

The proof of the tube lemma is given in Munkres.

Theorem. If $X$ and $Y$ are compact topological spaces, then $X\times Y$ is compact.

Proof. Let $\mathcal A$ be an open cover of $X\times Y$, and let $x_0\in X$. The "slice" $\{x_0\}\times Y$ is homeomorphic to $Y$, and so it is compact. Since $\mathcal A$ is also an open cover of $\{x_0\}\times Y$, there are sets $A_1,\dots,A_m\in\mathcal A$ such that $\{x_0\}\times Y\subseteq A_1\cup\dots\cup A_m$. Applying the tube lemma with $N=A_1\cup\dots\cup A_m$, we obtain an open neighbourhood $W$ of $x_0$ such that $\{x_0\}\times Y\subseteq W\times Y\subseteq A_1\cup\dots\cup A_m$. We have thus established the following claim:

For every $x\in X$, there is an open neighbourhood $W_x$ of $x$ such that $W_x\times Y$ can be covered by finitely many sets in $\mathcal A$.

The remainder of the proof is easy. Since $\mathcal W=\{W_x:x\in X\}$ is an open cover of $X$, by the compactness of $X$ there are finitely many sets $W_1,\dots,W_n\in\mathcal W$ such that $X=W_1\cup\dots\cup W_n$. The union of the sets $W_1\times Y,\dots,W_n\times Y$ is $X\times Y$, and as each $W_i\times Y$ can be covered by finitely many sets in $\mathcal A$, the whole of $X\times Y$ admits a finite subcover.

Answer 4

If $\mathscr{U}$ is an ultrafilter on $X\times Y$, then each projection $\pi_1\mathscr{U}$ and $\pi_2\mathscr{U}$ converges to at least one point $x\in X$ and at least one point $y\in Y$, respectively, since these spaces are compact. It is then easy to see that $\mathscr{U}$ converges to $(x,y)$; neighbourhoods of $(x,y)$ necessarily contain sets of form $V\times W$ where $V,W$ are open neighbourhoods of $x,y$ and these are contained in $\pi_{1,2}\mathscr{U}$, respectively, by definition of convergence; in other words, both $V\times Y$ and $X\times W$ are elements of $\mathscr{U}$ thus their finite intersection $V\times W$ is also in $\mathscr{U}$; therefore the original neighbourhood of $(x,y)$ is an element of $\mathscr{U}$, using the axioms of a filter.

Because we have established every ultrafilter on $X\times Y$ is convergent, it follows $X\times Y$ is compact. This generalises, almost verbatim, to a proof of the general Tychonoff theorem that any product of compact spaces is compact.