If A and B are compact, then AxB are compact.

Question

I am trying to prove the above.

That is, if $A \in \mathbb{R}^m$ and $B \in \mathbb{R}^n$ are compact, then $A \times B = \{(x,y)\in \mathbb{R}^{​n+m}:x\in A,y \in B\}$ is also compact. (x is the cross product).

I have already figured out this proof using sequences and subsequences. I am trying to prove an alternative proof by considering open covers and finite subcovers. I have attempted this proof, and hence I would like some feedback on it. (If it is correct, etc).

Proof.

We need to show that for every open cover, there exists a finite subcover that admits AxB.I let the collection of open covers, {$O_i\}_{i \in I}$ cover all of $\mathbb{R}^{n+m}$.

Since $A \subseteq \mathbb{R}^m$ is compact, and $B \subseteq \mathbb{R}^n$ is compact, then there exists finite subcovers such that $A \subseteq \{O_j\}_{j \in J}$ and $J \subseteq I$. Similarly, $B \subseteq \{O_k\}_{k \in K}$ with $K \subseteq I$.

Take an arbitrary m, and n, from A and B respectively. Then there exists $\mu = (m, n)$ with $\mu \in AxB$. Since m and n are inside their respective finite subscovers, $\mu$ is inside the cross-product of the finite subcovers. As a result, we have there exists a finite subcover for an arbitrary $\mu$, and since $\mu$ was arbitrary, this holds for any m, n.

So I am not sure if I did this right. I am still a bit new to these proofs so feedback is appreciated. I hope I am not too far off if I am... I guess my concern is my argument with the "cross product of open covers".

Thank you.

(And don't be too harsh please. :D)

Answer 1

If $((x_n,y_n))$ is a sequence in $A \times B$, then show that it contains a covergent subsequence with limit in $A \times B$. Use that $(x_n)$ is a sequence in $A$ and $A$ is compact and that $(y_n)$ is a sequence in $B$ and $B$ is compact

Answer 2

Here s a proof in terms of coverings. It is combinatorially demanding.

Assume that for each $(x,y)\in K:=A\times B$ an open box neighborhood $Q(x,y):=U_y(x)\times V_x(y)$ is given.

${\bf 1\quad}$Fix an $x\in A$. Then $\bigl(V_x(y)\bigr)_{y\in B}$ is an open cover of $B$. Since $B$ is compact we may choose finitely many of these $V_x(y)$ such that $$B\subset \bigcup_{k=1}^{N_x}V_x(y_k)\ .$$ ${\bf 2\quad}$The set $$U(x):=\bigcap_{k=1}^{N_x}U_{y_k}(x)\subset A\tag{1}$$ is an intersection of finitely many open sets, hence an open neighborhood of $x$. I claim that $$U(x)\times B\subset\bigcup_{k=1}^{N_x} Q(x,y_k)\ .\tag{2}$$ Proof. Let $(t,y)$ be an arbitrary point in the "strip" $U(x)\times B$. Then there is a $k$ with $y\in V_x(y_k)$. Now from $(1)$ it follows that $t\in U(x)\subset U_{y_k}(x)$, hence $$(t,y)\in U_{y_k}(x)\times V_x(y_k)=Q(x,y_k)\ .$$

${\bf 3\quad}$The family $\bigl(U(x)\bigr)_{x\in A}$ is an open cover of $A$. Since $A$ is compact we may choose finitely many of these $U(x)$ such that $$A\subset \bigcup_{j=1}^N U(x_j)\ .$$ Using $(2)$ we now have $$K\subset \bigcup_{j=1}^N\bigl(U(x_j)\times B\bigr)\subset\bigcup_{j=1}^N \bigcup_{k=1}^{N_{x_j}} Q\bigl(x_j,y_{jk}\bigr)\ ,$$ whereby $y_{jk}$ is the $y_k$ that had been chosen for $x:=x_j$ in ${\bf 1} \> $.