Product of open set is open?

Question

Show that $A\in \mathbb R^{n+m}$ is open IFF for each $(x,y)\in A$, with $x\in \mathbb R^n, y\in\mathbb R^m$, there exist open sets $U\in\mathbb R^n, V\in \mathbb R^m$ with $x\in U, y\in V$ such that $U \times V\subset A$.

I have done that "later implies former" part, how about the other direction?

Answer 1

HINT: You’ve fitted a ball inside a box, and now you need to find a box that will fit inside a ball. Use the Pythagorean theorem to show that if $x,y\in\Bbb R^n$ and $|x_k-y_k|<\epsilon$ for $k=1,\dots,n$, then $\|x-y\|<\epsilon\sqrt{n}$.

Answer 2

Note that $|x-x'|<\epsilon$ and $|y-y'|<\epsilon'$ implies $$|(x,y)-(x',y')|\le |(x,y)-(x',y)|+|(x',y)-(x',y')|<\epsilon+\epsilon'$$ and that $|(x,y)-(x',y')|<\epsilon$ implies $|x-x'|<\epsilon$ and $|y-y'|<\epsilon$. (Why?)