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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?

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What definition of open are you using? –  Brian M. Scott Feb 9 '13 at 15:59
    
There exist a positive radius,r, such that open ball center at that point with radius r is also in the set. –  JFK Feb 9 '13 at 16:03
    
You should have $U\times V\subseteq A$, not an element of. –  Clayton Feb 9 '13 at 16:04
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2 Answers

up vote 2 down vote accepted

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 $\|x-y\|<\epsilon\sqrt{n}$.

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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?)

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