Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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 $\|x-y\|<\epsilon\sqrt{n}$.

share|cite|improve this answer

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?)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.