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I need to prove that the set $C=\{ (x,y)\in \mathbb R^2|xy=1\} $ is closed in $\mathbb R^2$

I tried to prove it by proving that complement of $C$ in $\mathbb R^2$ is open ? Is it enough to show that for every $(x,y)$ not in $C$, $\exists r>0$ s.t. $B_r(x,y)\cap C=\emptyset$, where $B_r$ is a ball centered at (x,y) ?

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3 Answers 3

up vote 10 down vote accepted

Not only is it enough, one might say that is exactly what is required.


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Hint:

Let $f(x,y)=xy$ for all $(x,y)\in \Bbb R^2$. Note that $f$ is continuous.

$C=\{(x,y)\in \Bbb R^2:f(x,y)=1\}=f^{-1}[\{1\}\textbf{]}$.

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That's a proof, but it requires one to prove that $f$ is continuous, which is actually harder than proving that the complement of $C$ is open. –  Thomas Andrews Mar 11 '13 at 21:10
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@ThomasAndrews It really depends on what one can use and even if one must use the definition, is it really that much harder? Still it's good enough for an answer anyway and it's good that you pointed that out. –  Git Gud Mar 11 '13 at 21:26
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Yeah, I didn't say it was wrong, and it is a useful answer - I +1'd it. It reveals an underlying fact about the problem. But sometimes this sort of problem has as the goal something else, namely, a direct proof using the definitions of closed/open. These definitions don't even require a knowledge or definition of continuity, much less the fact that $xy$ is continuous. As such, it might be "jumping ahead" to later chapters in OPs learning. Still a useful answer. –  Thomas Andrews Mar 11 '13 at 21:34
    
@ThomasAndrews Agreed. –  Git Gud Mar 11 '13 at 21:38

Yes, what you describe is a direct proof that $C$ is closed. It can be proved in other ways too that employ special properties of $\mathbb R^2$, so just for the sake of giving a broader picture, here are some other approaches.

In $\mathbb R^2$ a set is closed if it contains all of its limit points. So, you can prove $C$ is closed by considering a sequence in $C$ and show that if it converges then the limit is in $C$.

More generally, if $f:\mathbb R^2\to \mathbb R$ is a continuous function then the set $\{(x,y)\in \mathbb R^2\mid f(x,y)=c\}$, for any constant $c\in \mathbb R$, is closed.

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Thanks, I see that now. –  user66306 Mar 11 '13 at 21:19

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