TITLE: Showing a set is closed in $\mathbb R^2$
QUESTION [5 upvotes]: 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) ?

REPLY [14 votes]: Not only is it enough, one might say that is exactly what is required.

REPLY [9 votes]: Hint:
Let $f(x,y)=xy$ for all $(x,y)\in \Bbb R^2$. Note that $f$ is continuous.
$C=\{(x,y)\in \Bbb R^2:f(x,y)=1\}=f^{-1}[\{1\}\textbf{]}$.

REPLY [3 votes]: 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.