Find the limit points of $A = \{ (x,y) \in\mathbb{R}^2,|\,xy = 1 \} \subset\mathbb{R}^2$ Find the limit points of $A = \{ (x,y) \in\mathbb{R}^2\,|\, xy = 1 \} \subset \mathbb{R}^2$ I need find $(x,y) \in \mathbb{R}^2$ such that $V_{r}(x,y)\cap A\neq\emptyset$. I don't know how start. Thanks in advance.
 A: In case you are not familiar with the concept of continuity:
If you graph the function $y=1/x,$ you will note that the set of all limit points of $A$ is in fact $A$ itself (whenever a closed set $E$ is equal to the set of all its limit points, $E$ is called a perfect set).  We shall prove that $A$ is perfect.
$A$ is closed.
Let $(x,y)$ be some arbitrary point of $\mathbb R^2\setminus A.$ Then either $xy>1$ or $xy<1.$ 
Case I. $xy>1.$ Since $\mathbb N$ is not bounded from above, there is some natural $n$ such that $n>1/(xy-1)\;\iff\;xy>1+(1/n).$ Now define $\delta:=\min\left\{1,\dfrac{1}{n(1+|x|+|y|)}\right\}.$ If $(z,w)$ is any point in the open ball $B_{\delta}((x,y))$ then
$$
\begin{aligned}
|xy-zw|&=|x(w-y)+(z-x)(w-y)+y(z-x)|\\\\&\leqslant|x||w-y|+|z-x||w-y|+|y||z-x|\\\\&\leqslant|x|\|(z-x,w-y)\|+ \|(z-x,w-y)\|^2+|y| \|(z-x,w-y)\|\\\\&< |x|\delta+\delta^2+|y|\delta\\\\&=(|x|+\delta+|y|)\delta\\\\&\leqslant(|x|+1+|y|)\delta\\\\&\leqslant(|x|+1+|y|)\dfrac{1}{n(|x|+|y|+1)}\\\\&=\dfrac{1}{n}
\end{aligned}
$$
and thus $|xy-zw|<1/n.$ Since $xy-zw\leqslant|xy-zw|$ then $xy<zw+1/n$ and since $1+1/n<xy$ then $1+1/n<zw+1/n,$ which implies that $1<zw$ and it follows that $(z,w)\in\mathbb R^2\setminus A.$ Therefore no point of the open ball $B_{\delta}((x,y))$ is a point of $A.$
Case II. $xy<1.$ Proceed as in the first case (with a slight change in the argument).
We conclude that every point of $\mathbb R^2\setminus A$ is an interior point of $\mathbb R^2\setminus A,$ which means that $\mathbb R^2\setminus A$ is open and hence $A$ is closed.   
Every point of $A$ is a limit point of $A.$
Let $(x,1/x)$ be any point of $A,$ with $x>0$ (you can adapt the argument for the case $x<0$), choose $\delta$ such that $0<\delta<\sqrt{2}x$ and put $\varepsilon:=\min\left\{\dfrac{\delta}{\sqrt2},\;\dfrac{x(x-\delta/\sqrt2)}{\sqrt2}\right\}.$ It follows that if $0<|y-x|<\varepsilon$ then 
$$
\begin{aligned}
\left|\dfrac{1}{x}-\dfrac{1}{y}\right|&=\dfrac{|x-y|}{|x||y|}\\\\&<\dfrac{\varepsilon}{|x||y|}\\\\&\leqslant\dfrac{x(x-\delta/\sqrt2)\delta}{xy\sqrt2}\\\\&<\dfrac{\delta}{\sqrt2}
\end{aligned}
$$
and hence the point $(y,1/y)$ is in the open ball $B_{\delta}((x,1/x))$ and since $(x,1/x)\neq(y,1/y)$ we conclude that $(x,1/x)$ is a limit point of $A$ and since $(x,1/x)$ was arbitrary, we conclude that every point of $A$ is a limit point of $A.$
Therefore, the set of all limit points of $A$ is $A.$
A: The map $(x,y) \mapsto xy $ is continuous and $A$ which is the inverse image of the closed set $\{1\}$ is closed.
Hence $A$ contains all of its limit points. As all points in $A$ are also a limit point of $A$, finally the set of limit points of $A$ is $A$ itself.
