Show that the set $\left\{(x_1, x_2) \in \mathbb{R}^2: x_1 x_2 = 1, x_1>0 \right\}$ is closed I would like to show that the set 

$$S = \left\{(x_1, x_2) \in \mathbb{R}^2: x_1 x_2 = 1, x_1>0 \right\}$$

is closed and the way I have been thinking about doing that is by showing that it contains its limit points. Hence, we consider a sequence $y_n =  \left(x_{n1}, x_{n2} \right) $ such that $ \lim y_n = y_0$ and try to show that $y_0 \in S$. Now, $ \lim y_n = y_0$ means that $\forall\ \epsilon>0$,  $\exists$ an $N$ such that for $n \geq N$ we have
$$\sqrt{\left(x_{n1} - x_{01}\right)^2 + \left(x_{n2} - x_{02}\right)^2 } < \epsilon$$
but since this implies that $|x_{n1}-x_{01}|<\epsilon$, it must be that $x_{01} \neq 0$ as otherwise we would have $x_{n1} < \epsilon$ and so the sequence would not be on the set. Then, as $x_{n2} = \frac{1}{x_{n1}}$, $x_{n1} >0$ and $x_{01} > 0$, in the limit we have 
$$ x_{02} = \frac{1}{x_{01}}$$
by the rules of the limits. Hence $\left(x_{01}, x_{02} \right) \in S$ and the set is closed. 
Could you please tell me if my proof is correct? Thank you. 
 A: Your argument that $x_{01} \neq 0$ is not good, since the first coordinates can indeed be approaching 0, take the set of points $(\frac{1}{n}, n)$, which all belong to your set. 
I think the easiest way to show that the set is closed, is to notice that the function 
$g: \{(x,y) \in \mathbb{R}^2: x >0\} \rightarrow \mathbb{R}: \;\;\; g(x,y) = xy$
is a continuous function, and the set in question that you want to show is closed is the inverse image of a closed set, namely the singleton $\{1\}.$
A: I can not commend so I am giving this answer. The idea of the proof is okay. An other way is to use the following result:

Let $X$ and $Y$ bet metric spaces and $f\colon X\to Y$a continuous function. Then is the graph of $f$ a closed set in $Y$.

A: Another way: the complement of $S$ is open.
In fact for all point $P\notin S$ the distance $d(P,S)=D\gt 0$ which implies that there is an open ball of radius $r\lt D$ contained out of $S$ i.e. in its complement. Hence the complement of $S$ is open.
A: Here's another approach. Let $f:(0,\infty)\times(0,\infty) \to \mathbb{R}$ be defined by $f(x_1,x_2) = x_1x_2$. This is a continuous function, so the preimage of any closed set is again closed. But $S = f^{-1}(\{1\})$ is the preimage of a closed set, so $S$ is closed.
