Euclidean spaces In a Euclidean space is the set $\{(x,y): xy=1\}$ a closed set? I am not sure, because  looking at the sequences where $x = n$ and $y = 1/n$ for all positive integers, it would appear the set is unbounded and thus does not contain its limit points, and therefore the set is not closed.
 A: The mapping $\phi \, : \, (x,y) \in \mathbb{R}^{2} \, \longmapsto \, xy$ is continuous. Let $\mathcal{P} = \lbrace (x,y) \in \mathbb{R}^{2}, \, xy=1 \rbrace$. Then, $\mathcal{P} = \phi^{-1}(\lbrace 1 \rbrace)$ where $\lbrace 1 \rbrace$ is closed in $\mathbb{R}$. Hence, $\mathcal{P}$ is closed in $\mathbb{R}^{2}$.
A: We need to show that the graph of $y=1/x$ contains all its limit points. Accordingly, suppose that $(a,b)$ is a limit point. Fix a sequence $(x_n,1/x_n)$ converging to it (this is a general fact in metric spaces: A limit point of a set $C$ is the limit of a sequence of points in $C$; in this case, with $C$ the graph of $y=1/x$, the points $(x_n,y_n)$ satisfy that $y_n=1/x_n$ for all $n$). 
First, note that $x_n\to a$ and $1/x_n\to b$. In particular, the sequences are bounded (since they converge). It follows that $a\ne 0$. Otherwise, the $x_n$ would eventually be arbitrarily small, and as a result the sequence $1/x_n$ would be unbounded. We then have that $b=1/a$, because
 $$ \left|\frac1{x_n}-\frac1a\right|=\frac{|x_n-a|}{|a||x_n|},$$ and $1/|x_n|$ is bounded. 
What we have shown is that any limit point of the graph has the form $(a,1/a)$ for some nonzero $a$, and therefore is also in the graph. Which is to say, the graph is closed.
This is the typical way we prove a closed graph theorem. The usual difficult point of ensuring convergence of a sequence is not an issue here, and the problem of finding useful bounds that can be applied is trivial in this case. Note in particular the argument does not require the notion of continuity.
A: Another method than that given by jibounet is to look at the map
$$f:\Bbb R-\{0\}\to \Bbb R\\
\qquad\quad x\mapsto \frac1x$$
It is continuous, and since its codomain $\Bbb R$ is Hausdorff, the graph $G(f)=\{(x,f(x))\mid x\in\Bbb R-\{0\}\}=\{(x,y):xy=1\}$ of $f$ is closed in $\Bbb R-\{0\}\times\Bbb R$, so it has no limit point in $\Bbb R-\{0\}\times\Bbb R$ outside of itself. And a point $(0,y)$ has the neighborhood
$$\left(\frac{-1}{y+1},\frac1{y+1}\right)\times (y-1,y+1)$$ if $y\ge0$, and 
$$\left(\frac1{y-1},\frac{-1}{y-1}\right)\times (y-1,y+1)$$ if $y<0$,
which does not intersect $G(f)$.
This shows that $G(f)$ has no limit points outside itself, so it is closed.
A: if any nbhd of a point in the given set contains infinite points of that set , then we say that point as its limit point. Thus tacing the curve taking an arbitrary point whose nbhd contains infinite points of that curve. So each point of this curve is its limit point and the curve contains all its limit points. Therefore the curve is closed and thus the set $\{ {(x,y): xy = 1 }\} $  is closed
