Why is the graph of $f(x) = 1/x$ when $x \in (0,\infty)$, $0$ else, closed? Consider the function 
$$f(x) = \begin{cases}\frac{1}x&&\text{if }x\in(0,\infty)\\0&&\text{else}\end{cases}$$
I know that this is the typical example of a discontinuous function whose graph is closed, but I can't see why it is closed. I understand that I'm probably just being dumb here, but if $x_n\to0^+$, we have $f(x_n) \to \infty$, so the graph doesn't seem closed to me...
I've also tried to reason this by proving that the complement of the graph is open, so that for any $(x,y) \not\in G(f)$, where $G(f)$ is the graph of $f$, we can find an open ball around $(x,y) \not\in G(f)$. The reasoning behind this approach makes sense to me, but I'm still really concerned about my conceptual misunderstanding behind the limit point definition proof. 
 A: More simply, consider that $[1,\infty)$ is closed in $\mathbb R$, despite the fact that the sequence $1,\,2,\,3,\,4,\,5,\ldots$ remains entirely within it, yet has no limit within $[1,\infty)$. The issue is that that sequence has no limit in $\mathbb R$ either, so we don't end up with a boundary point of $[1,\infty)$ not in the set - we just have a divergent sequence.
The same is true when we consider that $(x,f(x))$ traverses the curve as $x\rightarrow 0^+$, yet doesn't converge within the curve. It doesn't have a limit in $\mathbb R^2$, so we have no issue. That is, when we say $f(x)\rightarrow \infty$, this is a non-issue because we don't literally mean that $\infty$ is a point in the space and that $f(x)$ goes to it - we just mean that the limit diverges. (Of course, in the extended reals, or any compactification of $\mathbb R$, the graph of $f$ is not closed)
A: If we consider its complement into first quadrant than it is ${xy>1}$ union ${xy<1}$ which is easy to seen to be open.and now if also include other quadrants it still remains open.Hope this will help
