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Consider the well known example of a discontinuous function with closed graph: $f(x)=\frac{1}{x}$ for $x>0$ and $0$ otherwise.

I'm trying to convince myself that its graph is closed. I haven't been able to come up with a proof but I've been trying to show that $G_f$ equals the union of two closed graphs, namely $\{(x, \frac{1}{x}) : x>0\}\cup \{(x, 0) : x\leq 0\}$. Is this correct? How can I prove this?

In particular, how can I show that $\{(x, \frac{1}{x}) : x>0\}$ is a closed set, i.e., that it equals its closure?

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    That description of $G_f$ is certainly correct; but I suppose you would still have to prove that both of those sets are closed. – 2017-02-17
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    Exactly, that's where I need help. I can't get the proof. – 2017-02-17
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    What is the definition of "closed" you are using? – 2017-02-17
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    Basically containing its limit points / equaling its closure – 2017-02-17

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You are right:

$G_f=\{(x, \frac{1}{x}) : x>0\}\cup \{(x, 0) : x\leq 0\}$

Now try to proof that the sets $G_1=\{(x, \frac{1}{x}) : x>0\}$ and $G_2=\{(x, 0) : x\leq 0\}$ are closed.

Then $G_f$ is closed.

Let $i=1,2$ and let $(a_n)$ be a convergent sequence in $G_i$ with limit $a$.

Its your turn to show that $a \in G_i$

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    $G_2$ is easy, right? Since it has no limit points outside itself. And with $G_1$, is there a way I can prove closed-ness without using analytic ideas like convergent sequences? That is, just from the characterizations using limit points/closure? – 2017-02-17