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I want to show that the set $A:= \{(x, \frac{1}{x}) : x>0\}$ is closed in $\mathbb{R}^2$ in the usual topology. My strategy is to determine its limit points and show that it contains them.

I'm having a lot of trouble formalizing the argument. If we take an element $(a,b)\notin A$, then I want to show that there is some basic neighborhood $(p,q)\times (s,t)$ of it that doesn't intersect $A$. In this situation, how do I determine what form such a neighborhood should be in?

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    ? But the points $(a,1/a)$ are in $A.$ I think you want to start with such a point and show it can be approached by a sequence of points in $R^2,$ perhaps you need points other than $(a,1/a)$ in that sequence. – 2017-02-17
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    There is not much use in looking at $(a, \frac{1}{a}) \notin A$. This implies $a<0$. – 2017-02-17
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    One method you might try is by contradiction: Suppose $(x,y)$ is a limit point of $A$ and $(x,y) \notin A$. – 2017-02-17

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Your strategy will work, but it will not be the easier way to proceed. If you want to pursue in this direction, you should start by drawing the set $A$.

Let me propose you another strategy.

The function $f(x)=\frac 1x$ for $x>0$ is continuous.

So if you take a set of points $(x_n,\frac 1{x_n})\in A$ such that it converges in $\mathbb R^2$, then you get

$$(x_n,f(x_n)) \text { converges}$$

(because $f$ is continuous and $(x_n)$ converges) and it converges to $(x,f(x))\in A$ since, once again, $f$ is continuous.

(More details: we know that $x>0$ because from the convergence of $(x_n,1/x_n)$, we get both $\lim x_n<\infty$ and $\lim 1/x_n<\infty$, so $0<\lim x_n<\infty$)

So $A$ is closed.

Edit.

Since you asked in the comments, the sequential definition of continuity that I am using is:

A function $f$ is continuous at $a$ if, and only if, for all $(x_n)$ such that $\lim x_n=a$, we have $\lim f(x_n)=f(a)$.

Remark: it is a very useful definition when it comes to topological issues.

More generally.

We can show that if $\varphi \colon \mathbb R^2\to \mathbb R$ is a continuous function, then the set

$$\mathrm{Graph}(\varphi)=\{(x,\varphi(x)),\ x\in\mathbb R\}$$

is closed in $\mathbb R^2$.

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    Ah, I see. Could you remind me of the sequential definition of continuity you're using? How do you conclude that $(x_n, f(x_n))$ converges? – 2017-02-17
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    Thank you, but how do we know that $\lim (x_n)>0$? Can't we just conclude $\geq 0$? – 2017-02-17