How to show that $\{(x, \frac{1}{x}) : x>0\}$ is closed in $\mathbb{R}^2$ by showing equality to closure 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?
 A: 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 $\mathbb R^2$. 
