Show that the set $\{(x,y): xy=1, x>0\}$ is closed

Question

I am new to analysis and I'm afraid my approach is not detailed enough.

Here is my attempt:

"To show $S=\{(x,y)\in E^2: xy=1, x>0\}$ , $S\subset E^2$ is a closed set, we can show that $S^c$ is an open set.

Let $p\in S^c$. Then there exists an open ball of center $p$ and radius $\epsilon$, call it $B_\epsilon(p)$.

Let $q\in B_\epsilon(p)$. To ensure that $q\in S^c$, choose $\epsilon =\min\{d(p,z), z\in S\}$.

Is this enough?

Answer 1

This is not enough, because: (a) You don’t know that the minimum exists (though surely there is the infimum), and (b) Even if you replace the minimum with infimum, how are you sure that this infimum is nonzero?

The correct proof would take the point $(x,y)$ such that $x\gt 0, xy\ne 1$, then if you set $\epsilon=|1-xy|$, due to continuity of $(x,y)\to xy$ there is an open ball $B$ containing $(x,y)$ such that, for $(z,t)\in B$, $|zt-xy|\lt\epsilon$... but then $|1-zt|\ge|1-xy|-|zt-xy|\gt\epsilon-\epsilon=0$, i.e. $zt\ne 1$ on the whole $B$.

A more advanced way to rephrase the same proof here is to note that, in the continuous map $f:(x,y)\to xy$, we have the inverse image $f^{-1}(\{1\})$ of the closed set $\{1\}$, which is closed. Consequently, $S$ is closed as an intersection of that closed set with the closed half-plane $x\ge 0$.

Answer 2

Let $\vec{x}_n$ be a sequence of points in $S$ converging to some $\vec{x}=(x_1,x_2)$ in the ambient space $\mathbb{R}^2$. Then, we may rewrite $$ \vec{x}_n=\left(x_n,\frac{1}{x_n}\right) $$ If $\vec{x}_n$ is to converge to a this finite vector $\vec{x}$, we have the following condition $$ \lim_{n\to \infty}\vec{x}_n=\vec{x}\iff |x_1-x_N|+|1/x_N-x_2|<\epsilon $$ for given epsilon and corresponding $N$. Note immediately we can see that $x_n$ canot be converging to $0$, or the second term blows up. Note also that this means that $$ x_n\to x_1 $$ just ignoring the second positive term. Then $$ \frac{1}{x_n}\to \frac{1}{x_1} $$ since $x$ is nonzero and the function $f(x)=1/x$ is continuous at nonzero $x$. Thus $$ x_2=1/x_1\implies \vec{x}=(x_1,1/x_1)\in S $$ and $S$ contains its limit points.

Answer 3

Let $f:\mathbb{R}\times\mathbb{R}\to \mathbb{R}$ be the function that sends $(x,y)$ to $xy$. It is a continuous function hence the pre-image $A:=f^{-1}(\{1\})$ of the closed set $\{1\}$ is a closed subset of $\mathbb{R}\times\mathbb{R}$.

$B:= [0,+\infty)\times [0,+\infty)$ is also a closed subset of $\mathbb{R}\times\mathbb{R}$.

The intersection of two closed sets is closed so $S=A\cap B$ is closed.

Using this kind of techniques is often useful or sets defined by equations such as in your case.