How is $xy=1$ closed in $\Bbb{R}^2$?

Solution 1:

More simply, note that $(x,y)\mapsto xy$ is a continuous function $\Bbb R^2\to\Bbb R,$ and that $\bigl\{(x,y)\in\Bbb R^2:xy=1\bigr\}$ is the preimage of the closed set $\{1\},$ so is closed by continuity.

Solution 2:

In this case the graph is closed; and the graph of $y=f(x)$ will always be closed whenever $f$ is a continuous function.

However, for arbitrary $f : \mathbb{R} \to \mathbb{R}$, the graph might not be a closed subset of $\mathbb{R}^2$. Consider the indicator function $f$ of the set $\{0\}$, defined by $$f(x) = \begin{cases} 0 & \text{if}\ x \ne 0 \\ 1 & \text{if}\ x = 0 \end{cases}$$ Its graph is not a closed subset of $\mathbb{R}^2$: the point $(0,0)$ does not lie on the graph, yet all open balls about $(0,0)$ intersect the graph.

Solution 3:

Yes, exactly. For any point $(a,b)$ not on the hyperbola, there is an open disk with centre $(a,b)$ such that none of the points of the disk is on the hyperbola. So the complement of the set of points on the hyperbola is open. It follows that the set of points on the hyperbola is closed.

Solution 4:

Consider a point $(a,b)$ in the complement, so that $ab\neq 1$. This point has non-zero distance $d$ to $(x,y):xy=1 $ . Then the ball $B(a,b); d/2$ is contained in the complement.

Assume, wlg. that $ab>1$ . This means that $ab-1>0$ By density of Rationals in $\mathbb R$ , there is a rational q with $ab-1>q>0 $. Then the ball $B(a,b); q/2$ is contained in {$(x,y) in \mathbb R^2 : xy\neq 1$}. Another way: $ab>1$ means that :

$inf$ {${d((a,b),(x,y): xy>1}$}$=e>0$ , so that $B(a,b), e/2$ is strictly-contained in {$(x,y) in \mathbb R^2 : xy\neq 1$}