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I read somewhere that $xy=1$ is a closed set in $\Bbb{R}^2$.

A closed set is defined as the complement of an open set, or one which contains all its limit points. In metric spaces, it is defined as the complement of the union of balls $B(x,\epsilon)$, where $\epsilon>0$. For example, $(-\infty,0)$ is open in $\Bbb{R}$ as it is $\bigcup_{i=-1}^{-\infty}B(i,1)$, where $i\in \Bbb{Z}^-$.

Is $xy=1$, or any graph for that matter, closed because we can take any point $( x,y)$ outside of it and draw an open set centred on it such that $B((x,y),\epsilon)$ lies completely inside the complement of the graph, thus saying the complement is the union of open sets (hence open)?

Thanks in advance!

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4 Answers 4

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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.

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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.

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    $\begingroup$ This could probably use some clarification. For example, letting $A=\Bbb R\setminus\{0\},$ the function $f:A\to\Bbb R$ given by $f(x)=x$ is continuous, yet its graph is not closed in $\Bbb R^2.$ $\endgroup$ Jul 13, 2020 at 11:48
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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.

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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 {$(x,y) in \mathbb R^2 : xy\neq 1$}

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  • $\begingroup$ In second method, how can you say that infimum is strictly greater than zero? I think that is basically we have to prove. $\endgroup$
    – ramanujan
    Jul 3, 2018 at 18:22

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