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Given $A = \{(x,y) \in \mathbb{R^2}: y = \frac{1}{x}, x > 0\}$. Show (by considering convergent sequences or otherwise) $A$ is closed in $\mathbb{R^2}$.

Anyone able to give me some advice on how to approach this problem. I know that if every convergent sequence $(x_n, y_n) \in A$ converges in $A$ then $A$ is closed. But I don't know what process I should use to show this and I don't know how to deal with sequences that use two terms, ie $x_n, y_n$ instead of just one term.

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up vote 2 down vote accepted

HINT: Suppose that $\langle p_n:n\in\Bbb N\rangle$ is a sequence in $A$ that converges in $\Bbb R^2$. Then there are sequences $\langle x_n:n\in\Bbb N\rangle$ and $\langle y_n:n\in\Bbb N\rangle$ in $\Bbb R$ such that $p_n=\langle x_n,y_n\rangle$ for each $n\in\Bbb N$. Let $p=\langle x,y\rangle\in\Bbb R^2$ be the limit of the sequence $\langle p_n:n\in\Bbb N\rangle$; you want to show that $p\in A$.

For each $n\in\Bbb N$, $p_n=\langle x_n,y_n\rangle\in A$, so we know that $x_n>0$ and $y_n=\dfrac1{x_n}$, so that $$p_n=\left\langle x_n,\frac1{x_n}\right\rangle\;.$$

  • Use the fact that $\langle p_n:n\in\Bbb N\rangle\to\langle x,y\rangle$ to show that $\langle x_n:n\in\Bbb N\rangle\to x$ and $\langle y_n:n\in\Bbb N\rangle\to y$.

  • Use the continuity of the function $f(x)=\frac1x$ to conclude that $y=\frac1x$ and hence that $\langle x,y\rangle\in A$.


An alternative approach is to show that $\Bbb R^2\setminus A$ is open: show that if $p=\langle x,y\rangle\notin A$, then there is an $r>0$ such that $B(p,\epsilon)\cap A=\varnothing$, where $B(p,\epsilon)$ is the open ball of radius $\epsilon$ centred at $p$. Note that any $\epsilon$ less than or equal to the distance from $p$ to $A$ will work.

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As $p = (x,y)$ is the limit of $p_n = (x_n,y_n)$ can we just say it is clearly true that $x_n \to x$ and $y_n \to y$ as $p$ is composed of those two sequences and $p_n \to (x,y)$? I'm not sure why we need to use the continuity of $f(x)$ to show that $y=\frac{1}{x}$, aren't we given that fact in the question itself? The open ball approach also seems like a good idea but how do you take the distance from a point to a set? –  sonicboom Nov 7 '12 at 0:34
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@sonicboom: ‘Clearly true’? No, it has to be proved. One way to do so is to let $\pi_1:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto x$, the projection to the $x$-axis; prove that $\pi_1$ is continuous, and note that $\langle x_n:n\in\Bbb N\rangle$ is just $\langle\pi_1(p_n):n\in\Bbb N\rangle$, while $x=\pi_1(p)$. Then do something similar for the other coordinate. You’re certainly not given that $y=\frac1x$: that is in fact precisely what you need to prove! This is a specific $\langle x,y\rangle$, the limit of the $p_n$’s, and you want to prove that it’s in $A$, which means proving that $xy=1$. –  Brian M. Scott Nov 7 '12 at 0:40
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@sonicboom: The sequence approach is probably a bit easier than computing the distance from an arbitrary point to that hyperbola. –  Brian M. Scott Nov 7 '12 at 0:41
    
I need to think this one over in depth. Cheers for the pointers. –  sonicboom Nov 7 '12 at 1:09
    
I think I have a better understanding of this one now...I get what is happening with using the continuity of $f(x) = \frac{1}{x}$ to show $y = \frac{1}{x}$...ie by continuity of $f$, $(x_n) \to x$ implies $f(x_n) \to f(x)$, which is $y_n \to \frac{1}{x}$ and we already have $(y_n) \to y$ so $y=\frac{1}{x}$. I am 90% there on why we need to show $(x_n) \to x$ and $(y_n) \to y$ as opposed to it being clearly true, need to think that over a bit more. –  sonicboom Nov 7 '12 at 11:50
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Note that $A$ consists of all $(x,y)$ such that $xy-1=0$, so it is the preimage of $0$ under the continuous function $f(x,y)=xy-1$.

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...and since $\{0\}$ is closed in $\mathbb{R}$, the preimage of $\{0\}$ must be closed in $\mathbb{R}^2.$ –  Fly by Night Nov 6 '12 at 21:15
    
...actually, to get $A$ you also have to intersect the preimage with the (closed) set $\{(x,y)\mid x\ge 0\}$. –  nonpop Nov 6 '12 at 21:22
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I would start by simplifying the problem. Your graph is homeomorphic to the positive real numbers. Take the map $f : \mathbb{R}^2 \to \mathbb{R}^2$ given by $f : (x,y) \mapsto (x,y-\frac{1}{x}).$ This map is in fact a diffeomorphism for all $x \neq 0.$ (A diffeomorphism is a differentiable homeomorphism.)

This map was chosen because $ f : (t,\frac{1}{t}) \mapsto (t,\frac{1}{t}-\frac{1}{t}) = (t,0).$ Thus, your graph is sent to the positive real axis. Now you need only show that this is closed in $\mathbb{R}^2.$ Well, you should easily be able to show (using open discs) that the complement of this set is open, and so it itself if closed.

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