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Let $X=\left\{(x,1/x) : x\in(0,\infty) \right\}$ be a metric space with $$d((x,1/x),(y,1/y)) = \sqrt{(x-y)^2+(1/x-1/y)^2}.$$

Prove whether or not the space is complete.

I can't find a counter-example, and also my intuition is that it is indeed complete. But I might be wrong... I tried showing that it cannot be true that for every element in $X$ there exists an $\epsilon'>0$ such that for all $k\in\mathbb N$ $$\sqrt{(x_k-a)^2+(1/x_n-1/a)^2}>\epsilon'.$$ I also tried using triangular inequality to set an upper bound for the distance, but this approach is probably too cumbersome, and I can't find the right path...

Yoni
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  • In a complete metric space $(X,d)$, can you characterise the complete subspaces using some other (topological) property? – Daniel Fischer Dec 19 '16 at 00:00
  • This is just one branch of a hyperbola, with the usual Euclidean distance. Do you see any points that can be approached from within the curve that are not on the curve? If there aren't any, then the space is complete. – Michael Hardy Dec 19 '16 at 00:14

2 Answers2

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Let $(x_n,\frac1{x_n})$ be a Cauchy sequence. Then

$$\forall \epsilon>0 \exists N\in\mathbb{N}: n,m\ge N\implies d((x_n,\frac1{x_n}),(x_m,\frac1{x_m}))<\epsilon.$$ In particular we have that

$$|x_n-x_m|\le d((x_n,\frac1{x_n}),(x_m,\frac1{x_m})) <\epsilon$$ and $$\left|\frac1{x_n}-\frac1{x_m}\right|\le d((x_n,\frac1{x_n}),(x_m,\frac1{x_m})) <\epsilon.$$ Thus $(x_n)$ and $(1/x_n)$ are Cauchy sequences on $[0,\infty).$ Since $[0,\infty)$ is complete the sequences $(x_n)$ and $(1/x_n)$ must be convergent. Thus there exists $x_0\in(0,\infty)$ such that $x_n\to x_0.$ (Note that $x_0$ cannot be zero becaue in such a case $(1/x_n)$ is not a Cauchy sequence.) Now, it should be easy to show that $(x_n,\frac1{x_n})\to(x_0,\frac1{x_0}).$ Thus, the space is complete.

mfl
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HINT: If you know that closed subsets of complete metric spaces are complete in the inherited metric, you can get the desired result with almost no work.

  • Show that $X$ is a closed subset of $\Bbb R^2$.
  • Verify that the metric on $X$ is the one that it inherits from the usual Euclidean metric on $\Bbb R^2$.

If you don’t know the result about closed subsets of complete metric spaces, you might try proving it: it’s pretty easy.

This way you get to avoid dealing with messy details of specific metrics.

Brian M. Scott
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