Prove/Disprove Metric Space is Complete 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...
 A: 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.
A: 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.
