How to decide if the metric spaces $((0,1)$, $d(x,y)=|x^2-y^2|)$ and $((-\frac{\pi}{2},\frac{\pi}{2})$, $d(x,y)=|\tan x-\tan y|)$ are complete or not?

For the first metric, I let any Cauchy sequence $(x_n)$ in $((0,1),d)$, then by definition I have for any small number $s>0$, there is $N>0$ such that for all $n,m>N$, $d(X_n,X_m) = |(X_n)^2-(X_m)^2| < s$. Then I have the sequence $((X_n)^2)$ being Cauchy in $((0,1)$, $d(x,y)=|x-y|)$. Then what?

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    $\begingroup$ This is a must read. $\endgroup$ – Julien Apr 16 '13 at 18:45
  • $\begingroup$ You have to figure out if such a Cauchy sequence would necessarily have a limit or not. $\endgroup$ – ferson2020 Apr 16 '13 at 18:54

Looks like you've plunged into epsilons, deltas and abstract reasoning from the very start. But it may be a good idea to simply check out one or two concrete sequences first, and see if you can come up with a counterexample to completeness right away.

Hint: if you have an open interval, weird things will usually happen near its endpoints.

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  • $\begingroup$ Thank you for your hint. For the first metric, can I use (Xn)=1/n which is Cauchy in the metric but doesn't converge to anypoint in (0,1) beacuse 0 is not in the metric space to show that this metric is not complete? $\endgroup$ – Elvis Apr 17 '13 at 4:06
  • $\begingroup$ @Elvis Basically yes, although I would explain why $(x_n)$ diverges in different words, without mentioning $0$ explicitly. I have this feeling that whenever possible, properties of a metric space should be proved internally, i.e. using only points of that space. $\endgroup$ – Dan Shved Apr 17 '13 at 4:16
  • $\begingroup$ Cool, Thanks a lot. should the second metric space be complete? $\endgroup$ – Elvis Apr 17 '13 at 4:30
  • $\begingroup$ Hint for the second metric space: $((-\pi/2,\pi/2),d)$ is isometric to $(\mathbb{R}, d_{\mathbb{R}})$ where $d_\mathbb{R}$ is the standard distance in $\mathbb{R}$. $\endgroup$ – Dan Shved Apr 17 '13 at 4:33
  • $\begingroup$ For any s>0, there is N>0 such that for all n, m>N we have d(Xn,Xm)=|tan(Xn)-tan(Xm)|<s. Then I conclude that (tan(Xn)) is cauchy in (R,d(x,y)=|x-y|) which is complete. Then (tan(Xn)) converges to some X0 in R. Then I use this fact and get d((Xn),arctan(X0))=|tanXn-X0|<s. So the second metric is complete. Is this the right way to do it? $\endgroup$ – Elvis Apr 17 '13 at 4:40

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