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Let $X=(0,1)$ be a metric space with metric defined as $d(x,y):= \vert \tan x-\tan y \vert$ for all $x,y \in (0,1)$. The question is whether $X$ is complete with respect to the given metric or not. The answer is given to be "Yes", but I think the sequence $x_{n}=\frac{1}{n}$ serves as a counterexample. Can anyone tell me whether I am correct or not?

Martin Sleziak
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user92156
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  • If the question is really as you've written it, the you're right. – Ben Grossmann Jan 15 '14 at 13:13
  • Yes you are correct. The map $\tan$ provides an isometry from your space to the interval $(0, tan(1) )$, which is obviously not complete. – user68061 Jan 15 '14 at 13:13
  • Okay, thanks. I thought I might be wrong since $x$ going to $\tan x$ is a homeomorphism between $(0,1)$ and $(0,\infty)$ and it is not necessary that homeomorphism takes Cauchy sequences to Cauchy sequences. – user92156 Jan 15 '14 at 13:21
  • See http://math.stackexchange.com/questions/169706/which-of-the-following-metric-spaces-are-complete or http://math.stackexchange.com/questions/220813/is-this-metric-space-incomplete (I think it is basically the same question, so it can be closed as a duplicate.) – Martin Sleziak Jan 15 '14 at 14:13
  • @MartinSleziak I followed the hint provided by Brian Scott sir, he is not using the mathstackexchange for more than 6 months. $d(\frac{1}{2^n},\frac{1}{2^m})$ $=|tan(\frac{1}{2^n})-tan(\frac{1}{2^m})|$ $=|tan(\frac{1}{2^n}-\frac{1}{2^m})(1-tan(\frac{1}{2^n})tan( \frac{1}{2^m}))|$, which converges to $0$ as $m,n$ tends to $\infty$. hence it is cauchy Am I right? How to prove that it does not converges to a point in $X_1$? Please help me. –  Oct 21 '17 at 09:13
  • @ManeeshNarayanan I have tried to respond in chat. – Martin Sleziak Oct 21 '17 at 09:34
  • @MartinSleziak sorry sir. I was unable to connect at that time. can you please explain? –  Oct 21 '17 at 09:52
  • @ManeeshNarayanan The explanation is given in the chat transcript which is available from the link in my previous comment. Just in case, here is the link once again: https://chat.stackexchange.com/transcript/message/40676107#40676107 If there is some problem with the explanation, why not join that chat room. (Or, as an alternative, you can post new question on the main site.) – Martin Sleziak Oct 21 '17 at 09:57

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