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[NBHM_2006_PhD Screening test_Topology]

Which of the following metric spaces are complete?

  1. $X_1=(0,1), d(x,y)=|\tan x-\tan y|$

  2. $X_2=[0,1], d(x,y)=\frac{|x-y|}{1+|x-y|}$

  3. $X_3=\mathbb{Q}, d(x,y)=1\forall x\neq y$

  4. $X_4=\mathbb{R}, d(x,y)=|e^x-e^y|$

$2$ is complete as closed subset of a complete metric space is complete and the metric is also equivalent to our usual metric.

$3$ is also complete as every Cauchy sequence is constant ultimately hence convergent.

$4$ is not complete I am sure but not able to find out a counter example, not sure about 1.thank you for help.

Martin Sleziak
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Myshkin
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1 Answers1

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For (1), consider the sequence $\left\langle\frac1{2^n}:n\in\Bbb N\right\rangle$. Is it $d$-Cauchy? Does it converge to anything in $X_1$?

For (4), what about $\langle -n:n\in\Bbb N\rangle$?

Brian M. Scott
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  • for 1 $d(\frac{1}{2^n},0)=|\tan (\frac{1}{2^n})-0|\rightarrow \pi/2$ and for 4, $d(-n,0)=|1/e^n-1|\rightarrow 1$, both are cauchy seuence with respect to the specified metric. – Myshkin Jul 12 '12 at 00:10
  • @Patience: No: $\tan\frac1{2^n}\to 0$ as $n\to\infty$. But in both cases you’re missing the point: in order to see whether the sequences are Cauchy, you have to consider $|x_n-x_m|$. – Brian M. Scott Jul 12 '12 at 00:15
  • oops sorry, both are cauchy seuence but does not convergent in the space. as $0$ is missing. – Myshkin Jul 12 '12 at 00:23
  • Almost right: they’re both Cauchy, and the first doesn’t converge in the space because $0$ is missing; the latter doesn’t converge in $\Bbb R$ because $\Bbb R$ has no left endpoint. – Brian M. Scott Jul 12 '12 at 00:26
  • I do not understand your last line :( – Myshkin Jul 12 '12 at 00:28
  • @Patience: It was a non-rigorous explanation of why $\langle -n:n\in\Bbb Z^+\rangle$ doesn’t converge in $\Bbb R$: as you move to the left in $\Bbb Z^+$, the points are getting closer and closer together with respect to $d$, but there’s no $-\infty$ for them to approach. For a rigorous proof that $\langle -n:n\in\Bbb Z^+\rangle$ doesn’t converge in $\Bbb R$, just let $x\in\Bbb R$, and show that for each $\epsilon>0$ there are infinitely many $n$ such that $d(-n,x)>\epsilon$, thereby showing that $x$ isn’t the limit of the sequence. – Brian M. Scott Jul 12 '12 at 00:35
  • @BrianM.Scott, is not the distance here $d(x,y)=|e^x-e^y|$ Since $|e^{-x}-e^{-y}|=|1/{e^x}-1/{e^y}| \rightarrow 0$ as $x,y\rightarrow \infty$ It seems like the sequence converges to 0. – 007resu Jan 11 '13 at 12:26
  • @Freddy: What you’ve shown is that the sequence is Cauchy. But the sequence itself has no limit in $\Bbb R$, which is the point of the example. – Brian M. Scott Jan 11 '13 at 13:49
  • Sorry for my confusion and thank you! – 007resu Jan 11 '13 at 14:38
  • @Freddy: No problem: it’s always best to get confusions cleared up. You’re welcome! – Brian M. Scott Jan 11 '13 at 14:53
  • @BrianM.Scott can somebody please explain 4 again? its very hard. please – SighMath Dec 28 '15 at 17:15
  • @SighMath: The sequence $\langle -n:n\in\Bbb N\rangle$ has no limit in $\Bbb R$. For any $\epsilon>0$ there is certainly an $m_\epsilon\in\Bbb N$ such that $e^{-m_\epsilon}<\epsilon$. Suppose that $k,n\ge m_\epsilon$. Without loss of generality we may assume that $k\le n$. Then $$d(-k,-n)=|e^{-k}-e^{-n}|=e^{-k}-e^{-n}<e^{-k}\le e^{-m_\epsilon}<\epsilon;,$$ and we’ve shown that the sequence is $d$-Cauchy. That is, it’s a $d$-Cauchy sequence that does not converge, so $d$ is not complete. – Brian M. Scott Dec 29 '15 at 20:26
  • @BrianM.Scott Sir, I followed your hint, $d(\frac{1}{2^n},\frac{1}{2^m})$ $=|tan(\frac{1}{2^n})-tan(\frac{1}{2^m})|$ $=|tan(\frac{1}{2^n}-\frac{1}{2^m})(1-tan(\frac{1}{2^n})tan( \frac{1}{2^m}))|$, 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 09:06
  • @ManeeshNarayanan I have tried to respond in chat. – Martin Sleziak Oct 21 '17 at 09:33
  • @MartinSleziak Ok sir. i got the point. If any euclidean space with a function $d:X×X→\mathbb R$ is given. Isn't enough to show that given $d$ is metric and the subspace is complete? using the fact that all the norms are equivalent in finite vector spaces. Have I gone in wrong direction? –  Oct 21 '17 at 10:17
  • @MartinSleziak Sir Please help me. –  Oct 21 '17 at 10:31
  • I am not really sure how exactly I should help you. Why don't you join the chatroom I suggested @ManeeshNarayanan and explain clearly what the problem is? – Martin Sleziak Oct 21 '17 at 10:32
  • @MartinSleziak I tried several times. I am not able two type in the chat. I don't know why. –  Oct 21 '17 at 10:36