[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.


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 $\langle -n:n\in\Bbb N\rangle$?

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Marso Jul 12 '12 at 0:10
  • $\begingroup$ @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 $|x_n-x_m|$. $\endgroup$ – Brian M. Scott Jul 12 '12 at 0:15
  • $\begingroup$ oops sorry, both are cauchy seuence but does not convergent in the space. as $0$ is missing. $\endgroup$ – Marso Jul 12 '12 at 0:23
  • $\begingroup$ 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. $\endgroup$ – Brian M. Scott Jul 12 '12 at 0:26
  • $\begingroup$ I do not understand your last line :( $\endgroup$ – Marso Jul 12 '12 at 0:28

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