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I am stuck on the following problem that says:

Which one of the following subsets of $\Bbb R$ (with the usual metric)is NOT complete ?

  1. $[1,2] \cup [3,4]$

  2. $[0,\infty)$

  3. $[0,1]$

  4. $\{0\} \cup \{\frac 1n : n \in \Bbb N \}$

MY ATTEMPTS:

What I know is that the unit interval is a complete metric space. So, option 3 can be eliminated. We also know that a set then is compact if and only if it is closed and bounded. Since closed interval is a closed set,it must be compact. Also since we know Every compact metric space is complete, though complete spaces need not be compact,so the set given in option 1 is complete. Option 4 can be eliminated as the limit point of the set given in option 4 belongs to it. So, the answer should be option 2.

Am I going in the right direction? Is there any better ways to approach the problem. Feel free to comment. Thanks in advance for your time and regards to all.

Community
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learner
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2 Answers2

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The question is misleading: all of the sets are complete. (1), (3), and (4) are compact and therefore complete. As for (2), any Cauchy sequence in (2) is bounded and therefore a Cauchy sequence in some compact interval $[0,a]$, so it converges. (Alternatively, you can use the fact that every closed subset of a complete metric space is complete, and $\Bbb R$ is complete.)

Brian M. Scott
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  • Sir,I will be obliged if you explain a bit the line "As for (2), any Cauchy sequence in (2) is bounded.... " in your answer.Thanks and regards for the nice explanation. – learner Aug 05 '13 at 15:22
  • @learner: Every Cauchy sequence in any metric space is bounded. Let $\langle x_n:n\in\Bbb N\rangle$ be a Cauchy sequence in the metric space $\langle X,d\rangle$. There is an $m\in\Bbb N$ such that $d(x_k,x_n)<1$ for all $k,n\ge m$. Thus, $d(x_m,x_n)<1$ for all $n\ge m$. Let $$r=\max{1,d(x_m,x_0),d(x_m,x_1),\ldots,d(x_m,x_{m-1})};;$$ then $d(x_m,x_n)\le r$ for all $n\in\Bbb N$, so every term of the sequence is in the bounded set ${x\in X:d(x_m,x)\le r}$. – Brian M. Scott Aug 05 '13 at 15:34
  • thanks a lot sir. It is clear now. – learner Aug 05 '13 at 15:36
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    @learner: You’re welcome. – Brian M. Scott Aug 05 '13 at 15:38
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I claim they are all complete subspaces of $\Bbb R$. You've shown $1$ and $3$ are complete. Option $4$ is compact as a closed subspace of $[0,1]$, and you already know compact metric spaces are complete. The only doubt left is for Option $2$. However, if we take any Cauchy sequence in $[0,\infty)$, it is bounded (why?) hence has a convergent subsequence by the Bolzano-Weierstrass theorem. Now you can show that since the sequence is Cauchy, it converges to this limit point.

Clayton
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  • Thanks for the nice explanation. But I am finding it hard to understand the line " However, if we take any Cauchy sequence in [0,∞), it is bounded (why?) ". Can you explain it ? – learner Aug 05 '13 at 15:12
  • Maybe the easiest way to prove it is bounded is to suppose it isn't bounded. Then for every element of the sequence, there is something at least $1$ more than it. This shows the sequence can't be Cauchy, a contradiction. – Clayton Aug 05 '13 at 15:15