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Let $E$ be a metric space such that every bounded subset of $E$ has a least one accumulation point. What can we say about $E$?

Intuitively I know that $E$ is complete. To prove this I must prove that for every Cauchy sequence its limit is in $E$. However I cannot combine the data above. Especially that has at least one accumulation point confuses me.

May I have some hints?

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  • $\begingroup$ Pick a Cauchy sequence $\{a_n\}$. Then as a set $\{a_n\}$ is bounded and so has a accumulation point $a$. Now try to show $a_n \to a$. $\endgroup$
    – user99914
    Feb 19 '15 at 9:23
  • $\begingroup$ Ok.. thank you... Must I invoke the definition? Shall I say that "suppose it does not converge to $a$, thus $|a_n-a|>\epsilon$ and get to something it is not true? " $\endgroup$
    – Tolaso
    Feb 19 '15 at 9:25
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    $\begingroup$ You can do that, but it should also be possible to show that $a$ is the limit, by using the definition of a limit. $\endgroup$ Feb 19 '15 at 9:29
  • $\begingroup$ @Henrik The definition , as I know it, states that: $$(\forall \epsilon >0) (\exists n_0 \in \mathbb{R})(n>n_0)\implies d(a_n, a)<\epsilon$$ How could I use that? Actually I feel that I'm missing something here... $\endgroup$
    – Tolaso
    Feb 19 '15 at 9:33
  • $\begingroup$ Another hint: Let to find a subsequence $\{a_{n_k}\}$ of $\{a_n\}$ so that $a_{n_k} \to a$. Then (by the Cauchy property) show that also $a_n \to a$. $\endgroup$
    – user99914
    Feb 19 '15 at 9:38
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As already had been remarked in a comment, to show that $E$ is indeed complet, consider a Cauchy sequence $(a_n)$. Then the set $\{a_n\ |\ n\in\mathbb N\}$ is bounded and has an accumulation point $a$ by assumption. Thus, there exists a subsequence $(a_{n_k})_k$ converging to $a$. That means, for each $\varepsilon>0$ there is some $K\in\mathbb N$ such that $d(a_{n_k},a)<\varepsilon/2$ for each $k\geq K$. Moreover, since $(a_n)$ is a Cauchy sequence we find some $N\in\mathbb N$ such that $d(a_n,a_m)<\varepsilon/2$ for all $n,m\geq N$. Then $d(a_n,a)\leq d(a_n,a_{n_k})+d(a_{n_k},a)<\varepsilon$ for $n\geq\max\{N,K\}$. Note that $n_k\geq k$ for each $k\in\mathbb N$.

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  • $\begingroup$ I guess I need a good night's sleep! It was easy.. thank you all...!! I got badly stuck at this question... and I note that it was not something diffucult at all... $\endgroup$
    – Tolaso
    Feb 19 '15 at 9:53

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