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Claim

(M,d) be a metric space with a property that every bounded sequence has a convergent subsequence $\Rightarrow$ M is complete

Proof To be complete we need to verify $\forall$ cauchysequence in $M$ converges in $M$.

first what we know is that all cauchy sequence is bounded sequence. then by the assumption, bounded sequence in M has a convergent subsequence in M.

Thus, M is complete.

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Your proof is not complete. You must still prove that if a Cauchy sequence has a convergent subsequence, then the original sequence converges.

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  • $\begingroup$ Okay. Would you verify my statement ? Claim if sub-sequence of Cauchy-sequence converges then Cauchy-sequence converges (at the same point of the convergence point of sub-sequence) Proof Let the subseqeunce $(x_k)_i$ and it converges to x. Since sub-sequence converges to x, $\forall i>N$ $d((x_k)_i,x) < e \forall e>0$. Thus in Cauchy sequence, $\forall i>N$ $d(x_k, x)<e \forall e>0$ which is equivalent to the statement that Cauchy sequence converges to x. $\endgroup$ – Daschin Jun 1 '17 at 11:48
  • $\begingroup$ No, that is not correct. Suppose that the subsequence $(x_{k_n})_{n\in\mathbb{N}}$ converges to $x$. Take $\varepsilon>0$. There is a number $p_1\in\mathbb N$ such that $m,n\geqslant p\Longrightarrow|x_m-x_n|<\frac\varepsilon2$. And there is a number $p_2\in\mathbb N$ such that $n\geqslant p_2\Longrightarrow|x_{k_n}-x|<\frac\varepsilon2$. Take $p\in\mathbb N$ such that $k_p\geqslant p_1$ and that $p\geqslant p_2(\Longrightarrow k_p\geqslant k_{p_2})$. Then, if $n\geqslant p$, $|x-x_n|\leqslant|x-x_{k_p}|+|x_{k_p}-x_n|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon$. $\endgroup$ – José Carlos Santos Jun 1 '17 at 13:30

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