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Suppose $\{p_n\}$ is a Cauchy sequence in a metric space $X$, and some subseqeunce $\{p_{n_i}\}$ converges to a point $p\in X$. Prove that the full sequence $\{p_n\}$ converges to $p$.

Proof: $\{p_n\}$ is a Cauchy sequence then $\forall \varepsilon >0$ $\exists N$ such that $n,m\geqslant N$ implies $d(p_n,p_m)< \dfrac{\varepsilon}{3}.$ Then $\forall m, n_i\geqslant N$ implies $d(p_m,p_{n_i})< \dfrac{\varepsilon}{3}.$ By triangle inequality $$d(p_n,p_{n_i})\leqslant d(p_n,p_m)+d(p_m,p_{n_i})<\dfrac{2\varepsilon}{3}.$$ For this $\varepsilon$ exists $N'$ such that $n_i\geqslant N'$ implies $d(p_{n_i},p)<\dfrac{\varepsilon}{3}.$

Then $\forall n,n_i>\max \{N,N'\}$ we have $$d(p_n,p)\leqslant d(p_n,p_{n_i})+d(p_{n_i},p)<\dfrac{2\varepsilon}{3}+\dfrac{\varepsilon}{3}=\varepsilon.$$ Hence $\lim\limits_{n\to\infty}p_n=p$

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    $\begingroup$ Excellent approach,Cheers! $\endgroup$ – Arpit Kansal Sep 4 '15 at 8:17
  • $\begingroup$ OK! This is the proof. $\endgroup$ – Alex Sep 4 '15 at 8:42
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Another Approach: let $(X,d)$ be a metric space and let $\{p_n\}$ be a cauchy sequence with a convergent subsequence, say convergent to $L \in X$. Now consider the completion $\overline{X}$ of $X$: by definition every Cauchy sequence in $\overline{X}$ converges, so our sequence $\{p_n\}$ converges in $\overline{X}$, say to $M$. But then every subsequence also converges to $M$ and thus $M = L$. It follows that the original Cauchy sequence is convergent to $L$!

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  • $\begingroup$ @Arpit_Kansal, Nice solution with using completion of metric space! $\endgroup$ – ZFR Sep 4 '15 at 8:33
  • $\begingroup$ @Arpit_Kansal, Let me ask you question from my solution. We got that for any $\varepsilon>0$ $\exists M$ s.t. $n,n_i\geqslant M$ implies $d(p_n,p)<\varepsilon$. Right? But $\endgroup$ – ZFR Sep 4 '15 at 8:38
  • $\begingroup$ But we must get that "$\forall \varepsilon >0 \exists M$ s.t $n\geqslant M$ implies $d(p_n,p)<\varepsilon$" $\endgroup$ – ZFR Sep 4 '15 at 8:44

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