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 $\lim\limits_{n\to\infty}p_n=p$

  • 1
    $\begingroup$ 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

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 $L$!

  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.