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I know that a metric space is complete if every Cauchy sequence converges to an element of the space (we can call it $X$).
To prove that the metric space is complete, let $\left(x_{n}\right)$ be any Cauchy sequence.
I know thanks to the Th. of Bolzano-Weierstrass that every bounded sequence of real numbers has a convergent subsequence, so $\left(x_{n}\right)$ contains a subsequence converging to some point $a \in X$.
But then what proposition make the whole sequence $\left(x_{n}\right)$ converges to $a$ to show that $X$ is complete?

Curtosi
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  • If a Cauchy sequence has a convergent subsequence, it is convergent. Do you know how to prove this? – Anu Jul 29 '21 at 17:04
  • No, I don't know how to prove it. What is the name of this lemma/proposition? – Curtosi Jul 29 '21 at 17:17
  • It is a standard result, and a good exercise to try and prove it. I found this thread which may help in case you can stuck: https://math.stackexchange.com/questions/354965/cauchy-sequence-is-convergent-iff-it-has-a-convergent-subsequence – Anu Jul 29 '21 at 17:20
  • Thank you very much, I think I am remembering the remarkable property of the inverse, namely that if the elements of the subsequence converge then the sequence also converges. I will try to prove it as you suggested! – Curtosi Jul 29 '21 at 17:22

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