The question is the following:
Show that if a subsequence $\{x_{n_k} \}$ of a Cauchy sequence $\{ x_n\}$ is convergent, then $\{x_n\}$ is convergent.
I thought that all Cauchy sequences are convergent. At least in $\mathbb{R}$?
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But the question didn't specify $\mathbb R$. In $\mathbb Q$, for example, it's easy to find Cauchy sequences which do not converge. – lulu Mar 28 '17 at 19:41
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I am quite certain they meant $\mathbb{R}$. Hmm.. Ok. What would be an example of such a sequence? – naz Mar 28 '17 at 19:42
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Over $\mathbb Q$, take ${1,1.4,1.41,1.414,1.4142,\cdots}$ where $a_n$ is the first $n$ digits in $\sqrt 2$. Of course that converges in $\mathbb R$, but $\sqrt 2$ is not rational. – lulu Mar 28 '17 at 19:43
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The question does not presume the Cauchy criterion to be known. – Umberto P. Mar 28 '17 at 19:45
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The point here is that if the subsequence converges to $L$ then, since all the elements in the sequence approach each other, then they must approach $L$. – lulu Mar 28 '17 at 19:46
1 Answers
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One way of proving that every Cauchy sequence (in $\mathbb R$) converges is to follow these steps:
1) Prove every Cauchy sequence is bounded
2) Use Bolzano-Weierstrass to prove every Cauchy sequence thus has a convergent subsequence
3) Prove that if a Cauchy sequence has at least one convergent subsequence, then the full sequence converges to the same limit.
You're just being asked to prove the last step.
Umberto P.
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