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Problem: Let $(X,d)$ be a metric space and let $\{x_n\}_{n=1}^\infty$ a Cauchy sequence in $X$. Prove that if $\{x_n\}_{n=1}^\infty$ admits a convergent subsequence, then $\{x_n\}_{n=1}^\infty$ converges.

I am new to analysis, and am very much confused of the above problem because of the if part of the problem statement. Here is my attempt for the solution:

Since $\{x_n\}_{n=1}^\infty$ is Cauchy, for every $\epsilon > 0$, there exists a natural number $N$ such that for every $n,m \geq N$, $d(x_n, x_m) < \epsilon$. Since $\{x_n\}_{n=1}^\infty$ is a sequence in $X$, $x_k \in X$ for any $k$. Now, a sequence in $X$ is convergent if there exists $a \in X$ such that for every $\epsilon > 0$, there exists a natural number $N$ such that for every $n \geq N$, $d(x_n, a) < \epsilon$. Hence, combining the two definitions, there exists $N$ such that, if we fix $m \geq N$, then for any $n \geq N$, $x_n$ converges to $a = x_m \in X$.

So I didn't use the if part of the problem. It'd be great if anyone could point out what am I missing here, as well as how this if part of the statement is used to solve the problem statement.

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  • $\begingroup$ Since $\{x_n\}$ has a convergent subsequence, denote its convergent subsequence by $\{x_{n_k}\}$ and the value it converges to by $\hat{x}$. Show that $x_n\to\hat{x}$. $\endgroup$
    – Jimmy Yang
    Dec 15, 2023 at 0:16
  • $\begingroup$ Just parse your own answer. You are fixing $m$ and then you are saying that the sequence converges to $a = x_m$. But that means that you decide what the limit is, by fixing $m$. Somebody else could pick another $m$ and get a different limit. However limits of sequences are unique. So this answer makes no sense.. $\endgroup$ Dec 15, 2023 at 0:29
  • $\begingroup$ The problem is that $N$ depends on $\varepsilon$ and you are taking it as it was independant, so in order to prove for a $\varepsilon' <\varepsilon$ that $d(x_n,a) < \varepsilon'$ you must take anothe $N'$ and this can make $m < N´$ (and there for not enoght ) $\endgroup$ Dec 15, 2023 at 2:34
  • $\begingroup$ i belive this is a correct answer math.stackexchange.com/questions/662299/… $\endgroup$ Dec 15, 2023 at 2:39
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    $\begingroup$ Does this answer your question? If a subsequence of a Cauchy sequence converges, then the whole sequence converges. $\endgroup$ Dec 15, 2023 at 2:40

1 Answer 1

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I think that it's missing a "Cauchy" in the title. Anyway, let $(x_n)_n$ a Cauchy sequence in $X$ and let $(x_{n_{k}})_k$ a subsequence of $(x_n)_n$ s.t. $x_{n_{k}}\to x\in X$.

So, if $\varepsilon >0 $, then $\exists N_\varepsilon>0$ s.t. $d(x_{n_{k}},x)<\varepsilon$

and, because $(x_n)_n$ is a Cauchy sequence,

$d(x_n,x_m)<\varepsilon, \quad \forall n,m\geq N_\varepsilon$.

Moreover,

$d(x_n,x)\leq d(x_n,x_{n_{k}}) + d(x_{n_{k}},x) < \varepsilon + \varepsilon=2\varepsilon$

Therefore, $x_n \longrightarrow x$.

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