# Prove that if a cauchy sequence admits a convergent subsequence in a metric space, then the sequence converges to same limit

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.

• 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 $x_n\to\hat{x}$. Dec 15, 2023 at 0:16
• 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.. Dec 15, 2023 at 0:29
• 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 ) Dec 15, 2023 at 2:34
• i belive this is a correct answer math.stackexchange.com/questions/662299/… Dec 15, 2023 at 2:39
• Does this answer your question? If a subsequence of a Cauchy sequence converges, then the whole sequence converges. Dec 15, 2023 at 2:40

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, $$x_n \longrightarrow x$$.