Prove that if $\left ( x_{n} \right )$ is a Cauchy sequence in a metric space X then $\left ( x_{n} \right )$ is convergent if and only if $\left ( x_{n} \right )$ has a convergent subsequence.

Note: I realize that this question has been asked before and to many of you this is an embarrassingly easy question.... (Hopefully one day it will be the same for me) However, today I'm stuck on the second part of this proof..... :((( Any help would be greatly appreciated.... :)

Here is what I have so far:

($\Rightarrow $) Suppose $\left ( x_{n} \right )$ is a convergent Cauchy sequence in (X,d). Then $x_{n}\rightarrow x\in X$. Hence, by Lemma, $\left ( x_{n} \right )$ has a convergent subsequence such that $x_{n_{k}}\rightarrow x$.

[Lemma: Let $\left ( x_{n} \right )$ be a sequence in a topological space X. Then $x_{n}\rightarrow x\in X$ if and only if $x_{n_{k}}\rightarrow x$ for every subsequence $\left ( x_{n_{k}} \right )$ of $\left ( x_{n} \right )$.] - obtained this Lemma from my book.

($\Leftarrow $) Now suppose that $\left ( x_{n} \right )$ is a Cauchy sequence and it has a convergent subsequence $\left ( x_{n_{k}} \right )$ such that $x_{n_{k}}\rightarrow x\in X$. Since $\left ( x_{n} \right )$ is Cauchy, then by definition, for every $\varepsilon>0$ there exists an $N\in \mathbb{N}$ such that for all $i, j\geq \mathbb{N}$, $d(x_{i},x_{j})<\varepsilon$.

Since $\left ( x_{n_{k}} \right )$ s a covergent subsequence, then there is $K\in \mathbb{N}$ such that $x_{n_{k}}\in U$ for all $n_{k}\geq K$ where U is an open set in X containing x.

!!!!!!!And this is where I'm stuck.... :( I think I'm supposed to use Triangle Inequality, but I'm confused on how to incorporate it.... :(

  • $\begingroup$ Choose a limit point L -- compare distances to L using triangle ineq. I think you can manage the rest.. $\endgroup$ – Halil Duru Apr 8 '13 at 15:52
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    $\begingroup$ @ HalilDuru: Suppose $x_{n_{k}}\rightarrow L$. then by triangle inequality $d(x_{i}, x_{j})\leq d(x_{i}, L)+ d(L, x_{j})< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \varepsilon $ Does this look right? $\endgroup$ – Dome Apr 8 '13 at 15:58
  • $\begingroup$ posted a full solution... $\endgroup$ – Halil Duru Apr 8 '13 at 17:37
  • $\begingroup$ Alternatively, this is trivial by Bolanzo-Weierstrass and the triangle inequality. $\endgroup$ – Sanjoy Kundu Oct 11 '19 at 2:46

Certainly not the most elementary proof, but this one feels quite satisfying conceptually: let $(X,d)$ be a metric space and contemplate a Cauchy sequence $\{x_n\}$ 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 $\{x_n\}$ 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 $L$!

Added: Really though, the direct proof is also quite simple conceptually: you want to show that the sequence converges to $L$. For any fixed $\epsilon$ we know (i) all pairs of terms with sufficiently large index are within $\frac{\epsilon}{2}$ of each other, and (ii) there is at least one term of the sequence with sufficiently large index which is within $\frac{\epsilon}{2}$ of $L$. Apply the triangle inequality!

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Assume that there exists a subsequence $(x_{n_k})$ of $(x_n)$ such that $x_{n_k}\to x$ holds in $X$.Let $\varepsilon>0$ and choose $N$ such that $d(x_{n_k},x)<\varepsilon$ and $d(x_n,x_m)<\varepsilon$ for $n,m>N$. Now, if $n>N$, then $n_k>n>N$,and so $$d(x_n,x)\leq d(x_n,x_{n_k})+d(x_{n_k},x)<2\varepsilon,$$

i.e., $x_n\to x$ in $X$.

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Let $x_{n_k}$ $\to$ $L$ .

We need to show that $ \forall$$\varepsilon> 0 $ $\exists N $ such that $d(x_i,L)$ $ $< $\varepsilon$ for $i\geq N$ .

To choose $N$ -- use the Cauchyness of $x_n$ -- $\exists M$ such that $j \geq M$ implies $d(x_M,x_j) < \varepsilon/2$ .

Hence we also have $d(x_M,L)\leq \varepsilon/2$ .

Now take $ N=M$ .

$d(x_i,L) \leq d(x_N,L)+d(x_i,x_N) < \varepsilon/2 + \varepsilon/2 < \varepsilon$ $ \hspace{11mm} \blacksquare$

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    $\begingroup$ You needn't put a dollar sign between each "block" of code. Just put one, code code code, and close with another. $\endgroup$ – Pedro Tamaroff Apr 9 '13 at 2:42

Hint: Suppose $\{x_n\}$ converges to $a$. Is $\{x_n\}$ a subsequence of itself?

Hint 2: Suppose $x_n$ converges to $A$. Pick an arbitrary subsequence $\{a_n\}$ and prove
that $\{a_n\}$ must also converge to $A$. You need to pick an $\epsilon > 0$ and $n*$ such that

$|a_n - A|<\epsilon$ for all $n \ge n*$

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    $\begingroup$ Why not use $x_{n_j}$ instead of $a_n$? The notation $x_n$ is "shorthand" for $f(n)$ for some $f:\Bbb N\to\Bbb R$, so it makes more sense to talk about some sequence $n_1<\cdots <n_j<\cdots$ and "compose" $f(n_j)$ to get $x_{n_j}$. $\endgroup$ – Pedro Tamaroff Apr 9 '13 at 2:46
  • $\begingroup$ notation is whatever it's declared to be, yes? barring reserved variables of course. $\endgroup$ – user65384 Apr 9 '13 at 16:42

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