Cauchy sequence is convergent iff it has a convergent subsequence

Question

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.... :(

Answer 1

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 $X$.

Answer 2

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 $L$!

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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!

Answer 3

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$

Answer 4

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 $n \ge n*$

Answer 5

Here's an $(\epsilon,\delta)$-free proof. A sequence $(x_n)$ is Cauchy iff for all nonstandard integers $H,K$, the terms $x_H$ and $x_K$ are infinitely close (i.e., $x_H-x_K$ is infinitesimal): $$x_H\approx x_K.$$ A sequence is convergent to $L$ iff all terms at nonstandard rank are infinitely close to $L$. If a subsequence $x_{n_i}$ converges to $L$, then for nonstandard $n_i$ we have $x_{n_i}\approx L$. Choose such an $H=n_i$. Since for all nonstandard $K$, $x_H\approx x_K$, we get $x_K\approx x_H\approx L$ so that $x_K\approx L$. Hence $(x_n)$ is convergent.

Answer 6

I will prove both of the following statements, and we will see that the proofs are completely analogous:

1. If $X$ is a metric space and $(x_n)_{n \in \mathbb{N}}$ is a Cauchy sequence in $X$ that has a subsequence $(x_{n_k})_{k \in \mathbb{N}}$ with $x_{n_k} \to x$, then $x_n \to x$

2. If $X$ is a uniform space and $\mathcal{F}$ is a proper Cauchy filter on $X$ such that there exists a proper filter $\mathcal{G}$ with $\mathcal{F} \subseteq \mathcal{G}$ and $\mathcal{G} \to x$, then $\mathcal{F} \to x$

$$ \begin{array}{|l|l|} \hline \textbf{Metric spaces} & \textbf{Uniform spaces} \\ \hline \text{Take any } \varepsilon > 0 \text{.} & \text{Take any open neighbourhood } U \text{ of } x \text{.} \\ & \text{Then there is some entourage } V \text{ with } \\ & V[x] \subseteq U \text{. Since } X \text{ is a uniform space,} \\ & \text{there is some entourage } W \text{with} \\ & W \circ W \subseteq V \text{.} \\ \text{Since } (x_n)_{n \in \mathbb{N}} \text{ is a Cauchy sequence,} & \text{Since } \mathcal{F} \text{ is a Cauchy filter, there is} \\ \text{there is some } N \in \mathbb{N} \text{ such that} & \text{some } A \in \mathcal{F} \text{ such that } A \times A \subseteq W \text{.} \\ \forall n, m \geq N : d(x_n, x_m) < \frac{1}{2} \varepsilon \text{.} & \\ \text{Since } x_{n_k} \to x \text{, there is some } K \in \mathbb{N} & \text{Since } \mathcal{G} \to x \text{, there is some } B \in \mathcal{G} \\ \text{such that } \forall k \geq K : d(x_{n_k}, x) < \frac{1}{2} \varepsilon \text{.} & \text{such that } B \subseteq W[x] \text{.} \\ \text{It remains to show that} & \text{It remains to show that } A \subseteq V[x] \text{.} \\ \forall n \geq N : d(x_n, x) < \varepsilon \text{.} & \\ \text{For this, take any } n \geq N \text{. Then there} & \text{For this, take any } y \in A \text{. Now note} \\ \text{is some } k \in \mathbb{N} \text{ with } k \geq K \text{ and} & \text{that } C = A \cap B \in \mathcal{G} \text{. Since } \mathcal{G} \text{ is a} \\ n_k \geq N \text{.} & \text{proper filter, there is some } z \in C \text{.} \\ \text{Hence we have } n, n_k \geq N \text{, which} & \text{Hence we have } (y, z) \in A \times A \subseteq W \text{.} \\ \text{implies } d(x_n, x_{n_k}) < \frac{1}{2} \varepsilon \text{.} & \\ \text{We also have } k \geq K \text{, which implies} & \text{We also have } z \in B \subseteq W[x] \text{, i.e. } \\ d(x_{n_k}, x) < \frac{1}{2} \varepsilon \text{.} & (z, x) \in W \text{.} \\ \text{Therefore, we have:} & \text{Therefore, we have:} \\ d(x_n, x) \leq d(x_n, x_{n_k}) + d(x_{n_k}, x) & (y, z) \in W \text{ and } (z, x) \in W \\ < \frac{1}{2} \varepsilon + \frac{1}{2} \varepsilon = \varepsilon \text{.} \hspace{85pt} \square & \Longrightarrow (y, x) \in W \circ W \subseteq V \text{,} \\ & \text{i.e. } y \in V[x] \text{.} \hspace{98pt} \square \\ \hline \end{array} $$

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Metric spaces: Take any $\varepsilon > 0$. Since $(x_n)_{n \in \mathbb{N}}$ is a Cauchy sequence, there is some $N \in \mathbb{N}$ such that $\forall n, m \geq N : d(x_n, x_m) < \frac{1}{2} \varepsilon$. Since $x_{n_k} \to x$, there is some $K \in \mathbb{N}$ such that $\forall k \geq K : d(x_{n_k}, x) < \frac{1}{2} \varepsilon$. It remains to show that $\forall n \geq N : d(x_n, x) < \varepsilon$. For this, take any $n \geq N$. Then there is some $k \in \mathbb{N}$ with $k \geq K$ and $n_k \geq N$. Hence we have $n, n_k \geq N$, which implies $d(x_n, x_{n_k}) < \frac{1}{2} \varepsilon$. We also have $k \geq K$, which implies $d(x_{n_k}, x) < \frac{1}{2} \varepsilon$. Therefore, we have: $$ d(x_n, x) \leq d(x_n, x_{n_k}) + d(x_{n_k}, x) < \frac{1}{2} \varepsilon + \frac{1}{2} \varepsilon = \varepsilon \text{.} $$