# Completeness and Cauchy Sequences

I came across the following problem on Cauchy Sequences:

Prove that every compact metric space is complete.

Suppose $X$ is a compact metric space. By definition, every sequence in $X$ has a convergent subsequence. We want to show that every Cauchy sequence in $X$ is convergent in $X$. Let $(x_n)$ be an arbitrary sequence in $X$ and $(x_{n_{k}})$ a subsequence that converges to $a$. Since $(x_{n_{k}}) \to a$ we have the following: $$(\forall \epsilon >0) \ \exists N \ni m,n \geq N \implies |x_{n_{m}}-x_{n_{n}}| < \epsilon$$

Using this, we can conclude that every Cauchy sequence in $X$ is convergent in $X$? Or do we inductively create subsequences and use Cauchy's criterion to show that it converges?

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Hint: Prove that if a Cauchy sequence has a convergent subsequence then the limit is the limit of the entire sequence. –  Asaf Karagila Jun 25 '11 at 16:16

I think you aren't spending enough time thinking about what it is you want to prove and what it is you've proven. This is a great example of this.

Everything you write up to "Using this..." is correct. But all you are doing is expanding the definitions you have. It seems to me (and I confess that this is 100% guess) that at that point you hit a bit of a wall and didn't know how to proceed, so you just decided to "jump" straight into the conclusion.

What are you trying to show? You are trying to show that every Cauchy sequence in $X$ converges. So you take a Cauchy sequence, $(x_n)$; you want to prove that this sequence converges. That is, you need to show that there is an $a\in X$ such that for all $\epsilon\gt 0$ there exist $N\gt 0$ (which may depend on $\epsilon$) such that for all $m\geq N$, $d(x_n,a)\lt \epsilon$ (where $d(-,-)$ is the distance function of $X$). You do some stuff that you know is correct. At the end of all of this, what did you conclude? You concluded that you have a subsequence that converges.

But that's all you found: a subsequence that converges. Look at the two propositions; what you did: $$\exists\Bigl( (x_{n_k})\text{ subsequence such that}\forall\epsilon\gt 0 \Bigl(\exists N\gt 0\bigl( \forall m,k\geq N (d(x_{n_m},x_{n_k})\lt \epsilon)\bigr)\Bigr)\Bigr);$$ and what you are supposed to show: $$\exists a\Bigl(\forall\epsilon\gt 0\Bigl( \exists N\gt 0\bigl(\forall m\geq N (d(x_m,a)\lt \epsilon)\bigr)\Bigr)\Bigr).$$ Now: do they look anywhere near similar? Or equivalent? In one you are only saying things about a subsequence, and you don't even have a limit for that subsequence; in the other, you are saying things about the entire sequence and a limit. I think that some sober reflection should tell you that it is unlikely that the two propositions are equivalent (at least, not obviously so), so you cannot expect to be done.

Now, nothing wrong with getting stuck, but taking the point where you are stuck and simply "jumping the gun" to the conclusion you want is a recipe for disaster.

Let's try to think intuitively about this; it's best to have an idea of where you want to go and how you will get there to guide the formal proof, rather than just start shuffling symbols around in the hope that the proposition you want will magically appear (I'm not accusing you of this, but I see it all too often in students who are still groping their way through writing proofs).

You have a Cauchy sequence; that means that the points are getting close together: you specify how close you want them to be (specify $\epsilon$), and after a certain point (the $N$ that depends on $\epsilon$), any two terms of the sequence will be less than $\epsilon$ apart.

You also know that there is a subsequence that converges to some point $a$. That means that there are at least some terms (and you can find terms of the subsequence as far "down the line" as you want) that are getting as close as you want to $a$. If your original sequence converges, what does it have to converge to? Since a sequence converges to $L$ if and only if every subsequence converges $L$, that means that if the original sequence converges, it better converge to the same thing as the subsequence we already found.

Intuitively, what does it mean to say the sequence converges to $a$? That if you specify how close you want the terms of the sequence ot be, you can find a point after which (an $N\gt 0$) all terms will be at least that close to $a$. You know you can make the terms as close to one another as you want by going down far enough, and you know that you can find terms as close as you want to $a$ if you go down far enough (the subsequence). That suggests that if you insure that all terms are, say, $\epsilon/2$ close to one another, and the terms of the subsequence are $\epsilon/2$ close to $a$, then every term will be $\epsilon$-close to $a$: take a term that is sufficiently "down the line": you can find a term of the subsequence that is $\epsilon/2$-close to it, and the term in the subsequence is $\epsilon/2$-close to $a$, so by the triangle inequality the term you started with should be $\epsilon$-close to $a$.

That looks good; now it's a matter of writing it out carefully and mathematically. That's the thinking behind AlbertH's answer.

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Here's a proof without $\epsilon$'s. Let $X$ be a compact metric space, $\hat{X}$ be its completion, and $i:X \rightarrow \hat{X}$ be the inclusion map. Then $i$ is continuous and hence $i(X)$ is compact. It follows $i(X)$ is closed in $\hat{X}$. But $i(X)$ is dense in $X.$ Therefore $i(X) = \hat{X}.$ We conclude $X$ is complete.

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Very nice!${}{}$ –  Asaf Karagila Jun 25 '11 at 17:14
That's a very nice way to prove this, I've only ever seen compact metric space is sequentially compact so each cauchy sequence has a convergent subsequence. Therefore each cauchy sequence is convergent. –  JSchlather Jun 25 '11 at 19:13
Let $\epsilon > 0$. Since $(x_n)$ is Cauchy, exists $\eta_1\in \mathbb N$ such that $$\left\vert x_n - x_m\right\vert < \frac \epsilon 2$$ for each pair $n, m > \eta_1$.
Since $x_{k_n} \to a$, exists $\eta_2 \in \mathbb N$ such that $$\left\vert x_{k_n} - a\right\vert < \frac \epsilon 2$$ for each $n > \eta_2$.
Let $\eta = \max\{\eta_1, \eta_2\}$, if $n > \eta$ then $k_n \ge n > \eta$. Therefore we have $$\left\vert x_n - a\right\vert \le \left\vert x_n - x_{k_n}\right\vert + \left\vert x_{k_n} - a\right\vert < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$$