# Prove that a compact metric space is complete.

I'm reading Intro to Topology by Mendelson.

I'm in the section titled "Compact Metric Spaces".

The problem is in the title.

My attempt at the proof is as follows:

Let $\{a_n\}_{n=1}^\infty$ be a Cauchy sequence in $X$. We will show that $\{a_n\}_{n=1}^\infty$ converges to a point in $X$. Consider the set $S=\{a_n:n\in\mathbb{N}\}$. Then there are two cases to consider, $S$ finite and $S$ infinite. If $S$ is finite then there exists some $N\in\mathbb{N}$ such that $a_n=a$ for some $a\in S$ and so $\{a_n\}_{n=1}^\infty\to a$. Suppose now that $S$ is infinite. Then $S$ has at least one accumulation point in $X$, call it $a$. Thus, the neighborhood $B(a;\frac{1}{n})$ contains a point $a_n\in S$ and $\lim\limits_{n\to\infty} a_n=a$.

My concern with this proof is no where did I use the fact that the sequence was Cauchy, other than supposing it was. I know this is a flaw in my proof since I have to use the hypothesis some where.

I was also considering looking at the $\text{sup} S$, but I'm not sure how to go about using that fact or whether or not that's the right approach.

Thanks for any help or feedback!

• The main idea here is that if a Cauchy sequence $(a_{n})$ has a converging subsequence $(a_{\phi(n)})$, then $(a_{n})$ converges. Aug 14 '13 at 9:57
• @jibounet I see, so is that where I used the fact the sequence was Cauchy?. One thing though, that's something covered from an analysis course, but there isn't anything like that in the text I'm using, so would it be legal to use that in this context? Aug 14 '13 at 10:01
• I think that's legal since you have the following result : "$(X,d)$ is a compact metric space if and only if every sequence $(a_{n})$ of $X$ admits a subsequence $(a_{\phi(n)})$ which converges in $X$". Aug 14 '13 at 10:04
• @jibounet That actually wasn't covered in this section or in the section of limits. But I'll go ahead and try to prove that, that way I'll feel a bit better about using it. Thanks for the help. Aug 14 '13 at 10:05
• But, I might call the proof incomplete. The last sentence is way too short. Just because some subsequence of $a_n$ converges doesn't mean that the whole sequence does. So you should continue, and use Cauchy-ness to latch on to this one convergent subsequence and show that the rest of the sequence converges too.
– user18862
Aug 14 '13 at 14:03

In the cases you handle you only can construct a subsequence of the original sequence that converges to some $a$. In the case where $S$ is finite, so finitely many values $a_n$ occur, we can conclude (pigeon hole principle) that there exists $a \in S$ and infinitely many $n$ (say all $n \in M \subset \mathbb{N}$ that have $a_n = a$. This gives us a constant subsequence (all with value $a$) and thus trivially a convergent subsequence. But not yet convergence of the whole sequence (without using Cauchy).

Also, when $S$ is infinite, it has some limit point $a$, and then again all you can do at first is construct a subsequence of $a_n$ that converges to $a$: pick $n_1$ such that $d(a_{n_1}, a) < 1$, and having picked $n_1 < n_2 < \ldots < n_k$ such that $d(a_{n_i}, a) < \frac{1}{i}$ for all $i \le k$, we then pick $n_{k+1} > n_k$ such that $d(a_{n_{k+1}}, a) < \frac{1}{k+1}$, which can be done as there infinitely many points of the sequence in any open ball around $a$. And then $a_{n_m} \to a$ as $m \to \infty$.

Now where Cauchy is used is in the lemma: let $a_n$ be a Cauchy sequence in $(X,d)$ and let $a_{n_k}$ be a subsequence that converges to some $a \in X$. Then $a_n$ converges to $a$ as well.

Proof: let $\epsilon>0$. Pick $N \in \mathbb{N}$ such that for all $n,m \ge N$ we have $d(a_n, a_m) < \frac{\epsilon}{2}$, by Cauchyness. Also, pick $k$ such that $n_k > N$ and $d(a_{n_k}, a) < \frac{\epsilon}{2}$, by convergence of the subsequence to $a$.

Now for any $n \ge N$: $d(a_n, a) \le d(a_n, a_{n_k}) + d(a_{n_k}, a) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$. So having found $N$ for all $\epsilon>0$, $a_n \to a$ as $n \to \infty$, as required.

• So this problem actually requires coming up with and proving the lemma you stated about Cauchy sequences and their convergent subsequences? Aug 15 '13 at 2:04
• @ShantDanielian I suppose. It's not too hard to come up with: all points of a Cauchy sequence cluster together in tails, and a subsequence goes through all tails... It's often a separate lemma in many texts though. It helps to show compact == complete and totally bounded e.g., so it's bound to come up. Aug 15 '13 at 16:58
• Why not use the fact that compactness is equivalent to sequential compactness in a metric space, giving us that any cauchy sequence has a convergent subsequence, so by copying your final argument, the original sequence must converge. this just seems to require less work Apr 28 '14 at 13:03
• @ellya I wasn't sure the original poster knew this already. So I proved one way of it here. May 2 '14 at 17:17

Any sequence of points in a compact metric space has a convergent subsequence. We know that a sequence that has a convergent subsequence is in fact convergent. So this easily shows us that a compact metric space is complete.