# Proof that the metric space $M$ is complete if every closed ball of $M$ is complete.

Let $$M$$ be a metric space

I'm asked to prove the statement

"Every closed ball of $$M$$ is complete $$\implies$$ $$M$$ is complete".

My attempt at this is as follows:

Let $$\{y_i\}$$ be a cauchy sequence in $$M$$.

Since cauchy sequences are infinite(?) there exists a subsequence of the cauchy sequence which can be enclosed in a closed ball with diameter $$d=\text{diam} (x_i,...x_j)$$ such that $$d<\epsilon$$, where $$\epsilon$$ is arbitrarily close to $$0$$.

This subsequence converges to a $$y\in M$$ by our statement and if a subsequence of a cauchy sequence converges to $$y$$ then the whole cauchy sequence converges to $$y$$.

Is this correct? If it is not, can I modify it to be correct?. If I can't, how can I prove it?

## 2 Answers

Cauchy sequences are bounded, so eventually the Cauchy sequence is in some ball.

• So is my proof correct? – Heuristics Feb 23 at 23:01
• Yes. It's just that you can skip taking the subsequence part. A Cauchy sequence can always be enclosed in a ball of finite radius. – TheManWhoNeverSleeps Feb 24 at 0:16
• but isn't it true that some points of the cauchy sequence can be at the border of the metric space and thus can't be enclosed? – Heuristics Feb 24 at 17:01
• Usually the way you define a ball of some radius $R$ around a point $x$ in a metric space $M$ is by $B_R(x)=\lbrace y\in M, d(x, y)<R \rbrace$, so even if the points are at the border there is no problem. It might just be that, say you have a metric space where all points satisfy $d(x, y)<R$ for some $R$. Then the entire metric space is in $B_R(x)$ for any $x$. If you take $R_1>R$ then what will happen is $B_{R_1}(x)=B_R(x)$ – TheManWhoNeverSleeps Feb 25 at 2:16
• If you look at the answer the other person gave, he gives a proof of why a Cauchy sequence is bounded. – TheManWhoNeverSleeps Feb 25 at 2:23

It's quite simple really: let $$(x_n)$$ be a Cauchy sequence.

Appply the definition of Cauchy for $$\varepsilon=1$$ and find $$N$$ such that $$n,m \ge N$$ implies $$d(x_n,x_m) < 1$$. Now defining

$$R= 2 \max \{ d(x_1,x_N), d(x_2,x_N), \ldots , d(x_{N-1},x_{N}), 1 \}$$

we see that the sequence lies as a whole in the closed ball with radius $$R$$ and centre $$x_N$$. So the sequence converges by assumption (it's Cauchy in any subspace it lies in if we keep the metric) in this ball, and thus also in $$X$$.

If fond of subsequences: the tail (better than subsequence) $$(x_n)_{n \ge N}$$ lies in $$D(x_N, 1)$$ (the closed ball); hence converges to $$x$$ in that ball. And if a subsequence converges so does the whole sequence; so that idea of your original proof idea can be kept.

• But is my proof wrong? – Heuristics Feb 23 at 23:12
• @Heuristics it's sloppy and incomplete. – Henno Brandsma Feb 24 at 7:05