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?

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

• So is my proof correct? Commented Feb 23, 2019 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. Commented Feb 24, 2019 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? Commented Feb 24, 2019 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 $B_{R_1}(x)=B_R(x)$ Commented Feb 25, 2019 at 2:16
• If you look at the answer the other person gave, he gives a proof of why a Cauchy sequence is bounded. Commented Feb 25, 2019 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? Commented Feb 23, 2019 at 23:12
• @Heuristics it's sloppy and incomplete. Commented Feb 24, 2019 at 7:05