# Is this proof involving complete metric spaces correct?

Show that if every closed ball of a metric space $(X, d)$ is complete then $X$ is complete.

I thought the following: given $(x_n)$ a Cauchy sequence in $X$, we have that the set $A= \{x_{1}, x_{2},..., x_{n},...\}$ is bounded, that is, $diam(A) < M$ for some $M \in \mathbb {R}$ Therefore for any $n_{0}\in \mathbb {N}$, $A \subset B [x_{n_{0}}, M]$ From the hypothesis we know that every closed ball is complete and therefore $x_{n} \rightarrow x \in B [x_{n_{0}}, M] \subset X$ And so $X$ is complete.

For some reason I am not completely sure my proof is correct, could anyone tell me what you think?

• the proof is correct. Maybe you could change a bit your last sentence with something like: $(x_n)\subset B[x_{n_0},M]$ Cauchy and $B[x_{n_0},M]$ complete imply that $(x_n)$ converges to some $x\in B[x_{n_0},M]\subset X$.
– Surb
Commented May 11, 2015 at 19:06

As an $\epsilon$-improvement for clarity, I would change the last sentence with something like:
$(x_n)$ Cauchy, $B[x_{n_0},M]$ complete and $(x_n)\subset B[x_{n_0},M]$ imply that $(x_n)$ converges to some $x\in B[x_{n_0},M]\subset X$. It follows that $X$ is complete.
• $\epsilon$-improvement. Thats a great one. Yes I thought that was implicitly implied. Commented May 11, 2015 at 19:11