# Every closed ball in X is complete, then X is complete

1. Sep 21, 2013

### mahler1

The problem statement, all variables and given/known data.
Prove that if every closed ball in a metric space X is a complete subspace, then X is complete.

The attempt at a solution.

Let $\{x_n\}$ be a cauchy sequence in X. Then, for $ε=1$, $\exists$ $n_0 \in \mathbb N$: $\forall$ $m≥n_0$,$n≥n_0$ $d(x_m,x_n)<1$. Consider the closed ball centered at $x_{n_0}$ of radius 1. Then, the subsequence $\{x'_n\}$ such that $x'_n=x_{n_0 + n-1}$ is a subsequence of $\{x_n\}$ included in $K(x_{n_0},1)$ . By hypothesis, the ball is a complete subspace and the subsequence $\{x'_n\}$ is cauchy. But then $\{x'_n\}$ converges to some point $x \in K(x_{n_0},1)$ This implies $\{x'_n\}$ converges to the same point x in the metric space X. So the original sequence $\{x_n\}$ has a convergent subsequence that converges to the point x, this means that $\{x_n\}$ converges to x. I've proved that an arbitrary cauchy sequence in X is convergent, this implies every cauchy sequence in X is convergent, then X is a complete metric space.

Is my proof correct?

2. Sep 21, 2013

### Dick

Why are you having doubts? Eventually the tail of every cauchy sequence is contained in a closed ball. The closed ball is complete. So the sequence is convergent.

3. Sep 21, 2013

### mahler1

Thank you, I'm new to proofs and mathematics in general so I usually have strong doubts on whether I am doing the exercises right or not. I suppose it's a newbie's insecurity.

4. Sep 21, 2013

### Dick

Well, I thought that was a pretty fine and clear proof. I think you should start feeling less insecure.

5. Sep 22, 2013

### Jufro

The only thing is that you don't want to say ε=1.

Because your metric space may be "smaller" than the unit closed ball.

You could just say for some ε. But the proof seems fine

6. Sep 22, 2013

### Dick

A 'ball' in a metric space is just the set of all points x in the space such that |d(x,x0)|<=R. It doesn't have to be spherical. The proof really doesn't need any qualification.