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

Question

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?

Answer 1

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

Answer 2

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.