Let (X, d) be a metric space and, for $\displaystyle a \in X$ and r > 0, define the closed ball with centre a and radius r to be set B*(a, r) given by B*(a, r) = $\displaystyle (x \in X : d(x, a) \leq r)$.

Prove that B*(a,r) is closed.

Here's my thoughts... (not full proofs obviously, just rough ideas)

I thought about using the idea of sequences in closed sets to show that for every sequence $\displaystyle x_n$ in B*(a,r) that converge to some limit x, x is contained in B*(a,r).

And do i do this by saying let x some point contained in X\B*(a,r), then there is a $\displaystyle k \in \mathbb{N}$ such that $\displaystyle x_n$ is in X\B*(a,r) for all $\displaystyle k \geq n$. Hence $\displaystyle x_n \notin $ B*(a,r).

Is this looking like an actual proof or is it potentially a bit of a ramble..?

I also thought about just proving that its complement is open. Might be easier?