Given then so let
Then (considering the ball around b with radius e):
Therefore
Complement is therefore Open, hence A is closed.
[Sorry i renamed your original ball with A]
Let (X, d) be a metric space and, for and r > 0, define the closed ball with centre a and radius r to be set B*(a, r) given by B*(a, 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 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 such that is in X\B*(a,r) for all . Hence 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?