# Prove every open ball is both an open and closed set

I'm asked to prove that every open ball is both open an open and closed set

So far, I've managed to show it's open:

Given a ball $B=B(x,r)$

I made a new ball, $B=B(y,\delta)$

let z be an element of $B(y,\delta)$ and showed that $d(z,x)\leq d(z,y)+d(x,y)< \delta+d(x,y)= r$

Concluding that $B(y,\delta)$ is contained in $B(x,r)$ and therefore $B(x,r)$ is an open set Is this right?

As for showing that that the open ball is a closed set, I'm at a loss.

thank you

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In general an open ball is not a closed set. –  Michael Albanese Oct 12 '13 at 21:34
Under what metric? –  Pedro Tamaroff Oct 12 '13 at 21:41
As noted, what you are trying to prove is not true. Is it possible that you were actually asked to prove that you can 'inscribe' a closed ball inside an arbitrary open ball, and another open ball inside of that? That is a common technique in analysis, and I could see such a question being misinterpreted in the way you wrote. –  Aaron Taylor Oct 12 '13 at 22:09

You are trying to prove something that is not true. A counterexample: $X = \mathbb{R}$ and the open ball $B_1(0)$ with centre $0$ and radius $1$. Then $1 \in X - B_1(0)$, but for any $\epsilon > 0$ we have $1 - \frac{\epsilon}{2} \in B_\epsilon(1) \cap B_1(0)$ and so this intersection is not empty. So $X - B_1(0)$ is not open and hence $B_1(0)$ is not closed.