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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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2  
In general an open ball is not a closed set. –  Michael Albanese Oct 12 '13 at 21:34
2  
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

1 Answer 1

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.

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