A bounded set is contained in an open ball

Question

Given metric space $ (X, d) $ .

Statement : A set $ E \subset X $ is bounded iff $ \exists $ $ x \in E $ and $ r \in R^{+} $ such that $ E \subset B_{r}(x,d) $.

In the above statement, the set $ B_{r}(x,d) $ is an open ball.

Is $ B_{r}(x,d) = \{y \in X | d(x, y) < r \} $ or $ B_{r}(x,d) = \{ y \in E | d(x, y) < r \} $ , in context of the above statement ?

the first one, of course $E$ is then also contained in the second set. — Peter Melech, Jan 19 '21 at 14:24
The assertion $E\in B_r(x,d)$ makes no sense. It should be $E\subset B_r(x,d)$. — José Carlos Santos, Jan 19 '21 at 14:24

score 0 · Accepted Answer · answered Jan 19 '21 at 14:26

The correct characterization of a bounded set in a metric space $(X,d)$ is:

$E \subset X$ is bounded iff there exists an $r>0$ and a point $x \in E$ (can be also just $x \in X$) such that $E \subset B_r(x) = \{ y \in X : d(x,y)<r \}$. I.e. $E$ must be contained in some open ball.

A bounded set is contained in an open ball

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