TITLE: Is this statement true: a set is open if every point has a closed ball contained inside of the set
QUESTION [10 upvotes]: Is this statement true:

Set $S$ (on a metric space) is open if $\forall x \in S$, $\exists
 \delta > 0,$ s.t. $\thinspace \overline B_\delta(x) \subset X$

I am a little bit thrown off by the closed ball instead of open ball definition of open set. Can someone verify the above statement and show how it is same as open ball.

REPLY [2 votes]: Both other answers are correct, but they are not getting to the metric content of the problem...
Let $X$ be a metric space, $S \subset X$, and $x \in S$.  If $S$ is open, then $X \setminus S$ is closed, so $\mathrm{dist}(x, X \setminus S)$ exists and is positive.  Otherwise, if $x \in \partial S$, $\mathrm{dist}(x, X \setminus S)$ is only nonnegative.  Set $\delta= \frac{1}{2} \mathrm{dist}(x, X \setminus S)$.  Then $\overline{B_\delta (x)} \subset S$.
Note that $\delta = \frac{1}{2}\mathrm{dist}(\dots)$ is not essential.  We can choose and $\eta \in (0,1)$ and $\delta = \eta \mathrm{dist}(\dots)$ works as well.  In any event, if $x$ may be chosen on $\partial S$, $\delta$ is forced to $0$ and the hypothesis ("$\forall x \in S, \exists \delta > 0 \dots$") does not hold.