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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.

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    $\begingroup$ This statement is true but do not confuse it with the analogue for general topological spaces: it is not true that every point in an open set $S$ in a topological space has a closed neighborhood inside $S$. $\endgroup$ – Colin McLarty Feb 28 '16 at 18:02
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Hint: show that $\overline{B}_r(x) \subseteq B_\delta(x)$ for every $r < \delta$.

And clearly $B_\delta(x) \subseteq \overline{B}_\delta(x)$ for all $x$ and all $\delta$.

So every closed ball contains a smaller open ball with the same centre, and every open ball contains a smaller closed ball with the same centre as well.

So for openness we can both use open balls and closed balls.

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  • $\begingroup$ Additionally, it doesn't matter if we take the closure of $B_r(x)$ as closed ball or the set $\{\,y\in X\mid d(x,y)\le r\,\}$. $\endgroup$ – Hagen von Eitzen Feb 29 '16 at 7:05
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Pick $x\in S$. Then there's a closed ball around it that's contained in S. There's an open ball with a slightly smaller radius that's centered around $x$, so it's also contained in S. So $x$ is an interior point of $S$.

More formally, for $x\in S$, there exists some $\delta>0$ so that $\overline{B_\delta(x)} \subset D$. Then for $0<d<\delta$, $B_d(x)\subset S$.

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  • $\begingroup$ Could there be a case where closed balls fails to be contained in $S$ but open balls are contained in $S$? Intuitively since closed balls are slightly larger than open balls then there is a chance where closed balls fails to be contained but open balls are contained $\endgroup$ – Shamisen Expert Feb 28 '16 at 17:31
  • $\begingroup$ Be careful with using the plural expressions "closed balls" or "open balls". To prove that $S$ is open you need only find one open ball centered on $x$ contained in $S$. And similarly, you need only find only one closed ball, as this answer shows. $\endgroup$ – Lee Mosher Feb 28 '16 at 17:36
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    $\begingroup$ Nope. As Henno says, there's a closed ball inside every open ball- just shrink the radius. $\endgroup$ – manofbear Feb 28 '16 at 17:36
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Observe that you can write the set

$$S = \bigcup_{x \in S} B_{\delta/2}(x) $$

as a union of open sets, which is therefore open, as desired. QED.

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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.

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