The definition of interior-point says "point $p$ in a set $S$ is interior point of $S$ if $\exists \delta \gt 0 : \mathcal B \left( p,\delta \right) \subseteq S $ but then as the open Ball $\mathcal B \left( p,\delta \right)$ is entirely contained in S and open ball have always infinite number of points, can we say $\exists \delta \gt 0 : \mathcal B \left( p,\delta \right) \subsetneq S \implies \exists \frac{\delta}{2} \gt 0 : \left( \mathcal B \left( p,\frac{\delta}{2} \right) \subseteq \mathcal B \left( p,\delta \right) \right) \cup \left( limit \space points \space of \space \mathcal B \left( p,\frac{\delta}{2} \right) \subseteq \mathcal B \left( p,\delta \right) \right) $
thus implying that thereis a closed ball containing $p$ . So will all interior points will have a closed ball containing it?
Best Answer
General facts: $B(p,r) = \{x\mid d(x,y) < r\}$ is an open set in any metric space (any $r>0$). This justifies the name "open ball" (where open = every points is interior in the above sense).
$D(p,r)=\{ x\mid d(x,y) \le r\}$ is a closed set for any $r>0$, justifying the name closed disk for such a set. The proof of this is most easily done by noting that if $y \notin D(x,r)$: $$B(y,d(x,y) - r) \subseteq X\setminus D(x,r)$$ so that every point of $X\setminus D(x,r)$ is an interior point of it etc.
This indeed implies that if $x$ is an interior point of $S$ witnessed by $B(x,r)$ indeed $D(x,\frac{r}{2}) \subseteq B(x,r)$ shows that $x$ also has a closed disk neighbourhood sitting inside $S$.
From a general topology point of view this shows that $X$ in its metric topology is regular (it has local bases of closed sets). Of course, much more is true (such topologies are even perfectly normal).