True/ False: all interior points of $\bar{S}$ are points of $S$?
Claim: False.
Consider the set $[a,b]$. The closure of this set can be defined as $\overline{[a,b]} = (a,b) \cup \{a,b\}$ as in the frontier points are explicitly $a$ and $b$ individually. So if $x \in \overline{[a,b]}$ then it is in either $(a,b)$ or $\{a,b\}$. If $x \in (a,b)$ it is trivially an interior point. So now examining the case for the set of frontier points $\{a,b\}$, going back to the definition of an interior point of a set, it says that a point $x$ is an interior point if there exists an $\epsilon > 0$ such …