Proof that if a point $P$ is not in the interior of $S^c$, then any ball in $E$ of center $P$ contains points of $S$

Question

I have a very short proof and am questioning one of the logical steps I made.

Let $S$ be a subset of a metric space $E$.

I want to prove that if $p$ is not an interior point of $S^c$, then any ball in $E$ contains points of $S$.

Proof:

Suppose that $\forall\epsilon >0, B_{\epsilon}(P)\nsubseteq S^c$

$\implies \forall\epsilon>0, \exists q\in B_{\epsilon}(P)$ such that $q\notin S^c$.

$\implies \forall \epsilon >0, \exists q\in B_{\epsilon}(P)$ such that $q\in S^c$.

Can I conclude that since its not in the complement it must be in the original set?

Answer 1

Observe that the $\operatorname{int}(S^c)$ is the complement of $\bar{S}$.

So your point P lies in $\bar{S}$. By definition of the closure, that means all balls centered on $P$ intersect $S$.

Answer 2

Yes. This is, because the compliment is taken with respect to $E$, (the 'relative complement' as described on Wikipedia: https://en.m.wikipedia.org/wiki/Complement_(set_theory).
So, $S^C$ here is the set of all elements of $E$ that are not in $S$. This is implicit information one would expect you to infer, as the absolute complement wouldn't make any sense in the given context.

As mathstudent_101 pointed out, you probably meant to write $q \in S$.