Needing help with a proof in topology of sets

Question

Given $S\subset R^n$ and $P$ any point of $S$. Proof that $P$ is either an interior point or a boundary point

Edit: A point $x \in R^n$ is said to be an interior point of $A \subset R^n$ in case there is a neighborhood $V$ of $x$ such that $V \subseteq A$.

A point $x \in R^n$ is said to be a boundary point of $A \subset R^n$ in case every neighborhood $V$ of $x$ contains points in $A$ and points in $A^c$.

Answer 1

If $p \in S$, there are two logically mutually exclusive options:

1. There exists a neighbourhood $N$ of $p$ such that $N \subseteq S$.

This means by definition that $p$ is an interior point of $S$.

2. All neighbourhoods $N$ of $p$ obey $N \nsubseteq S$.

This means that every neighbourhood $N$ of $p$ contains a point of the complement $S^c$ of $S$ (there is a witnessing point for $N \subseteq S$), and every neighbourhood $N$ of $p$ contains a point of $S$ (as $p \in S \cap N$, trivially). So then $p$ is a boundary point of $S$.

Answer 2

I'm assuming it's the first time you deal with general topology, so I'll try to provide details.

Suppose that $p\in S$ is not an interior point. That means that there's no neighborhood $U$ which satisfies $p \in U \subset S$.

But $p \in S$, thus for every neighborhood $U$ of $p$ we have $S \cap U \neq \emptyset $. So in order to have $U \not\subset S$, there must be some element $q \in U$ but $q\not\in S$; so by definition $q \in S^c$. So every neighborhood $U$ of $p$ has points in $S$ and in $S^c$, thus $p$ is a boundary point.

Such abstract reasoning from first principles must be applied when learning point-set topology. Rely heavily in set theory and definitions, and you'll derive almost all basic results.