# Proving that an accumulation point of a set is either an interior point or a boundary point of the set.

Prove that an accumulation point of a set $$S$$ is either an interior point of $$S$$ or a boundary point of $$S$$.

Proof:

Let $$x \in S\prime$$.

Suppose $$x \notin intS$$

Then for all $$\epsilon \gt 0$$, $$N(x;\epsilon)\backslash\{x\} \cap S \neq \emptyset$$

This must mean $$N(x;\epsilon) \cap S \neq \emptyset$$ since $$N(x;\epsilon)\backslash\{x\} \subseteq N(x;\epsilon)$$.

Since $$x \notin int S$$, we cannot have $$N(x;\epsilon) \subseteq S$$ which implies $$N(x;\epsilon) \cap \Bbb R \backslash S \neq \emptyset$$.

$$\therefore$$ $$x$$ is a boundary point of $$S$$.

Okay so I figured out the proof with some help.

• Check that interior points and boundary points are indeed accumulation point...than what happens if $x$ does not belong to the closure of your set? – Giuseppe Bargagnati Mar 12 at 23:58
• In any topological space $X,$ metrizable or not, if $A\subset X$ then (by definition) $\partial A=\partial (X$ \ $A) =\overline A \cap \overline {X\setminus A}.$ The sets int($A$), int ($X$ \ $A), \partial A$ are pair-wise disjoint and their union is $X$ . And $\overline X=$int($X)\cup \partial X,$ and similarly $\overline {X \setminus A}=$int ($X$ \ $A)\cup \partial (X$ \ $A$)=int ($X$ \ $A)\cup \partial A.$ – DanielWainfleet Mar 13 at 5:34

Take $$x$$ and accumulation point of $$S$$. By definition, for all neighborhood $$V$$ of $$x$$ holds $$V\setminus\{x\}\cap S\neq\emptyset$$. If there exist a neighborhood $$V_0$$ of $$x$$ such that $$x\in V_0\subseteq S$$ then clearly $$x$$ is an interior point of $$S$$. In the other way, if for all neighborhood $$V$$ of $$x$$ holds that $$V\not\subseteq S$$ then $$V\cap (X\setminus S)\neq\emptyset$$. Therefore, $$x$$ is a boundary point.