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


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.

  • $\begingroup$ 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? $\endgroup$ – Giuseppe Bargagnati Mar 12 at 23:58
  • $\begingroup$ 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 ($X$ \ $A)\cup \partial A.$ $\endgroup$ – 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.


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