# boundary points, isolation points, accumulation points.

For a fixed subset $S$ in $R^n$ and a point $p$ in $R^n$ How do you prove that if a point $p$ is a non isolated boundary point of $S$, then it must be an accumulation point in $S$?

Thinking about the solution: I get that a boundary point must either be isolated point or an accumulation point, so if it's not an isolated point it must be an accumulation point, but I'm not sure why it has to be one or the other. I know isolated points are never accumulation points, because accumulation points must have other points in every neighborhood. But I can't quite grasp my head around it or know where to begin in writing a logical proof.

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What definition of Isolate point you have? –  P.. Oct 31 '12 at 21:08
It was defined to me that an isolation point is a point p that lies in some open ball that contains no other points of the set S. And every isolated point of S in R^n is a boundary point of S –  george wyatt Oct 31 '12 at 21:14
@georgewyatt Why exactly are you deleting all your questions? Especially when they all have very nice responses. –  EuYu Nov 1 '12 at 7:03

HINT: Let $p$ be a boundary point of $S$.

1. Show that if $p\notin S$, then $p$ is an accumulation point of $S$.

2. Show that if $p\in S$, then either $p$ is an accumulation point of $S$, or $p$ is an isolated point of $S$.

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OH thank you so much this really helped me get started –  george wyatt Oct 31 '12 at 21:32
@george: You’re welcome. If you have questions later, feel free to ask. –  Brian M. Scott Oct 31 '12 at 21:35
Assume that $p \in S$. Then since $p$ is not an isolated point, for every open ball $B$ centered at $p$ you can find elements $s \in S \cap B$ different from $p$. So $p$ is a limit point of $S$.
If $p \not \in S$ since $p$ is a boundary point by definition is a limit point of $S$.