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For a fixed subset $S$ in $\mathbb{R^n}$ and a point $p$ in $\mathbb{R^n}$ 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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    @georgewyatt Why exactly are you deleting all your questions? Especially when they all have very nice responses.2012-11-01

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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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    @george: You’re welcome. If you have questions later, feel free to ask.2012-10-31
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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 $S$.