Show that every accumulation point of a set that does not itself belong to the set must be a boundary point of the set.

  • $\begingroup$ why you cannot prove this yourself? no disparaging but this is elementary enough. $\endgroup$ – Bombyx mori Nov 7 '12 at 4:02

Call the set $S$. Denote this accumulation point as $p \not\in S$. Every neighborhood around $p$ contains a point of $S$ (because it's an accumulation point of $S$) and a point not in $S$ (namely, itself).

Therefore, $p$ is in the boundary of $S$. QED.

Of course, your proof will look different depending on the version of 'boundary' you're using. I was using the third definition here.


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