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(For metric spaces in Rudin, related to this answer)

Let $E'$ denote the set of all limit points of $E$.

If $y \notin E'$, then there must be an $r\gt 0$ such that $B(y,r)$ does not contain any element of $E$

How do I prove this?

Definitions:

A point $p$ is a limit point of the set $E$ if every neighbourhood of $p$ contains a point $q\ne p$ such that $q\in E$

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Understand
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1 Answers1

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If $y \notin E'$, then there must be an $r\gt 0$ such that $B(y,r)$ does not contain any element of $E$

Proof:

$y$ is not a limit point, and hence we fail to meet the criteria:

  • A point $p$ is a limit point of the set $E$ if every neighbourhood of $y$ contains a point $q\ne y$ such that $q\in E$

and hence there is some neighborhood of $y\in (E')^c$ that contains no point $q\ne y$ such that $q\in (E')^c$

And since $E\subset (E')^c = X/E'$ where $X$ is the universe, the conjecture is proven. $\blacksquare$

Understand
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  • The notation is a bit weird, but it's the right idea. – Moya Feb 24 '15 at 03:24
  • @Moya Would you use $\complement E'$ for complement of $E'$? and something else for the set of limit points of $E$? – Understand Feb 24 '15 at 03:25
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    Really the statmeent is this: $y\in (E')^c$ means there is $N$ a neighborhood such that $y\in N$ and $N\cap E=\emptyset$. Thus $y\in N\subset E^c$. So there is $r>0$ such that $y\in B(y,r)\subset N\subset E^c$, so $B(y,r)\cap E=\emptyset$. – Moya Feb 24 '15 at 03:36
  • @Moya Thanks :). – Understand Feb 24 '15 at 04:05