QUESTION : Let E be a non-empty subset of R. Let E' be its derived set i.e. the set of all limit points of E. Prove that E' is a closed set.
THANK YOU.
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No. There is no reason to assume that $E'$ is finite. For example, consider the set ${k + \frac{1}{n} : k, n \in \Bbb{N}}$ which has the set ${0, 1, 2, ...}$ as its limit points. – Sep 15 '13 at 17:16
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This is incorrect. $E'$ does not have to be finite. (Think about $E=[0,1]$.) – mrf Sep 15 '13 at 17:16
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@mrf every point in E is a limit point right ? including 0 and 1. – user93765 Sep 15 '13 at 17:19
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Do you know that a set is closed iff its complement is open? Can you prove that for $E'$? – Prahlad Vaidyanathan Sep 15 '13 at 17:34
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is that a theorem? – user93765 Sep 15 '13 at 17:40
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It's more a definition than a theorem, but you can see that the 'contains all its limit points' definition is equivalent. – Ian Coley Sep 15 '13 at 17:59
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@PrahladVaidyanathan so the answer would be : the complement of E' is a set that doesn't contain limit points, hence all its points are interior points so this set is open so we conclude that E' is closed since its complement is open. is that okay? Thank you. – user93765 Sep 15 '13 at 19:11
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@IanColey are you here? – user93765 Sep 15 '13 at 19:15
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Sounds right to me. – Ian Coley Sep 15 '13 at 19:16
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@IanColey so a point is either a limit point or interior point? Thank you – user93765 Sep 15 '13 at 19:21
3 Answers
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This actually holds in any $T_1$ space $X$. Let $p\in X\setminus E'$. Then there is an open set $U$ about $p$ such that $U\cap E\setminus\{p\} = \varnothing$. Suppose for the sake of contradiction that $p$ is a limit point of $E'$. Then $U$ contains an element $q$ of $E'$. Since $p\notin E'$, $p\ne q$. Thus $U\setminus \{p\}$ is an open neighborhood of $q$ that does not contain an element of $E$, contradicting the fact that $q\in E'$.
dfeuer
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You want to show that $E'$ is closed, so it is equivalent to showing that $U = (E')^c$ is open. For any $x\in U$, $x$ is not a limit point of $E$. Hence, no sequence in $E$ converges to $x$. In other words, (why?), there is a neighbourhood $V$ of $x$ such that $$ V\cap E = \emptyset $$ Now for any point $y\in V$, $y$ cannot be a limit point of $E$, and hence $y \in U$. Hence, $$ x\in V \subset U $$ Thus, every point $x\in U$ has a neighbourhood $V$ such that $V\subset U$. In other words, $U$ is open.
Does this help?
Prahlad Vaidyanathan
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You said that $E'^c$ is a set that "doesn't contain limit points". Perhaps you meant "doesn't contain any limit points of $E$"? – Prahlad Vaidyanathan Sep 15 '13 at 19:28
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Or more directly you can show that $(E')'=E'$.
So let $a\in (E')'$ and $\epsilon>0$.
Then there is a $b\in E'$ such that $|a-b|<\frac\epsilon2$ and there is a $c\in E$ such that $|b-c|<\frac\epsilon2$ (why?).
Can you see that $|a-c|\lt\epsilon$ and therefore $a\in E'?$
P..
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