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Let $(X, \mathcal{T})$ be a topological space. I came across a question asking for the proof of the fact that the set of limit points $S'$ of any subset $S \subseteq X$ is closed assuming the space is Hausdorff. However, is this always true for a general topological space that is not necessarily Hausdorff?

I am expecting the answer is no. To find a counter example, I tried to use the topology $\mathcal{T} = \{\emptyset, \{1\},\{1,2\},\{1,2,3\}\}$ on $X = \{1,2,3\}$, however I have not been able to find an example of a set whose set of limits points is not closed in this case.

I would appreciate a hint to find a counter example/a proof that none exists rather than a full solution. Thank you very much.

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    Since usage is varied, by $S'$ do you mean $\{p: p\in \overline {S\backslash \{p\} }\}$? – 2017-01-27
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    Yes, that is correct. – 2017-01-27

3 Answers 3

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Try any set with cardinality at least $2$, equipped with the indiscrete topology.

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If $X$ is a $T_1$ space then $S'$ is closed.

Suppose by contradiction that $p\in \overline {S'}$ \ $S'.$ Since $S'\subset \bar S$ we have $p\in \bar S.$

Now $p\not \in S'$ implies there is an open $U$ with $p\in U$ and $U\cap S\subset \{p\}.$

But $U$ must contain $q\in S'$ because $p\in \overline {S'}$, and by hypothesis, $p\ne q.$ Then there is an open $V$ with $q\in V$, and $p \not \in V$, because $X$ is $T_1.$ The open set $U\cap V$ contains $q\in S'$, so there exists $r\in U\cap V\cap S$ with $r\ne q.$ BUT then $p\ne r\in S\cap U,$ contrary to $U\cap S \subset \{p\}$.

As Opn Ball has already pointed out, the property does not hold for the coarse topology on $X$ if $X$ has at least 2 points: For $S=\{p\}\subset X$ we have $S'=X$ \ $\{p\}$ and $\overline {S'}=X\ne S'.$

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If $X$ is not $T_1$, it might not work as the following example shows. $X=\{a,b,c\}$ and $\tau=\{\emptyset, \{a,b\},X\}$. If we take $A=\{a\}$ then $A'=\{b,c\}$ which is not closed since its complement $\{a\}$ is not in $\tau$.