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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.

user334263
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3 Answers3

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

In detail, let $X=\{a,b\}$ with topology $\tau=\{\emptyset,X\}$. If we take $A=\{a\}$, then $b$ is a limit point of $A$. But $a$ is not a limit point of $A$ since $a$ is an isolated point in $A$. (A nbhd of $a$ cannot contain a point of $A$ different from $a$, as there are no other points.) This shows that the derived set is $A'=\{b\}$, which is not closed in $X$.

PatrickR
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    Are you sure? As far as I can see, any set with the indiscrete/trivial topology has a limit set of any non-empty set to be the whole space, which is closed. For an indescrete topological space $X$, every point's only (open) neighbourhood is $X$, which trivially contains any sequence within any (non-empty) subset of $X$, for all $N \geq 1$, and hence every point is in set of limit points. – Alex W Jun 29 '19 at 19:35
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    @AlexW the comments below the question say that $x$ is a limit point of $A$ iff $x\in\overline{A\setminus{x}}$. If ${a,b}$ has the indiscrete topology put on it and $A={a}$ then $a\not\in\overline{A\setminus{a}}=\emptyset$ so $a$ is not a limit point of ${a}.$ Note that ${a,b}$ could have been $\mathbb{R}$ with the indiscrete topology put on it and you'd still end up with a non-closed set of limit points of ${a},$ namely $\mathbb{R}\setminus{a},$ since $x\in\overline{{a}\setminus{x}}=\mathbb{R}$ for all $x\neq a.$ – mathematrucker May 06 '23 at 15:38
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    @mathematrucker You're right, I must (four years ago) have been using a different definition for 'limit points' than the one suppled. (Presumably one that didn't remove the point when evaluating its limit set.) – Alex W May 07 '23 at 16:52
  • @AlexW fair enough. Those are "adherent points" but I don't see that term used very often. – mathematrucker May 07 '23 at 20:05
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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 $\overline {S'}=X\ne S'.$

DanielWainfleet
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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 $\tau$.

Leo
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