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Let $(X, \mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then $A'$ is a closed set. ($A'$ is the set of all limit points)

I originally thought this was a true statement based on the research I have done on the internet. I however am now beginning to doubt myself based on my definition of limit points.

My definition of limit points is "Let $(X, \mathfrak T)$ be a topological space with $A \subseteq X$ A point $x$ in X is said to be a limit point of $A$ provided that every open set containing $x$ contains a point of $A$ different from $x$.

I have been trying to work on some examples in the usual topology and I have come up with this: Let $X = \mathbb R$ in the usual topology and let $A = \mathbb Q$ then $A' = irrationals$ but I not sure if the set of irrationals is closed? I have looked through my notes and I think it is neither open nor closed therefore this is false.

I have also tried to come up with some very basic examples but I am unsure about the limit points: Let X = {a, b, c} and $\mathfrak T = \{X, \emptyset, \{a\}, \{b\}, \{a,b\}\}$. Let $A = \{b\}$ but I don't think this works as a counterexample .

Can anyone clarify my thinking? We have not talked about metric spaces and I am familiar with the usual, half-open line, half-open interval , discrete and indiscrete topologies.

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  • $\begingroup$ Have you carefully proven that if $A=\mathbb Q$, then no rational number belongs to $A'$? (You're right that the set of irrationals is neither open nor closed) $\endgroup$ – Karl Kronenfeld May 14 '15 at 22:39
  • $\begingroup$ I am now thinking for my example $A'= \mathbb R$ which open? $\endgroup$ – user219081 May 14 '15 at 22:42
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Here is a counterexample. Let $X = \{a, b, c\}$ and $\mathfrak T = \{\varnothing, \{a, b\}, X\}$. Then the set of limit points of $\{a\}$ is $\{b, c\}$, which is not closed.

If you assume that the space $X$ is $T_1$ (singletons are closed), then the statement you have can be proved.

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  • $\begingroup$ Do you mean let $A = \{a\}$ and $A' = \{b,c\}$ which is not closed? Just want to be sure I am reading this correctly $\endgroup$ – user219081 May 14 '15 at 22:50
  • $\begingroup$ @Alyssa Exactly. $\endgroup$ – Ayman Hourieh May 14 '15 at 22:51
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Your counterexample is wrong. You have $\mathbb Q^\prime=\mathbb R$ in the topology of $\mathbb R$. Note you find for each rational number $q$ a sequence of other rational number converges to $q$. Thus each rational is also a limit point of $\mathbb Q$.

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  • $\begingroup$ I just commented that I am now thinking $\mathbb Q' = \mathbb R$ and $\mathbb R$ is open? $\endgroup$ – user219081 May 14 '15 at 22:45
  • $\begingroup$ $\mathbb R$ is open and closed in the topology of $\mathbb R$ (the whole space is by definition always open and closed in any topology) $\endgroup$ – Stephan Kulla May 14 '15 at 22:48
  • $\begingroup$ Would the fact that it is open AND closed be enough to have a counterexample... I am not sure... $\endgroup$ – user219081 May 14 '15 at 22:49
  • $\begingroup$ No, that's no problem... $\endgroup$ – Stephan Kulla May 14 '15 at 22:51
  • $\begingroup$ I didn't think so $\endgroup$ – user219081 May 14 '15 at 22:51

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