[Math] The set of accumulation points of A is closed

general-topology

I'm having a bit of a hard time proving or disproving the following claim in general topology:

Let X be a topological space, A $\subseteq$ X, and B the set of accumulation points of A. Is B necessarily closed?

Best Answer

no. you need $X$ to be $T_1$ as mentioned in the comments, and if it is $T_1$ so you also presented by @mookid with a proof of that case.

here is a counter example for the general case: let $X=\{a,b,c,d\}$ and $\tau =\{\{a,b,c\},\{b,c,d\},\{b,c\},X,\emptyset \} $ note that $\{b,c\}$ is the set of accumulation of the set's $ \{a,b,c\},\{b,c,d\}$ and $\{a,b\}$ is not colsed.

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