If $X$ is a Hausdorff topological space and $A$ is a non-empty subset of $X$. Prove that $(cl(A))'=A'$.
Denote $A'$ as the set of accumulation points of $A$. Denote $cl(A)$ as the closure of $A$.
In a Hausdorff topological space, $(cl(A))'=A'$.
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general-topology
1 Answers
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Let $x \in (cl(A))'$ and $O$ be an open nbhd of $x$. Then $(O - \{x\}) \cap cl(A) \neq \varnothing$, so there is a $y \in (O - \{x\}) \cap cl(A)$. As $X$ is $T_1$, $O-\{x\}$ is an open nbhd of $y$, so as $y \in cl(A)$, $(O - \{x\}) \cap A \neq \varnothing$. Hence $x \in A'$. The other inclusion is evident.
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0I can prove that $A'$ is closed. But I don't know how to continue. – 2017-01-19
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0@Kenneth.K why would you want to prove $A'$ closed? Start with what you have to prove and proceed. – 2017-01-19
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0The other inclusion follows as $x \in A' \rightarrow x \in cl(A)$? – 2017-01-19