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derived set = closure of derived set, implies T0 ?

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hbghlyj Posted 2022-12-17 20:06 |Read mode
Does it hold that, if every set $A$ in a topological space $X$ satisfies $A'=\bar{A}'$, then $X$ is a T0-topological space?

The equality doesn't imply T1:
Sierpiński space $S$ is T0 but not T1 since $A=\{1\}$ is not closed.
$A$ is the only non-closed set, and we have $\bar{A}=S$ and $\bar{A}'=\{0\}=A'$.

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2025-6-6 10:28 GMT+8

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