How do I do the proof? My "proof" is more like a list of to me known facts.

Lemma. In a $T_1$ space, $\bar{S}=S'$.

Proof. We know that $\bar{S}=S\cup S'$ (already proven), thus $S'=\bar{S}$ iff $S\subseteq S'$. $x\in S'$ iff $\forall U_x\in N(x)$, $U_x\cap S-\{x\}\ne\varnothing$. Given any $x\in S$, suppose $x\notin S'$, then, since $S'$ is closed (already proven, valid only in $T_1$), there is $U\in N(x)$ such that $U\cap S'=\varnothing$. $x$ is an isolated point, thus there is $V\in N(x)$ such that $S\cap V=\{x\}$. I've also proved that in $T_1$ singletons are closed.


The lemma is not true.

Let $S=\{x\}$.

Then $S'=\varnothing$ so cannot equalize the non-empty set $\overline S$.

  • $\begingroup$ You are welcome. $\endgroup$ – drhab Nov 21 '18 at 9:45

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