# In a T1 space, derived set = closure

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 $$\overline S$$.

• You are welcome. – drhab Nov 21 '18 at 9:45