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This is an equivalence to normality in topology

Let $X$ be a topogloical space. Then $X$ is normal $\iff \forall H$ open containing closed set $F$, we have an open set $G$ so we have the containments $$F \subset G \subset \bar{G} \subset H$$

I will spare you all from the details of the proof. I am only going to ask about $(\Longleftarrow)$. Let $F_1, F_2$ be disjoint closed sets. The proof hinges on taking $X - F_2$ (say) as the open set $H$ and applying the hypothesis and exhibiting the existences of $G$ and its closure $\bar{G}$

Eventually we reach a point where we have to argue $X -\bar{G}$ is the open (disjoint) set containing $F_2$ and thus completing the proof.

However the disjointness is not fully justified as $G \cap \bar{G}$ might not be empty if $G$ is a clopen set, but the hypothesis of the theorem is not violated. Am I right?

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The open disjoint sets are $G$ and $X - \bar{G}$. It is clear that they are disjoint since $X - \bar{G} \subset X - G$.

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    $\begingroup$ lol i misread the proof. I am stupid. $\endgroup$ – Hawk Sep 9 '17 at 2:59
  • $\begingroup$ Don't worry, it happens! $\endgroup$ – João Caminada Sep 9 '17 at 3:01

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