"Separation of disjoint closed set by open neighborhood" implies perfect normality?

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Topospace gives equivalent definitions for a normal space:

Def 1 Disjoint closed sets are separated by open neighborhood

Def 2 Separation of disjoint closed subsets by continuous functions

From what I learned, "Separation of disjoint closed set by open neighborhood" means normal space; the definition 2 means perfectly normal space. So the two definitions seem not equivalent.

Is topospace wrong or I missed something?

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Topospace correctly states the definitions are equivalent. If you take Definition 1 as the primary one, then the separation of disjoint closed subsets by continuous functions is the contents of the Urysohn lemma. The converse is obvious: If for two closed subsets $A,B \subseteq X$ such that $A \cap B = \varnothing$ there exists a continuous map $f:X \to [0,1]$ such that $f(x) = 0 \ \forall x \in A$ and $f(x) = 1 \ \forall \ x \in B$, then $U = f^{-1}([0,1/2))$ and $V = f^{-1}((1/2,1])$ are disjoint open neigborhoods of $A, B$.

Perfectly normal means normal plus every closed subset being a $G_\delta$- subset. It can be characterized via functions as follows:

A space $X$ is perfectly normal iff every closed $A \subseteq X$ admits a continuous $f : X \to [0,1]$ such that $A = f^{-1}(0)$. It is then easy to see that each pair $A,B$ of disjoint closed subsets admits a continuous $f : X \to [0,1]$ such that $A = f^{-1}(0)$ and $B = f^{-1}(1)$.

In a normal space we in general only find $f$ such that $A \subseteq f^{-1}(0)$ and $B \subseteq f^{-1}(1)$.