Relationships between a few strong separation axioms without Choice Disambiguation: Let $\langle X,\tau\rangle$ be a topological space.


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*I will say that two subsets $A,B$ of $X$ are separated if they are disjoint from each other's closures (their closures needn't be disjoint).

*I say that $\langle X,\tau\rangle$ is normal if for any two disjoint closed subsets $E,F$ of $X$ there exist disjoint open sets $U,V$ such that $E\subseteq U$ and $F\subseteq V$.

*I will say that $\langle X,\tau\rangle$ is completely normal if for any two disjoint closed subsets $E,F$ of $X$ there exists a continuous function $f:X\to\Bbb R$ such that $E\subseteq f^{-1}\bigl[\{0\}\bigr]$ and $F\subseteq f^{-1}\bigl[\{1\}\bigr]$.

*I will say that $\langle X,\tau\rangle$ is hereditarily normal if for any two separated subsets $A,B$ of $X$ there exist disjoint open sets $U,V$ such that $A\subseteq U$ and $B\subseteq V$.

*I will say that $\langle X,\tau\rangle$ is perfectly normal if for any two disjoint closed subsets $E,F$ of $X$ there exists a continuous function $f:X\to\Bbb R$ such that $E=f^{-1}\bigl[\{0\}\bigr]$ and $F=f^{-1}\bigl[\{1\}\bigr]$.



As mentioned in my earlier post, completely normal spaces are always normal, and the converse holds with sufficient Choice, but may fail to hold without it. Since disjoint closed sets are separated, then hereditarily normal spaces are normal, and it is readily apparent by definition that perfectly normal spaces are completely normal.
I'm given to understand (i.e.: Wikipedia says) that perfectly normal spaces are hereditarily normal, but I believe that we may need some Choice in proving this. Does anyone know whether perfectly normal spaces are necessarily hereditarily normal in $\mathsf{ZF}$? Also, need a hereditarily normal space be completely normal in $\mathsf{ZF}$?
 A: The property of perfect normality is hereditary. Any subspace of perfectly normal space is also perfectly normal and hence normal, so the original space is hereditary normal (since equivalent definition is that any subspace is normal). See $T_6$ implies $T_5$ for details. I think none of these need AC.
I also played with separation axioms. I like to think about separation axioms as (at least) two dimensional table (although $T_i$ suggests some kind of linerity). I have four groups of axioms: Level 1 axioms are $T_0$, symmetric and $T_1$. These form a diamond (since $T_1$ is the same thing as $T_0$ plus symmetric). Level 2 axioms are: Hausdorff ($T_2$), Urysohn ($T_{2\frac{1}{2}}$), completely Hausdorff, totally separated. Corresponding level 3 axioms are: regular, “Urysohn-regular”, completely regular, zero-dimensional and their $T_0$ variants ($T_0$ is what you need to level 3 imply level 2). And corresponding level 4 axioms are: normal, “Urysohn-normal”, comletely normal and strongly zero-dimensional and their symmetric and $T_1$ variants (symmetric is what you need to level 4 imply level 3 and $T_1$ to level 4 imply level 2).
Note that “Urysohn-regular” and “Urysohn-normal” are not needed since they are equvialent to regular resp. normal. When you have needed choice, then also completely normal and normal degenerate, but I think this is the only choice-related degeneration.
You may also want to include “precisely separated by continuous function to $[0, 1]$” to get perfect normality and level 2, 3 variants. Also note, that all level 1, 2, 3 axioms are hereditary so hereditary variant might be needed just for some level 4 axioms like normality.
Back to your question, the proof in the first paragraph actually proves that perfectly normal space is hereditarily completely normal which is trivially stronger than being herediraty normal and completely normal.
Update: For your question in comments. Claim: $X$ is perfectly normal iff any closed set is zero-set. For $\impliedby$, let $F$, $H$ be the closed disjoint sets, let $f, h: X \to [0, 1]$ such that $F = f^{-1}(0)$, $H = h^{-1}(0)$, then you can just take $f / (f + g)$ to precisely separate $F, H$, no need for $T_1$. For $\implies$, to show that $F ≠ ∅, X$ is a zero set, we just need some other closed set disjoint with $F$. As you pointed out, we can take just empty set.
