$T_6$ implies $T_5$ I have a question about $T_6$ (perfectly normal) spaces. I first want to prove that the $T_6$ property is hereditary. I only know the Lemma from Urysohn and Tietze. Then i also want to conclude that every $T_6$ space is also a $T_5$ (completely normal) space, thus that every subspace is normal. Can someone help me?! Thank you for help :)

For my own work: a topologic space is perfectly normal if for every $A,B\subset X$ closed and disjoint, there exists a function $f\in C(X,[0,1])$ such that $f^{-1}(0)=A$ and $f^{-1}(1)=B$, but i can not see how to conlcude that than every subspace is normal.
 A: To show that perfect normality implies hereditary normality, it suffices to show that perfect normality is hereditary (since all perfectly normal spaces are normal).
Note that a T$_1$ space $X$ is perfectly normal iff for every closed $F \subseteq X$ there is a continuous $f : X \to [0,1]$ such that $F = f^{-1} \{ 0 \}$.
Assume $X$ is perfectly normal, and let $Y \subseteq X$.  Then $Y$ is clearly T$_1$.  Suppose that $F \subseteq Y$ is closed.  Then there is a closed $F_0 \subseteq X$ such that $F = F_0 \cap Y$.  As $F_0$ is closed in $X$ there is a continuous  $f : X \to [0,1]$ such that $F_0 = f^{-1} \{ 0 \}$.  But then $g = f \restriction Y : Y \to [0,1]$ is also continuous, and $g ^{-1} \{ 0 \} = f^{-1} \{ 0 \} \cap Y = F_0 \cap Y = F$.  Therefore $Y$ is perfectly normal.

As this question apparently comes from Munkres's text, I'll add a bit more detail to match his terminology a bit better.  According to Munkres, a space is perfectly normal if it is normal and every closed subset of $X$ is G$_\delta$.  (And normal implies T$_1$).
A subset $A$ of a topological space $X$ is called a zero-set if there is a continuous $f : X \to [0,1]$ such that $A = f^{-1} \{ 0 \}$.  It is quite easy to see that zero-sets are always closed G$_\delta$.
Claim: The zero-sets in normal spaces are exactly the closed G$_\delta$ sets.
proof.  By the above observation it suffices to show that closed G$-\delta$ subsets of normal spaces are zero sets.  Let $X$ be normal, and let $F \subseteq X$ be closed G$_\delta$.  Then there is a sequence $( U_n )_n$ of open subsets of $X$ such that $F = \bigcap_n U_n$.  For each $n$ the sets $F$ and $X \setminus U_n$ are disjoint closed subsets of $X$, and so by Urysohn's Lemma there is a continuous $f : X \to [0,1]$ such that $F \subseteq f_n^{-1} \{ 0 \}$ and $X \setminus U_n \subseteq f_n^{-1} \{ 1 \}$.  It follows quite easily that the function $f : X \to [0,1]$ defined by $$f(x) = \sum_n 2^{-n} f_n (x)$$ is continuous, and $F \subseteq f^{-1} \{ 0 \}$.  But note, also, that if $x \in X \setminus F$ then there is an $n$ such that $x \notin U_n$, and therefore $f_n(x) = 1$, meaning that $f(x) > 0$.  It follows that $F = f^{-1} \{ 0 \}$. $\quad\Box$
Therefore, according to Munkres's definition we have that perfectly normal spaces are precisely those normal spaces such that every closed set is a zero-set.  The following claim shows that the assumption of normality in this characterisation can be replaced with T$_1$.
Claim: Every T$_1$ space in which all closed subsets are zero-sets is normal.
proof.  If $F , E \subseteq X$ are disjoint closed sets, then there are continuous $f , g : X \to [0,1]$ such that $F = f^{-1} \{ 0 \}$ and $E = g^{-1} \{ 0 \}$.  Define $h : X \to [0,1]$ by $$h(x) = \frac{f(x)}{f(x) + g(x)}$$  As either $f(x) > 0$ or $g(x) > 0$ for all $x \in X$, it follows that $h$ is continuous.


*

*Given $x \in F$ we have that $g(x) = \frac{f(x)}{f(x) + g(x)} = \frac{0}{0+g(x)} = 0$.  Therefore $F \subseteq h^{-1} \{ 0 \}$.

*Given $x \in E$ we have that $g(x) = \frac{f(x)}{f(x) + g(x)} = \frac{f(x)}{f(x)+0} = 1$.   Therefore $E \subseteq h^{-1} \{ 1 \}$.


Therefore, in particular $h^{-1} [ [0,\frac{1}{2} ) ] \supseteq F$ and $h^{-1} [ ( \frac{1}{2} , 1 ] ] \supseteq E$ are disjoint open sets. $\quad \Box$
A further inspection of the above proof gives the following:


*

*If $h(x) = 0$, then $f(x) = 0$, and so $x \in F$.

*If $h(y) = 1$ then $f(y) = f(y) + g(y)$, and so $g(y) = 0$, meaning that $y \in E$.


That is, $h^{-1} \{ 0 \} = F$ and $h^{-1} \{ 1 \} = E$, and so $X$ is actually completely normal (or, as some call the property, hereditarily normal).
