In some books the terms: regular and $T_3$; normal and $T_4$; completely normal and $T_5$; perfectly normal and $T_6$ are synonyms, but in some books, the difference is that regular, normal, completely regular and perfectly regular spaces are not $T_1$ and $T_3, T_4, T_5, T_6$ spaces are $T_1$.
If we take the second definitions (that regular and $T_3$ and other pairs are not synonymous), than $T_6 \implies T_5 \implies T_4\implies T_3$, but completely normal $\implies$ normal but normal spaces do not include regular spaces. I know that perfectly normal $\implies$ completely normal.
But I am interested in: Does there exist
- A perfectly normal space that is $T_0$ but not $T_1$.
- A perfectly normal space that is not $T_0$ and is not regular.
- A perfectly normal space that is not $T_0$, is regular but is not completely regular.
- A perfectly normal space that is not $T_0$ and is completely regular.
Assuming the definitions in my comment: if $X$ is perfectly normal and $T_0$ then let $x \in X$, and suppose it were not closed, so we have $y \in \overline{\{x\}}$ with $y \neq x$. Then $T_0$ tells us we have an open $U$ containing $x$ but not containing $y$ (the other way around is ruled out in this situation). Apply perfect normality and we get a continuous $f: X \to [0,1]$ that $f^{-1}[\{0\}]=U^\complement$, so in particular $f(y)=0$ and $f(x)=p > 0$. But then $f^{-1}[[0,p)]$ is an open neighbourhood of $y$ missing $\{x\}$, contradiction. So $\{x\}$ must be closed and so $X$ is $T_1$ (and hence $T_2$, etc.). So then a space as in 1. does not exist.
The indiscrete space on $\{0,1\}$ is not $T_0$ but is perfectly normal: it is normal for trivial reasons (no non-trivial disjoint closed sets to consider) and it is a $G_\delta$ space. This is an example for 2, and possibly for 4.?