Why say "completely regular T1" when completely regular T0 implies T1?

Question

Everywhere I look, on-line and in books, about separation axioms, the phrase "completely regular-$T_{1}$" is used. (That's using the weak meaning of "completely regular", namely, that any closed set and any point not in that set can be separated by a continuous map to $[0, 1]$.)

But surely completely regular-$T_{0}$ implies $T_{1}$, in fact, directly implies $T_{2}$ — and the proof is essentially the same as that completely regular (with no other separation assumption) implies regular: Let $x$ and $y$ be distinct points in the completely regular-$T_{0}$ space $X$. Then one of the two points has an open neighborhood not containing the other, say $x$ has an open neighborhood $W$ with $y \notin W$. Define $E = X \setminus W$. There is a continuous $f : X \to [0, 1]$ with $f(x) = 0$ and $f(z) = 1$ for all $z \in E$. Define $U = f^{-1}\bigl([0, 1/2)\bigr)$ and $V = f^{-1}\bigl((1/2, 1]\bigr)$. Then $U$ and $V$ are disjoint neighborhoods of $x$ and $y$, respectively (of course, $V$ is a neighborhood of $E$, too).

So why do all those sources seem to insist on saying "completely regular-$T_{1}$" instead of (the seemingly weaker) "completely regular-$T_{0}$"?

Or am I missing something here?

Answer 1

You could (like some texts do) call it $T_{3 {1 \over 2}}$, and normal plus $T_1$ is called $T_4$. And regular and $T_1$, $T_3$. (T stands for "Trennungsaxiom", separation axiom in German)

Then we have a nice hierarchy $T_4 \implies T_{3{1 \over 2}} \implies T_3 \implies T_2 \implies T_1 \implies T_0$, so that it's clear from the numbers which implies which. Because regular and normal are often overused, I think (normal subgroups, normal covers, etc.) I prefer to use it in the bare "closed sets separation" term, which is used less often. The real power of normality comes with $T_1$ added, or it could be voidly fulfilled (no disjoint closed sets to be separated, like in included point topologies). And complete regularity by itself is pretty useless too , without at least Hausdorff.

One could prove as a lemma that (completely) regular + $T_0$ is equivalent to (completely) regular + $T_1$ but then call the resulting property $T_{3{1\over2}}$ or $T_3$, and I've seen it done this way (I think Herrlich's German books do this).