# Why say “completely regular T1” when completely regular T0 implies T1?

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?

• One possible reason is that $T_1$ is a more familiar separation axiom than $T_0$. – Andreas Blass Jul 13 '17 at 18:32
• Do all these sources use $T_0$? If $T_1$ is the separation axiom with lowest index used in such a source, it's natural to not use $T_0$ there. – Daniel Fischer Jul 13 '17 at 18:50
• Yes, most of these sources do start at $T_{0}$. Some use the term "Tychnonoff" to mean completely regular and $T_{1}$, again with $T_{1}$ there being strong than what's actually needed. – murray Jul 13 '17 at 19:59
• The separation terminology is a mess. I use T0, T1, Hausdorff, regular, normal, regular T0, normal T1, Tychonov, Tychonov T0 and for clarity, nothing else. None of that confusing T4,T5 stuff which may or may not include T1 depending upon the author or text. – William Elliot Jul 13 '17 at 20:09
• $T_4$ is shorter than normal $T_1$, e.g. $T_{3{1\over 2}}$ also beats "completely regular $T_0$ / $T_1$. If you're using $T$-notation, go all the way ,I say! – Henno Brandsma Jul 13 '17 at 21:48

## 1 Answer

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).