Tychonoff space

Question

I am a bit confused about tychonoff space and a completely regular space. As far i have understood is that, every completely regular space is regular. And every regular space is also Hausdorff. Then can we say that a tychonoff space is a completely regular space rather than saying it is a completely regular hausdorff space?

Answer 1

There is the notion of "functionally separating a point from a closed set", i.e. for every closed set $C\subseteq X$ and every $x \in X\setminus C$ there is a continuous $f:X \to [0,1]$ such that $f(x)=0$ and $f[C]=\{1\}$, or some minor equivalent variation thereof. This is what most texts would call completely regular. It implies regular (for $x$ and $C$ as before, we have disjoint open sets separating them, and we can use the inverse images of $[0,\frac13)$ and $(\frac23,1]$ under $f$ e.g. to achieve that)

To ensure that we actually have enough closed sets to apply this too, and to make it fit into the hierarchy $T_0,T_1,T_2,T_3,T_4$ of separation axioms, we can additionally demand that $X$ be $T_1$ (or $T_0$) and get a notion we can abbreviate as $T_{3\frac12}$ or $T_{3.5}$ and which is implied by $T_4$ (normal plus $T_1$) and implies $T_3$ (regular plus $T_1$) just as its numbering suggests. This latter property is often called Tychonoff (French transliteration) (or Tikhonov, in the offical English transliteration) as well.

An indiscrete space is trivially completely regular and regular but not $T_0$ or $T_1$ or Hausdorff. In practice we use these functions to embed Tychonoff spaces so we need "lots" of functions so $T_1$ is almost always assumed. E.g. all locally compact Hausdorff spaces are Tychonoff, all separated uniform spaces are too (this includes most topological groups and vector spaces); so it turns out to be quite a ubiquitous class in analysis.