Completely regular spaces and Tychonoff ($T_{3\frac{1}{2}}$) spaces

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $[0,1]\subseteq \mathbb{R}$ carry the subspace topology of the usual topology of $\mathbb{R}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

• $X$ is completely regular when a point can be separated from a closed set not containing it by a continuous function: for every closed $C\subseteq X$ and every $x_0\in X\setminus C$ there is a continuous $f:X\to [0,1]$ (Continuity of a map of topological spaces at a point and globally) with $f(x_0)=1\text{and}f(y)=0\text{ for every }y\in C\mathrm{.}$
• $X$ is Tychonoff, also written $T_{3\frac{1}{2}}$ and completely regular Hausdorff, when it is completely regular and $T_1$ ($T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces).

The case $C=\mathrm{\varnothing }$ is allowed and is satisfied by the constant function $1$, which is continuous (Zero sets and cozero sets of continuous real-valued functions); so the condition hides no nonemptiness hypothesis.

The same condition in the vocabulary of zero sets. With $f$ as displayed, $C\subseteq Z(f)$ and $x_0\in \mathrm{coz}(f)$ (Zero sets and cozero sets of continuous real-valued functions), so complete regularity says: for every closed $C$ and every $x_0\mathrm{\notin }C$ there is a continuous $f$ whose zero set contains $C$ and whose cozero set contains $x_0$. In particular $\mathrm{coz}(f)$ is an open set containing $x_0$ and disjoint from $C$; that alone is weaker than regularity, and the passage from the function to two disjoint open sets is the next item.

The values $0$ and $1$ are a normalisation, not a restriction. If $g:X\to \mathbb{R}$ is continuous with $g(x_0)=a$, $g[C]=\{b\}$ and $a\mathrm{\ne }b$, then the condition above is met by a function built from $g$ by an affine change of variable followed by truncation into $[0,1]$; this page never needs that construction, because every function it builds is already normalised. The direction of the normalisation is a genuine convention and is fixed here as $f(x_0)=1$ and $f[C]=\{0\}$, following the most common usage; some texts write the reverse, and a reader must check which is meant before quoting a formula.

The convention fork over $T_1$ is the same one as for regularity. Completely regular names the function-separation condition alone, and Tychonoff names the conjunction with $T_1$ (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly). The indiscrete topology on a two-point set (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies) is completely regular, its only closed set disjoint from a point being $\mathrm{\varnothing }$, and it is not $T_0$; so the two halves are independent here as well.

• Complete regularity is a strong hypothesis in disguise. It asserts the existence of many continuous real-valued functions, and a space may have almost none; producing such functions is what Urysohn's lemma does for normal $T_1$ spaces, and that lemma is not available at this point in the reading order (Conventions on this page, and the one implication of the classical chain that is not available at this point in the reading order).

• Why the numeral is $3\frac{1}{2}$. Complete regularity implies regularity, as the next item proves, and every normal $T_1$ space is completely regular, which this page does not prove; so the axiom sits between $T_3$ and $T_4$, and the fractional numeral records that position and nothing more.

• Both names are in use for the conjunction. Tychonoff, completely regular Hausdorff and $T_{3\frac{1}{2}}$ denote the same class; this library writes Tychonoff.