"internal" definition of complete regularity?

Question

There is something strange (I think) about the complete regularity separation axiom. Consider the definitions.

• T0 means for every two distinct points there is an open set containing exactly one of them.
• T1 means for every two distinct points there is an open set containing one (specified) but not the other.
• Hausdorff means for every two distinct points there are disjoint open neighborhoods around them.
• Regular means for every point and closed set not containing it there are disjoint open sets containing them.
• Normal means for every two disjoint closed sets there are disjoint open sets containing them.

However when we get to complete normality we have the following definition:

• Completely regular means for every point and closed set not containing it there is a continuous real-valued function which is 0 at the point, and 1 at every point in the closed set.

It seems odd that all the other separation axioms are easily defined in terms of the open (and maybe closed) subsets of that space but for complete regularity you need some "external object" (the real line with the metric topology). Is there a way to define complete regularity without using an "external object" like this?

Answer 1

I don't think there is a nice clean characterisation of completely regular spaces (which I will take to include the T₁ property) in terms of just the open (or closed) subsets of the space. However, in the following article

Orrin Frink, Compactifications and semi-normal spaces, Amer. J. Math. vol.86 (1964), pp.602–607, MR0166755, JSTOR link

the following internal characterisation of competely regular spaces was proved:¹

Theorem. A T₁-space $X$ is completely regular iff it has a base $\mathcal{B}$ satisfying the following:

1. Given any $x \in X$ and any $U \in \mathcal{B}$ containing $x$ there is a $V \in \mathcal{B}$ such that $x \notin V$ and $U \cup V = X$;
2. Given $U , V \in \mathcal{B}$ with $U \cup V = X$ there are $U^\prime , V^\prime \in \mathcal{B}$ such that $X \setminus V \subseteq U^\prime$, $X \setminus U \subseteq V^\prime$, and $U^\prime \cap V^\prime = \varnothing$.

In one direction, it is relatively easy to show that if $X$ is completely regular, then the family of all co-zero sets (sets of the form $f^{-1} [\;(0,1]\;]$ where $f : X \to [0,1]$ is continuous) forms a base as desired.

In the other direction, you use such a base to mimic the usual proof of Urysohn's Lemma.

(sketch) Enumerate $[0,1] \cap \mathbb{Q}$ as $\{ q_ i \}_{i \geq 0}$ with $q_0 = 0$ and $q_1 = 1$. Given $x \in X$ and a closed $F \subseteq X$ not containing $x$, first pick $U_1 \in \mathcal{B}$ such that $x \in U_1 \subseteq X \setminus F$, and then pick $V_0 \in \mathcal{B}$ such that $x \notin V_0$ and $U_0 \cup V_1 = X$. Then inductively pick $U_2, V_2, U_3, V_3, \ldots \in \mathcal{B}$ so that for all $i, j \geq 0$

1. If $i \neq 0$ and $j \neq 1$ and $r_i > r_j$, then $U_i \cup V_j = X$;
2. If $i \neq 0$ and $j \neq 1$ and $r_i \leq r_j$, then $U_i \cap V_j = \varnothing$;
3. If $i, j \neq 0$ and $r_i \leq r_j$, then $U_i \subseteq U_j$;
4. If $i, j \neq 1$ and $r_i \leq r_j$, then $V_i \supseteq V_j$.

Denoting $U_i$ as $U^{q_i}$, define $f : X \to [0,1]$ by $$f(x) = \begin{cases} \inf \{ q \in [0,1] \cap \mathbb{Q} : x \in U^q \}, &\text{if }x \in U_1 = U^1 \\ 1, &\text{otherwise.} \end{cases}$$

It can be shown that this function is continuous, $f(x) = 0$ and $F \subseteq f^{-1} [ 1 ]$.

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¹_{This is also an exercise in Engelking's General Topology, where I learned about it. The phrasing of the theorem comes from there.}