Question

Metrizability and complete regularity are topological properties that are, in a sense, different from the Hausdorff condition because they are not defined purely in the terms of the open sets, but rather using some external object, namely $\mathbb R$. Now a space is metrizable if it has the weak topology induced by a function $d : X \times X \rightarrow \mathbb R$ satisfying the properties of a metric, but metrization theorems tell us that an equivalent condition is that $X$ is regular and has a sigma-discrete basis (and this is purely topological). Is there a similiar characterization of Tychonoff spaces that makes no reference to $\mathbb R$ whatsoever?

Answer 1

A $T_1$ space $X$ is completely regular if and only if it has a basis $\mathcal{B}$ such that

1. For every $x \in X$ and every $U \in \mathcal{B}$ that contains $x$, there exists $V \in \mathcal{B}$ such that $x \notin V$ and $U \cup V = X$.
2. For any $U, V \in \mathcal{B}$ satisfying $U \cup V = X$, there exist $U', V' \in \mathcal{B}$ such that $X \setminus V \subset U'$, $X \setminus U \subset V'$ and $U' \cap V' = \varnothing$.

The proof can be found in

Frink, Orrin Compactifications and semi-normal spaces. Amer. J. Math. 86 1964 602–607.

The statement I've given above is actually dual to the statement of this paper. I took it directly from Engelking's General topology, which is the book I strongly recommend for finding references to this kind of questions.

Answer 2

Another answer was given by Aarts and De Groot in Complete regularity as a separation axiom Canadian Journal of Mathematics 21 1969 96–105. Frink's condition sets things up for a proof a la Urysohn's Lemma; Aarts and De Groot's condition is in terms of subbases and the proof proceeds by constructing a Hausdorff compactification.