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?



Characterization of Tychonoff spaces in terms of open sets
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.


