Characterization of Tychonoff spaces in terms of open sets

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.

Answer 3

Suppose that $(X,\mathcal{T})$ is a topological space. Then define a relation $\prec$ on $\mathcal{T}$ by letting $U\prec V$ if and only if $\overline{U}\subseteq V$. Define a relation $\ll$ on $\mathcal{T}$ by letting $U\ll V$ if and only if there is some $(U_{r})_{r\in[0,1]\cap\mathbb{Q}}$ such that $U_{0}=U,U_{1}=V$ and $\overline{U_{r}}\subseteq U_{s}$ whenever $r<s$. Then

1. the space $(X,\mathcal{T})$ is regular if and only if for each $U\in\mathcal{T}$, we have $U=\bigcup\{V\in\mathcal{T}|V\prec U\}$, and

2. the space $(X,\mathcal{T})$ is completely regular if and only if for each $U\in\mathcal{T}$, we have $U=\bigcup\{V\in\mathcal{T}|V\ll U\}$.

This characterization of regular and completely regular spaces is the way that one typically defines regularity and complete regularity in point-free topology. In fact, in point-free topology it is much more convenient to use this characterization of complete regularity than it is to use the characterization of complete regularity in terms of frame homomorphisms from the real numbers to your frame.

Perhaps the most illuminating characterizations of complete regularity are the following characterizations.

Let $X$ be a topological space. Then the following are equivalent:

1. $X$ is completely regular.

2. $X$ has a compatible proximity.

3. $X$ has a compatible uniformity.

Of course, proximities and uniformities impose additional external structure on topological spaces.