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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?

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Here's one $\mathbb{R}$-free characterization: a space is Tychonoff iff it has a Hausdorff compactification. Is that the sort of thing you want? –  Chris Eagle May 8 '11 at 10:31
    
I don't think, because it still makes a reference to an external object. –  mathahada May 8 '11 at 10:53
    
I don't necessarily buy the distinction you're making between open sets and "external objects." After all, talking about open sets is the same thing as talking about functions to the Sierpinski space. Does the Sierpinski space count as an external object? –  Qiaochu Yuan May 8 '11 at 11:35
    
There's a natural bijection between the set of continuous functions from a $T_0$ space to the Sierpinski space and the topology of the space, but there's no such bijection between the topology of a regular space and the the set of continuous functions to the unit interval. I don't know category theory so I don't know how to make this precise but I think you should get the intuitive feeling that the Sierpinski space is in some sense canonical and minimal. Anyway the answer given is what I've been looking for (I actually read the book but managed to miss that part) –  mathahada May 8 '11 at 12:52
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2 Answers

up vote 5 down vote accepted

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.

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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.

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