The separation axioms have exploded a little since the original list of four! Amongst them can be found "completely regular" spaces and "perfectly normal" spaces. The former is well-known: a point can be separated from a disjoint closed subset by a continuous real-valued function. The latter may be less well-known, but is fairly simple: given two disjoint closed sets then there is a continuous real-valued function such that the first closed set is the preimage of {0} and the second closed set is the preimage of {1}. This is equivalent to the condition that every closed set be the set of zeros of some continuous real-valued function.

Urysohn's lemma says that there's little point in looking for a completely normal space: in a normal space, any disjoint closed sets can be separated by a continuous real-valued function. Similarly, if we imposed the obvious exactness condition from perfectly normal onto completely regular then we'd actually end up with T6 (perfectly normal and T0).

However, there's room in the middle for something else and that's what I'm interested in: spaces which are completely regular and are such that the function separating a point from a closed set can be chosen so that the point is the preimage of its value. No assumption is made on the interaction of the closed set and the function (beyond that already given by the "completely regular" condition).

So my question is simple:

Has anyone encountered this notion before, and if so, where?

If it helps, I think that this condition is equivalent to the space being completely regular and singleton sets being Gδ sets (intersection of a countable number of open sets). Note that this isn't the same as first countable.

Motivation: As one might expect, my motivation comes from Froelicher spaces. Any Froelicher space comes equipped with two topologies: generated by either the curves or the functions. I'm trying to find conditions under which they are equal. Since the identity is continuous from the curvaceous topology to the functional topology, theorems like "a continuous bijection from a compact space to a Hausdorff space is a homeomorphism" spring to mind. Hausdorff is no problem - I can easily assume that both topologies are Hausdorff. In another context, I decided that I quite liked it when the curvaceous topology was sequentially compact. I think that I can prove that a continuous bijection from a sequentially compact space to a "perfectly regular" space is a homeomorphism, so it's a bit like a weakening of one condition at the expense of strengthening the other.

For a purely topological motivation, consider the following. In a completely regular space, one can test whether or not a net converges to a particular point by looking at its image under all continuous functions. In a "perfectly regular" space, one can test whether a convergent net converges to a particular point by using just one function and that function depends only on the proposed limit and not on the net.

So for both of these reasons, it's a nice condition to have and I wondered if it was known about so that, hopefully, I could build on others' work rather than having to invent it myself (hey, I've invented this really cool round thing!).


1 Answer 1


An answer is: completely regular plus countable pseudocharacter, the latter means that points are $G_\delta$-sets. In completely regular spaces a point is a $G_\delta$-set iff it is the zero-set of a continuous function, the proof is just like that for arbitrary closed $G_\delta$-sets in normal spaces: if $\lbrace x\rbrace=\bigcap_u U_n$ then pick continuous functions $f_n:X\to[0,1]$ that are zero at $x$ and $1$ outside $U_n$; then $\sum_n2^{-n}f_n$ has $\lbrace x\rbrace$ as its zero-set. See Engeling's General Topology, proof of Corollary 1.5.12.

  • $\begingroup$ Thanks for the answer. "Countable pseudocharacter" - now why didn't I think of that? That gives me 49 hits on MathSciNet so plenty to read there! And I'm pleased to hear that the proof I figured out is the same as you indicated; is there a standard reference for that? $\endgroup$ Jun 22, 2010 at 14:19

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