tl;dr: Is there an accepted or proposed term for a topological space whose $T_0$ quotient is sober?

The condition that a topological space be sober (and therefore equivalent to a locale) may be broken into two parts:

- there are enough points,
- and there aren't too many.

The condition that there are enough points is somewhat complicated: we note that every point gives rise to a completely prime filter in the lattice of open subsets, and require that every such filter arise in this way. I don't know any simpler way to say that, and I don't know any simple term for it.

The condition that there aren't too many points may be treated similarly: if two points give rise to the same filter, then they must be equal. But this can be more simply stated: if two points have the same (open) neighbourhoods, then they are equal. And in this guise, the condition is well known: the $T_0$ separation axiom.

I'd like to know if there's an accepted or proposed term for the first condition, of having enough points, simpler than "with a sober space as $T_0$ quotient". (The $T_0$ quotient is the quotient space found by identifying points with the same neighbourhoods, and it's easy to prove that this is equivalent.) Another obvious possibility is "with enough points", but this seems rather vague without context. (It could also lead to confusion, as it's not quite symmetric to locales; a locale with enough points is equivalent to a topological space, while a topological space with enough points is only equivalent to a locale if it's also $T_0$.)

Does anybody know terminology for this?

canbe phrased in another way: every irreducible closed subset is the closure of precisely one point. (A closed subset isirreducibleif it cannot be expressed as the union of two proper closed subsets. That's "union", not "disjoint union". The closure of a point is always irreducible.) Of course, whether you call this "simpler" than the characterization in terms of completely prime filters is subjective. – Tom Leinster Aug 6 '11 at 1:11