# Sober except not $T_0$?

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?

-
As I'm sure you know, Toby, the definition of sobriety can be phrased in another way: every irreducible closed subset is the closure of precisely one point. (A closed subset is irreducible if 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
Also, I belatedly learned from Prakash Panangaden, yesterday, the reason for the term "sober": it's that if you're drunk, you see double. Inebriation might cause you to see two points where there should only be one, i.e. where they have the same closure. But this story is somewhat lacking, because it only really motivates the condition that "there aren't too many points". It misses the condition you're actually looking for: that "there are enough points". So, your task is to think of some drug that causes you to see 0-tuple rather than double. – Tom Leinster Aug 6 '11 at 1:18
Re Tom's point about terminology, I have always understood "sober" as carrying both halves of the definition: if you're sober you don't see double and you don't hallucinate. – Alex Simpson Aug 8 '11 at 10:24
@Tom. I was repeating a story I have heard from several sources. The point is exactly that a sober person doesn't hallucinate. This does makes sense. The completely-prime filter is what one "sees" - a set of compatible "observations" (= opens) that one interprets as a point. This is a hallucination if there is no point there. – Alex Simpson Aug 8 '11 at 18:48
@Alex: hmm, I think we're operating this analogy in different ways, but I'm still not convinced. Maybe it's easiest if you explain this to me in person, preferably over a beer. – Tom Leinster Aug 9 '11 at 12:26

Well, I've decided to go ahead and use ‘with enough points’. There are a lot of reasons to restrict to $T_0$ spaces, over and above reasons to restrict to sober spaces, and at least within that context having enough points is perfectly symmetric between topological spaces and locales.

-