Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A coauthor and I have stumbled upon a useful topological property -- namely, we are interested in the property that every nonempty closed set contains a closed point. However, neither of us are topologists, so we don't know whether this exists in the literature yet. Does it? If so, what is it called, and where can I find information on it? If not, I'm happy to just call such a space "pearled" (intuition: the closed sets are the oysters), but I thought I'd ask here before publishing an existing definition under a new name.

By the way, every T$_1$-space is pearled, as is every finite T$_0$-space and every spectral space, but the property of being pearled is independent of the T$_0$-property. However, I would be interested even in a name for a pearled T$_0$-space. Is this the same as a T$_0$-space whose lattice of closed sets is atomic?

share|cite|improve this question
Another important example: The topological space underlying a quasi-compact scheme is pearled (and $T_0$). But I'm sure that you already know that :). – Martin Brandenburg Nov 6 '11 at 15:07
closed point? Are you talking about a Scattered space? – Michael Blackmon Nov 6 '11 at 17:00
a related notion, which I've only heard for schemes though, is that a space is said to be Jacobson if every closed subset is the closure of the subset of its closed points. – Laurent Berger Nov 6 '11 at 18:44
Any quasicompact $T_0$ space is pearled, since you can just keep intersecting closed sets until you reach a minimal closed set which must be a closed point. – Eric Wofsey Nov 6 '11 at 20:36
In answer to your last question, yes, $T_0$+pearled is equivalent to $T_0$+atomic lattice of closed sets. $T_0$ implies that the atomic closed sets are exactly the closed singletons. – David Milovich Nov 8 '11 at 16:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.