Timeline for Name for a topological space where every closed set contains a closed point
Current License: CC BY-SA 3.0
12 events
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| Nov 11, 2011 at 12:22 | comment | added | Neil Epstein | @de la Vega: Looking in the literature, apparently T_{1/4} already exists. It means that every point is either closed or an intersection of open sets. I doubt this is equivalent to T_0 pearled (i.e. to T_0 + atomic lattice of closed -- thanks David), but it certainly isn't implied by it. For instance, in Spec k[x,y] (k a field; x,y indeterminates) and p=(x), the set {p} isn't closed (since (x,y) is in its closure), and any open set that contains p contains (0) as well. | |
| Nov 9, 2011 at 20:11 | comment | added | Ramiro de la Vega | I have heard the term $T_{1/2}$ for a space in which every singleton is either closed or open. This condition implies $T_0+$pearled, so perhaps the latter should be called $T_{1/4}$. | |
| Nov 8, 2011 at 16:59 | comment | added | David Milovich | 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. | |
| Nov 6, 2011 at 20:36 | comment | added | Eric Wofsey | 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. | |
| Nov 6, 2011 at 19:30 | comment | added | Neil Epstein | @Berger: That's an interesting property, and it clearly implies mine. It's a lot stronger, though. For instance, the Spec of a semilocal ring will never be a Jacobson space unless the ring is zero-dimensional. I expect the notion of a "Jacobson space" was invented to generalize the idea of a Jacobson ring. | |
| Nov 6, 2011 at 19:09 | comment | added | Neil Epstein | @Brandenburg: I suppose this follows from the fact about spectral spaces. @Blackmon: I call a point $x$ closed if the singleton subset $\{x\}$ that it defines is a closed subset of $X$. A pearled space need not be scattered. For example, every T_1 space (hence every Hausdorff space, hence every metric space) is pearled, since every point in such a space is closed. But a connected metric space with at least two points is never scattered, if I understand the definitions correctly. | |
| Nov 6, 2011 at 18:59 | history | edited | Neil Epstein | CC BY-SA 3.0 |
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| Nov 6, 2011 at 18:44 | comment | added | Laurent Berger | 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. | |
| Nov 6, 2011 at 17:06 | comment | added | Not Mike | planetmath.org/encyclopedia/ScatteredSet.html (wiki seems to be missing it) | |
| Nov 6, 2011 at 17:00 | comment | added | Not Mike | closed point? Are you talking about a Scattered space? | |
| Nov 6, 2011 at 15:07 | comment | added | Martin Brandenburg | 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 :). | |
| Nov 6, 2011 at 12:13 | history | asked | Neil Epstein | CC BY-SA 3.0 |