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This property seems like it should have a nice name, but I can't find one anywhere. Does anyone know a name for this?

For each non-empty open set $U$, there exist proper open subsets $\{U_i\}_{i\in I}$ such that $U=\cup_i U_i$.

I suppose this could also be formulated as each nonempty open set having an open cover of proper subsets, or being the colimit of its open subsets.

(Also, apologies if this is something obvious I should have thought of.)

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    $\begingroup$ Each nonempty open set? $\endgroup$
    – user2734
    Commented Mar 19, 2010 at 16:41
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    $\begingroup$ For what it's worth, there's no need to add "nonempty". The property always holds when U is the empty set, since you can take I to be empty. $\endgroup$
    – Tom Leinster
    Commented Mar 19, 2010 at 17:09
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    $\begingroup$ @ Ketil Tveiten, re: Charlie Frohman's comment. For what it's worth, I think this is a fine question. I don't think the answer would be obvious to every "research-level" mathematician, although I'm only a first-year graduate student myself... $\endgroup$
    – Vectornaut
    Commented Mar 19, 2010 at 19:52
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    $\begingroup$ @Charlie: Are you saying that serious researchers only study T1 spaces?!? $\endgroup$
    – François G. Dorais
    Commented Mar 19, 2010 at 23:25
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    $\begingroup$ I agree with Charlie, and here's why: I don't see off hand any good reason for caring what the name of such a space is. I mean, if you have examples of some of these spaces, and some result that says that this precise property is what you need for some application, then by all means, it should have a name, and knowing the conventional name will help you look up the appropriate literature. But as it is, I'd like some motivation before I'll like the question. $\endgroup$
    – Theo Johnson-Freyd
    Commented Mar 20, 2010 at 1:43

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In spaces where singleton points are closed, your property is equivalent to saying that the space has no isolated points. Or in other words, that it is perfect.

Clearly, no space with an isolated point can have your property. Conversely, when singletons are closed, then you can subtract one point from any open set and thereby have a proper open subset. So if U has at least 2 points x,y, then U = U-{x} union U-{y}, giving an instance with I of size 2.

However, your property does not imply that points are closed, since the space on reals R, where open sets have the form (-infty, a), has your property, but points are not closed in this space.

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  • $\begingroup$ Note that in the example space I give, no finite I would suffice, in contrast to the T1 perfect space situation, where I of size 2 suffices. $\endgroup$
    – Joel David Hamkins
    Commented Mar 20, 2010 at 2:26
  • $\begingroup$ Sort of going the other way (you gave a condition that implies the OP's condition), the condition implies that any local system, ordered by set inclusion, cannot have a smallest element. I think this is saying that local base cannot be finite? (This distinguishes your two examples of R with (-infty,a) and Z with (-infty,n). ) $\endgroup$
    – Willie Wong
    Commented Mar 20, 2010 at 14:01
  • $\begingroup$ Willie, this is ,clearly, saying that every local base can not be finite. $\endgroup$
    – Alexei Fedotov
    Commented Mar 20, 2010 at 15:01
  • $\begingroup$ Thanks for verifying my question (I am a bit rusty here). $\endgroup$
    – Willie Wong
    Commented Mar 20, 2010 at 15:44
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Are you just saying that the topology is an atomless lattice? I'd call it "a space with atomless topology".

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    $\begingroup$ Consider the integers Z, with open sets (-infty,n). This is a topology and is an atomless lattice, but it doesn't have the desired property. $\endgroup$
    – Joel David Hamkins
    Commented Mar 20, 2010 at 11:08
  • $\begingroup$ Hmm, good point. I would like to hear where these spaces come from. $\endgroup$
    – Andrej Bauer
    Commented Mar 20, 2010 at 20:00
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Isn't this just the Base of the topology?

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    $\begingroup$ Huh? Take the set {0,1} with the discrete topology. The subsets { {0}. {1}} form a base of the topology, and they don't satisfy the conditions given in the question. Perhaps you are thinking of something else? $\endgroup$
    – Willie Wong
    Commented Mar 20, 2010 at 13:39
  • $\begingroup$ Stupid me. I missed the word 'Proper'. $\endgroup$
    – Undergrad
    Commented Mar 20, 2010 at 13:59

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