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An arbitrary union, or a finite intersection, of open sets in a topological space is again open. What name is given to the hypothetical property that an arbitrary intersection of open sets is open?

As an example, consider a partially ordered set $X$. Call a subset $U\subseteq X$ open if $y\le x\in U$ implies $y\in U$. (Bonus question: Are there other interesting examples?)

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It suffices in your example for the order relation to be only a pre-order. That is, you can allow x <= y <= x for distinct x, y. With this addition, the property is fully equivalent. Define x <= y if x is in every open set that y is in. –  Joel David Hamkins Mar 16 '10 at 16:14
Right; I noticed that in the Wikipedia article linked to from the answer. –  Harald Hanche-Olsen Mar 16 '10 at 16:32
Does "arbitrary" intersection should include "empty" intersection? Then the whole space must be "open". Similarly, arbitrary union means the empty set is "open". This is what I would call a "complete lattice of sets". –  Gerald Edgar Mar 16 '10 at 16:33
@Gerald: Yes. But the whole space, and the empty set, are already open per the definition of topology. Surely, “complete lattice of sets” is correct, but not so good when I wish to emphasize the topology aspect. So I'll stick with Alexandrov space. –  Harald Hanche-Olsen Mar 16 '10 at 16:44
Harald: the reason I asked was that your original example usually does not have the whole space open. –  Gerald Edgar Mar 16 '10 at 17:43
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2 Answers

up vote 14 down vote accepted

Alexandrov spaces.

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Well, that was fast. Thanks. –  Harald Hanche-Olsen Mar 16 '10 at 16:12
Alexandrov topology and Alexandrov-discrete space are OK, but "Alexandrov spaces" is mostly used for something else... –  Anton Petrunin Mar 16 '10 at 17:40
Peter May uses ‘Alexandrov space’ (really, ‘A-space’): math.uchicago.edu/~may/MISC/FiniteSpaces.pdf. –  L Spice Mar 31 '10 at 1:22
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Harald, as Peter May observes in his notes on finite topological spaces (but as must already be standard), your pre-ordered sets provide the only examples. Indeed, if $X$ is an Alexandrov space (or Alexandrov-discrete space, whatever the terminology is), then we may impose a pre-order on it by demanding for $x, y \in X$ that $x \le y$ if and only if $x$ lies in every neighbourhood of $y$. This becomes a genuine partial order exactly when $X$ is $T_0$.

(Sorry; I didn't realise until after posting that Joel had already said this.)

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In general, this is not a partial order, but only a partial pre-order, since it may be that x and y are distinct, but lie in all the same open sets together. But with pre-orders, what you say is true (and was observed previously in the comments to the question). –  Joel David Hamkins Mar 31 '10 at 1:33
Joel, thanks; I'm sorry that I missed your comment. I also forgot that we weren't assuming the spaces were $T_0$. Should I delete this answer? –  L Spice Mar 31 '10 at 1:38
Oh, its no problem. Probably its fine to keep the answer, if you correct it to explain about the pre-order business. But I think this issue also appears on the Wikipedia page. –  Joel David Hamkins Mar 31 '10 at 1:45
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