Timeline for What do you call a topology that is closed under arbitrary intersections?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Mar 31, 2010 at 1:28 | answer | added | LSpice | timeline score: 2 | |
| Mar 16, 2010 at 17:50 | comment | added | Harald Hanche-Olsen | @Gerald: How so? I am confused. | |
| Mar 16, 2010 at 17:43 | comment | added | Gerald Edgar | Harald: the reason I asked was that your original example usually does not have the whole space open. | |
| Mar 16, 2010 at 16:44 | comment | added | Harald Hanche-Olsen | @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. | |
| Mar 16, 2010 at 16:33 | comment | added | Gerald Edgar | 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". | |
| Mar 16, 2010 at 16:32 | comment | added | Harald Hanche-Olsen | Right; I noticed that in the Wikipedia article linked to from the answer. | |
| Mar 16, 2010 at 16:14 | comment | added | Joel David Hamkins | 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. | |
| Mar 16, 2010 at 16:12 | vote | accept | Harald Hanche-Olsen | ||
| Mar 16, 2010 at 16:12 | history | edited | Harald Hanche-Olsen |
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| Mar 16, 2010 at 16:09 | answer | added | Mariano Suárez-Álvarez | timeline score: 16 | |
| Mar 16, 2010 at 16:07 | history | asked | Harald Hanche-Olsen | CC BY-SA 2.5 |