Timeline for Do the empty set AND the entire set really need to be open?
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Mar 25, 2010 at 0:16 | comment | added | Harry Gindi | Yes, I meant what Qiaochu just said I meant. | |
| Mar 25, 2010 at 0:04 | comment | added | M.G. | As zeb pointed out, I do not forbid the entire set and the empty set to be open. I do not require it either. | |
| Mar 24, 2010 at 23:50 | comment | added | Qiaochu Yuan | I think fpqc is trying to correct zeb's example, although I'm not sure; it is a reasonable "topology" in which the empty set is not open. | |
| Mar 24, 2010 at 23:31 | history | edited | KConrad | CC BY-SA 2.5 |
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| Mar 24, 2010 at 23:27 | comment | added | KConrad | fpqc, how does finiteness play a role here? In most natural examples the (nonempty) finite sets are not open, so are you suggesting this could a reason not to consider the empty set as open, so it's on the same footing as the other finite subsets? | |
| Mar 24, 2010 at 23:25 | comment | added | KConrad | Andrea, I was giving as an example one subset by itself. That would fit the conditions of a "topology" without the first axiom, but then the space doesn't admit an open covering in that "topology". It's hard to do much at all without open coverings. | |
| Mar 24, 2010 at 23:19 | comment | added | Harry Gindi | The empty set is vacuously a finite set by any definition of finite... | |
| Mar 24, 2010 at 23:18 | comment | added | Andrea Ferretti | I also think this does not answer the question. You are just saying that for a space admitting two disjoint open sets, this axiom follows from the intersection axiom. So only irreducible topological spaces remain. For these, calling the empty set open or not more is a matter of conventions. But I guess this could change the notion of continuos function for non surjective functions. | |
| Mar 24, 2010 at 23:16 | history | edited | KConrad | CC BY-SA 2.5 |
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| Mar 24, 2010 at 23:06 | comment | added | zeb | I don't think he wants to require that the empty set is never open, but rather to lift the restriction that it must be open. For instance, in algebraic geometry - say the theory of algebraic curves - a lot of statements might become simpler if we said that the closed sets are exactly the finite sets of points. This would be consistent, since the intersection of any two nonempty open sets on an irreducible topological space is nonempty. | |
| Mar 24, 2010 at 22:55 | comment | added | Qiaochu Yuan | Similarly, it can happen that two open sets have union the entire set, and we don't want to say that the union of open sets is open or the entire set... | |
| Mar 24, 2010 at 22:49 | history | answered | KConrad | CC BY-SA 2.5 |