Timeline for Sigma Algebra that is not a topology [closed]
Current License: CC BY-SA 3.0
12 events
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| Jul 12, 2011 at 21:55 | comment | added | Joel David Hamkins | Clinton, why not ask the question explicitly as an MO question: Is it consistent with ZF that every $\sigma$-algebra is a topology? François's answer with Gitik's model nearly answers it, and may very well provide a full answer, if we dig a bit deeper into it. | |
| Jul 12, 2011 at 20:48 | comment | added | François G. Dorais | @Joel: Yes, my old answer only covers the case when the $\sigma$-algebra contains all the singletons. I don't see what could go wrong in the general case, but my vision is often blurry without my AC goggles... | |
| Jul 12, 2011 at 20:45 | comment | added | François G. Dorais | @Clinton: I just added an update to my old answer. The wellordered case is much easier to prove, but the fact holds for all sets. | |
| Jul 12, 2011 at 20:28 | comment | added | Joel David Hamkins | Clinton, I was working (habitually) in ZFC, and thinking about the usual $\sigma$-algebra of Borel sets, which are a counterexample once you know there is a non-Borel set. But you are right that this cannot be proved in ZF, and so your comment makes an interesting question! Namely, is it consistent with ZF that every $\sigma$-algebra is a topology? I'm not sure if François's answer in his linked question provides the answer, but it is surely very relevant. | |
| Jul 12, 2011 at 20:24 | comment | added | Clinton Conley | @François: Oh, and I forgot to say thanks for the link! | |
| Jul 12, 2011 at 20:23 | comment | added | Clinton Conley | @François: Of course I'm sure $\mathtt{ZFC}$ is the assumed framework of the question (or at least some countable choice principle, since $\sigma$-algebras get hideous without them), but the comments below the answer got me thinking in this direction. Anyway, the answer to that question seems only to cover wellorderable sets, right? | |
| Jul 12, 2011 at 20:09 | comment | added | François G. Dorais | @Clinton: I'm pretty sure Joel (like most of us) was working in ZFC. However, you are correct: see this older MO question - mathoverflow.net/questions/33028/… | |
| Jul 12, 2011 at 19:31 | comment | added | Clinton Conley | @Joel: I'm probably being silly, but is it clear that in $\mathtt{ZF}$ this gives the requested example? More precisely, can you pin down a set $A$ such that the $\sigma$-algebra $\sigma(\{\{a\} : a \in A\})$ generated by singletons of elements of $A$ forms a proper subset of $\mathcal{P}(A)$? Things seem to get tricky in models where all uncountable cardinals are singular. | |
| Jul 12, 2011 at 18:18 | history | closed |
Joel David Hamkins Gerald Edgar user9072 Qiaochu Yuan Qfwfq |
off topic | |
| Jul 12, 2011 at 17:25 | answer | added | Marc Palm | timeline score: 3 | |
| Jul 12, 2011 at 16:13 | comment | added | Joel David Hamkins | This question might be better suited to math.stackexchange. Note: if your $\sigma$-algebra includes singletons, as many do, then if it were a topology, it would have to include all subsets of the space. | |
| Jul 12, 2011 at 16:08 | history | asked | Claudia Brave | CC BY-SA 3.0 |