Timeline for Wanted: chain of nowhere dense subsets of the real line whose union is nonmeagre, or even contains intervals
Current License: CC BY-SA 2.5
10 events
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| Jan 18, 2011 at 2:46 | comment | added | Joel David Hamkins | Andres, well, chains of meager sets must be meager if the cofinality of the chain is less than the additivity of the meager ideal (a cardinal characteristic appearing in Cichon's diagram), and this is clearly sharp, but not saying much. I'm not sure about the determinacy angle, but that is very interesting. | |
| Jan 18, 2011 at 2:13 | comment | added | Andrés E. Caicedo | @Joel, @George: Yes. But there ought to be a result there. The reason I ask is that under determinacy, one can easily check that well-ordered unions of meager sets are meager. Determinacy imposes a kind of definability restriction, but this usually translates into theorems in ZFC under appropriate assumptions (but I am curious whether we can say something that is not simply a use of, say, Borel determinacy in disguise) | |
| Jan 18, 2011 at 2:04 | comment | added | George Lowther | @Andres: The union of a chain of meagre sets does not have to be meagre. If it was, Zorn's lemma would imply the existence of a maximal meagre set, which can't exist. Also, assuming CH, the real line is a union of a chain of countable (hence, meagre) sets, as my comment to the original question showed. I'm not sure how far you can go without CH. | |
| Jan 18, 2011 at 1:55 | comment | added | Andrés E. Caicedo | @Joel: Your nice argument is very general. What can you say if the sets are meager (rather than nowhere dense)? | |
| Jan 18, 2011 at 1:17 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
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| Jan 18, 2011 at 0:52 | comment | added | Joel David Hamkins | Yes, George, you are right! I will edit. | |
| Jan 18, 2011 at 0:27 | comment | added | George Lowther | Nice. Doesn't this give a full answer to the question? That is, the union of a chain of nowhere dense sets is always meagre? I.e., suppose it isn't, so it's union is not nowhere dense, and replace $\mathbb{Q}$ in your argument by any countable dense subset of the union. | |
| Jan 18, 2011 at 0:21 | vote | accept | Michael | ||
| Jan 18, 2011 at 0:18 | comment | added | Andrés E. Caicedo | Oh, that's nice! | |
| Jan 18, 2011 at 0:07 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |