Timeline for Does every set of reals contain a measure-zero set of the same cardinality? Does it contain a meager set of the same cardinality?
Current License: CC BY-SA 3.0
16 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 31, 2015 at 4:41 | answer | added | Boaz Tsaban | timeline score: 4 | |
| Oct 29, 2013 at 13:43 | vote | accept | Joel David Hamkins | ||
| Oct 12, 2013 at 15:56 | answer | added | Goldstern | timeline score: 12 | |
| Oct 12, 2013 at 3:29 | answer | added | Mohammad Golshani | timeline score: 17 | |
| Oct 11, 2013 at 0:27 | answer | added | Miha Habič | timeline score: 28 | |
| Oct 11, 2013 at 0:22 | comment | added | Joel David Hamkins | @Miha, your Luzin set idea also provides a negative answer to Ashutosh's question under CH or MA. Namely, let $A$ be any Luzin set of size continuum and let $B$ be an interval. These are both non-meager, but there can be no meager-preserving bijection between them, since $A$ has no uncountable meager subsets but $B$ does. I would encourage you to post the solution there. And using the measure-theoretic analogue of Luzin sets, it seems that one will also get a negative answer there in the measure case. | |
| Oct 11, 2013 at 0:12 | comment | added | Joel David Hamkins | Regarding Miha's answer: en.wikipedia.org/wiki/Luzin_space. Thus, CH or MA implies a negative answer to question 2. (Miha, please post.) And that Wikipedia page mentions that there is a measure analogue of Luzin sets, which will mean a negative answer also to question 1 under CH or MA. But could someone explain this analogue? | |
| Oct 11, 2013 at 0:04 | comment | added | Joel David Hamkins | Yes, I see. On my post on Ashutosh's question, I mention that when the original set has the perfect set property, then we can easily do it via Cantor sets. | |
| Oct 11, 2013 at 0:03 | comment | added | Asaf Karagila♦ | Joel, thank you for the encouragement, but this is really just an artifact of the perfect set property. So it's not that interesting after all. | |
| Oct 11, 2013 at 0:00 | comment | added | Joel David Hamkins | @Asaf, yes, I am thinking of the ZFC situation, but I would encourage you to post an answer with the $\neg\text{AC}$ situation if it is interesting. | |
| Oct 10, 2013 at 23:58 | comment | added | Joel David Hamkins | Miha, please post your comment as an answer. But I don't agree that there are only continuum many nowhere dense sets, since every subset of the Cantor set is nowhere dense, and there are $2^{2^{\aleph_0}}$ many such subsets. Perhaps you are speaking only of Borel sets? Does this idea still answer the question? | |
| Oct 10, 2013 at 23:58 | comment | added | Asaf Karagila♦ | It is consistent without choice (in Solovay's model) that this is the case, although that's not very helpful here I suppose. :-) | |
| Oct 10, 2013 at 23:55 | comment | added | Miha Habič | As a quick comment, MA (and, in particular, CH) imply that there are Luzin sets, which are sets of size continuum whose intersection with every meager set has size less than continuum, showing that Question 2 has a negative answer in this case. You build such sets by diagonalizing against all nowhere dense sets, of which there are continuum many. | |
| Oct 10, 2013 at 23:47 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Improved exposition
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| Oct 10, 2013 at 23:06 | history | asked | Joel David Hamkins | CC BY-SA 3.0 |