Timeline for Perfect subset of a non-null set
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Aug 11, 2021 at 14:25 | comment | added | Haim | One more comment about Shelah's proof: it's not hard to see that the non-measurable set that he constructs (under the assumption $\omega_1=\omega_1^{L[x]}$) does contain a perfect subset (the easiest way is to note that it's a set of reals closed under finite changes). So I suspect that a straightforward modification of Shelah's proof will not be enough. | |
| Aug 11, 2021 at 5:35 | comment | added | Haim | I didn’t claim that we’re guaranteed to have a $y$ with $A_y$ being non-null (indeed, this is false). I was rather trying to imitate the beginning of Shelah’s proof, where the sets $A_y$ are replaced with certain unions of Borel null sets from models of the form $L[x]$. The easy case is when one of them is non-null and the difficult case is when all unions in question turn out to be null. Unfortunately, I don’t see a straightforward modification of the rest of his proof. | |
| Aug 11, 2021 at 5:19 | comment | added | 喻 良 | It seems unlikely, unless I misunderstand your idea. By c.c.c forcing over $L$, we may have Martin's axiom, $\neg CH$, and preserve $\omega_1^L$. Then every $\Sigma^1_2$-set is measurable. So $A_y$ is null for every $y$. | |
| Aug 11, 2021 at 4:40 | comment | added | Haim | It's not clear to me how to modify Shelah's proof to answer the above question. A possible way to start is the following: assuming that $\omega_1=\omega_1^{L[x]}$ for some real $x$, I wonder what can be said about the measure of the set $A_x=\{ z\in WO \cap L[x] : z$ is the $<_{L[x]}$-minimal real with rank $||z|| \}$. If the existence of $x$ such that $\omega_1=\omega_1^{L[x]}$ implies the existence of some $y$ such that $\omega_1=\omega_1^{L[y]}$ and $A_y$ is non-null, then we're done. Otherwise, we are in a situation quite analogous to the non-trivial case in Shelah's proof. | |
| Aug 9, 2021 at 5:20 | comment | added | 喻 良 | @JeremyRickard too lazy | |
| Aug 8, 2021 at 13:47 | comment | added | Jeremy Rickard | @喻良 Why not? $\,$ | |
| Aug 8, 2021 at 10:20 | comment | added | 喻 良 | I understand that there are many typos and misleading statements, but I will not correct them. | |
| Aug 8, 2021 at 7:36 | comment | added | Jason Zesheng Chen | @AndreasBlass right, that was a silly reading I made. Thanks | |
| Aug 8, 2021 at 7:24 | comment | added | YCor | Should "set" be interpreted as "subset of the reals" in this question? | |
| Aug 8, 2021 at 3:34 | comment | added | Andreas Blass | @JasonZeshengChen In view of the context and standard usage in set theory, I assume "non-null set" just means a set that doesn't have Lebesgue measure zero. So it might be a non-measurable set. | |
| Aug 8, 2021 at 1:28 | history | edited | 喻 良 | CC BY-SA 4.0 |
deleted 65 characters in body
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| Aug 8, 2021 at 0:18 | history | asked | 喻 良 | CC BY-SA 4.0 |