Timeline for Sets of Vitali's type in models of $\mathsf{ZF}+\mathsf{GCH}$ where $V \neq L$
Current License: CC BY-SA 3.0
14 events
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| S May 24, 2016 at 7:12 | history | edited | Michael Renardy | CC BY-SA 3.0 |
fixed some weird MathJax stuff; retagging; included defn of "set of Vitali type" as seems to be intended from the comments.
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| S May 24, 2016 at 7:12 | history | suggested | user642796 | CC BY-SA 3.0 |
fixed some weird MathJax stuff; retagging; included defn of "set of Vitali type" as seems to be intended from the comments.
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| May 24, 2016 at 6:32 | review | Suggested edits | |||
| S May 24, 2016 at 7:12 | |||||
| Dec 31, 2014 at 12:16 | comment | added | Thomas Benjamin | (cont.) satisfy this criterion your argument will work. | |
| Dec 31, 2014 at 12:14 | comment | added | Thomas Benjamin | $V$ is a selector with respect to translation by $G$ (which you indicated in your comments), selecting one element from each equivalence class, then $V$ will be nonmeasurable (at least it is my understanding now, thanks to Prof. Hamkins' answer to the mathoverflow question "Generalizing Vitali Sets to uncountable dense subgroup selectors..."). Of course $\mathscr N$ is the null ideal and the additivity number $add$$($$\mathscr N$$)$ is the largest cardinal such that the union of of fewer than $add$$($$\mathscr N$$)$ many measure zero sets still has measure zero, so as long as your subgroups | |
| Dec 31, 2014 at 11:12 | comment | added | Thomas Benjamin | @Avshalom: Thanks, that is very helpful. It is my understanding that the classical Vitali argument shows that if $G$ is any subgroup of $\mathbb R$ of size less than the additivity number $add$$($$\mathscr N$$)$ and | |
| Dec 31, 2014 at 0:26 | comment | added | Avshalom | If so, then the countable subgroup $\langle G, \tau \rangle$, where $\tau$ is a generic real, should give rise to new Vitali sets that are not in the ground model too. | |
| Dec 30, 2014 at 22:26 | comment | added | Avshalom | Let $G$ be a countable subgroup. If $x_E$ is the equivalence class of $x \in \mathbb{R}$, then $\mathbb{R} = \bigcup_{g \in G}(x_E + g)$; now apply $\sigma$-additivity and translation invariance of Lebesgue measure. Will that work? The argument will apply more generally. | |
| Dec 30, 2014 at 22:11 | comment | added | Thomas Benjamin | (With respect to $\mathbb Q$, that is....) | |
| Dec 30, 2014 at 22:04 | comment | added | Thomas Benjamin | @Avshalom: Exactly so. What you wrote is the definition of sets of Vitali's type, relative to $G$. As you point out, $\mathbb Q$ is the usual subgroup of $($$\mathbb R$,$+$$)$ used to define "Vitali sets" . My question to you is, does any choice for $G$ make the set of representatives under the equivalence relation $x$$-$$y$$\in$$G$ nonmeasurable with respect to the Lebesgue measure? The phrase "sets of Vitali's type" is used as a synonym for "Vitali sets". | |
| Dec 30, 2014 at 18:52 | comment | added | Avshalom | Could you clarify what is meant by a set of Vitali type? For example, I think if $G$ is any countable subgroup of $(\mathbb{R}, +)$, then choosing representatives under the equivalence relation $x - y \in G$ will give a "Vitali set relative to $G$". The rationals are one instance. | |
| Dec 30, 2014 at 15:18 | vote | accept | Thomas Benjamin | ||
| Dec 30, 2014 at 14:24 | answer | added | Joel David Hamkins | timeline score: 5 | |
| Dec 30, 2014 at 11:26 | history | asked | Thomas Benjamin | CC BY-SA 3.0 |