Timeline for Translates of null sets
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 27, 2015 at 3:20 | answer | added | Paul Larson | timeline score: 8 | |
| Feb 23, 2015 at 21:42 | answer | added | fedja | timeline score: 22 | |
| Feb 22, 2015 at 10:09 | comment | added | Pietro Majer | If the countable set A is given, that's equivalent to: "there exists a maximum null set", which is obviously false... | |
| Feb 22, 2015 at 8:42 | comment | added | Mario Carneiro | @bof Ah, now I see I misread Null's question as "there exists a countable set $A$ of translations of $N$ such that every null set is a subset of $N+x$ for some $x\in A$", which is much weaker but also not obvious. Indeed "covered by" implies that you are taking unions, and I guess the countable set is not specified in advance, so actually it is the same question. | |
| Feb 22, 2015 at 7:01 | comment | added | bof | @MohammadGolshani: Was that comment addressed to me? If so, what is your point? It suffices that every $G_\delta$ null set is covered by the union of countably many translates of $N$, and there are only continuum many $G_\delta$ null sets. The fact that there are many more null sets seems irrelevant. | |
| Feb 22, 2015 at 6:54 | comment | added | bof | @MarioCarneiro: I am mystified. What's the difference between "every null set is covered by countably many translates of N" (Null's question) and "every null set is a subset of a countable union of translates of N" (the question I asked)? | |
| Feb 22, 2015 at 5:55 | comment | added | Mario Carneiro | @bof It is not obvious to me that a nullset $N$ whose closure under subsets, countable unions, and all (uncountably many) translates induces a nullset $N'$ such that the same is true with only subsets and countably many translates. | |
| Feb 22, 2015 at 2:23 | comment | added | bof | @MarioCarneiro Looks like the same question to me. What is the difference? | |
| Feb 22, 2015 at 1:44 | comment | added | Mario Carneiro | @Ashutosh The idea I had in mind was to use that $aC+b$ has measure zero in the measure on $C$ for almost all $a$. (I'm sure the esteemed folks here can much more easily tell if this is false or useless, though.) | |
| Feb 22, 2015 at 1:40 | comment | added | Mario Carneiro | @bof This question is actually quite a bit weaker than the version you asked me at MSE, which allows all translates of $N$, and allows countable unions as well. My proof sketch should apply here, if the gaps are fixable. | |
| Feb 21, 2015 at 22:25 | comment | added | Pietro Majer | I was vaguely thinking to Hausdorff measures w.r.to gauge functions. One needs to know that, given $N$, there is $\phi=o(t)$ (for $t\to0$) such that $H^\phi(N)=0$. So there is still room for a $\psi$, $\phi(t)<\psi(t)<t$ such that there are strict inclusions of the classes of null sets of $H^\phi\subset H^\psi\subset H^1$ (Some close claim is made here en.wikipedia.org/wiki/Hausdorff_measure#Generalizations) | |
| Feb 21, 2015 at 21:56 | comment | added | Ashutosh | Hi Pietro, Could you elaborate a little on how you intend to construct N'? Thanks. | |
| Feb 21, 2015 at 21:55 | comment | added | bof | This answer (to a somewhat different question) on Math Stack Exchange sketches what is claimed to be a proof that no such $N$ exists. | |
| S Feb 21, 2015 at 21:41 | history | suggested | Sheman |
The answer may well be independent of ZFC
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| Feb 21, 2015 at 21:31 | review | Suggested edits | |||
| S Feb 21, 2015 at 21:41 | |||||
| Feb 21, 2015 at 21:14 | comment | added | Pietro Majer | I don't think so. Given a Lebesgue null $N$, I think there is another Lebesgue null set $N'$ and a translation invariant measure $\mu$ such that $\mu(N)=0$ and $\mu(N')>0$. | |
| Feb 21, 2015 at 21:01 | history | edited | Joonas Ilmavirta | CC BY-SA 3.0 |
This question is not about set theory.
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| Feb 21, 2015 at 20:42 | review | First posts | |||
| Feb 21, 2015 at 20:46 | |||||
| Feb 21, 2015 at 20:40 | history | asked | Null | CC BY-SA 3.0 |