Timeline for Measure of rational hyperplanes of $\mathbb{R}$
Current License: CC BY-SA 4.0
15 events
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| Apr 17, 2019 at 11:57 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Apr 8, 2019 at 21:51 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Apr 8, 2019 at 18:12 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Apr 8, 2019 at 18:12 | comment | added | Pietro Majer | @Skeeve since you made me think, I got a simpler reason (added) | |
| Apr 8, 2019 at 18:06 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Apr 8, 2019 at 17:12 | comment | added | Andreas Blass | @AsafKaragila If the omitted basis vector, called $v_0$ in the problem, happens to be a rational number, then the subspace spanned by the other vectors is a set of representatives of $\mathcal R/\mathcal Q$, a Vitali set. If the omitted basis vector is something else, then you get a set of representatives of $\mathcal R/(v_0\mathcal Q)$, which I think still qualifies as a version of the Vitali set. Shrink it by a factor $v_0$ and it becomes a genuine Vitali set again. | |
| Apr 8, 2019 at 16:51 | comment | added | Skeeve | @PietroMajer could you please add some more details on why $V$ has infinite outer measure? | |
| Apr 8, 2019 at 16:43 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Apr 8, 2019 at 16:42 | comment | added | Asaf Karagila♦ | Nobody imposed a set of representatives to be bounded. I am simply claiming that there is a bounded one. And I'd be happy to see any reference to Vitali set meaning anything other than the set of representatives. | |
| Apr 8, 2019 at 16:41 | comment | added | Pietro Majer | I don't think there exists a standard definition of Vitali set. As far as I know, usually people say informally "Vitali set" alluding to a construction via choice of representatives. In any case I don't see why one should impose by definition that "a Vitali set" needs to be bounded, but if you prefer, let's say this is "a version of the Vitali set" then. | |
| Apr 8, 2019 at 16:33 | vote | accept | Arno | ||
| Apr 8, 2019 at 16:19 | comment | added | Asaf Karagila♦ | Of course, it also depends on your definition of $\Bbb R$, set, an "measure". But the standard definition of a Vitali set is a set of representatives for $\Bbb{R/Q}$. Which yes, it spans the vector space you mention, and you can argue that for that reason it is somehow equivalent. But that also changes some of its fundamental properties (e.g. being inside some small interval). | |
| Apr 8, 2019 at 16:18 | comment | added | Pietro Majer | That's depend on your definition of "Vitali set"; of course there are non measurable sets even into any measurable set with positive measure. A standard way to exhibit a non-Lebesgue measurable set is, a rational hyperplane of $\mathbb{R}$, which is essentially the original quotient construction. | |
| Apr 8, 2019 at 16:05 | comment | added | Asaf Karagila♦ | It's not a Vitali set, since it is closed under sums and products by a rational scalar, and a Vitali set is not (a priori) closed under such sum and products. Moreover, Vitali sets can be taken from arbitrarily small intervals so they can have arbitrarily small outer measure. | |
| Apr 8, 2019 at 16:02 | history | answered | Pietro Majer | CC BY-SA 4.0 |