Timeline for How complicated is the formula expressing that a set is non-measurable?
Current License: CC BY-SA 3.0
11 events
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| Dec 8, 2013 at 0:03 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Dec 7, 2013 at 22:06 | comment | added | Joel David Hamkins | In any case, the property "$X$ is not measurable" is expressible using quantifiers only over natural numbers and reals, and so it has your desired form. So the family of non-measurable sets can be picked out by formulas of your favored type. | |
| Dec 7, 2013 at 22:05 | comment | added | Joel David Hamkins | @user38200, in my answer I am using the usual notion of projectively definable sets, where we have a projective property $\varphi(x)$ that picks out the members of the set $X$ that is being defined. You have used an idiosyncractic notion, where $\varphi(X)$ is a property of the set of reals itself, and for this, I'm not yet sure about the situation, whether one can define such a set this way that is provably non-measurable. The argument from large cardinals does not seem directly to answer it. | |
| Dec 7, 2013 at 19:50 | comment | added | Joel David Hamkins | Carlo, yes, of course, but this is still really only using reals as parameters. My point was that if the formula allows only quantification over $\omega$ and over $\mathbb{R}$, then the formulas simply cannot distinguish between any two infinite ordinals as parameters, since they have all the same elements below $\omega$ and in $\mathbb{R}$. So it is not sensible to have ordinal parameters directly. Rather, one wants to use codes for ordinals in the manner you are suggesting, with real parameters. | |
| Dec 7, 2013 at 19:48 | comment | added | user38200 | If $\varphi(x)$ is not a definition of a set of reals, but rather a property shared by a family of sets of reals, does the same reasoning applies? | |
| Dec 7, 2013 at 19:46 | vote | accept | user38200 | ||
| Dec 7, 2013 at 19:44 | comment | added | Rachid Atmai | In your first paragraph, concerning quantification over ordinals for projective formulae, one could allow quantification for $\alpha < \omega_1$ because countable ordinals are coded by reals (The $WO$ set). If $x^{\#}$ exists for every $x \in \mathbb{R}$ then one could also code ordinals up to $u_{\omega}$, the $\omega$-th uniform indiscernible, by reals (under $AD, u_{\omega}=\aleph_{\omega}$. | |
| Dec 7, 2013 at 18:27 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Dec 7, 2013 at 18:23 | comment | added | Joel David Hamkins | You are completely right, and I have now edited to correct this. | |
| Dec 7, 2013 at 18:19 | comment | added | Emil Jeřábek | You cannot assume $U$ and $V$ to consist of only finitely many intervals. This would in fact define Jordan–Peano measurability (consider $X=\mathbb Q$). | |
| Dec 7, 2013 at 18:05 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |