Timeline for Non-measurable sets and Determinacy...
Current License: CC BY-SA 3.0
16 events
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| Jun 10, 2011 at 19:14 | comment | added | Ali Enayat | There is an old beautiful argument of Sierpinski that shows that the axiom of choice for PAIRS $AC_2$ implies the existence of a nonmeasurable set; which in turn shows that $AD$ fails as soon as $AC_2$ is true. So a model in which $AC$ fails but $AC_2$ holds is yet another way of seeing that the negation of $AD$ does not imply $AC$. Cohen's so-called "first (symmetric) model" in which $AC$ fails is such a model (since every set has can be linearly ordered in that model, but the reals cannot be well-ordered there). | |
| Jun 10, 2011 at 12:12 | comment | added | Joel David Hamkins | That seems natural. Or perhaps $L(P(\mathbb{R}))$ in a suitable forcing extension, such as after adding a subset to $\omega_1$ by initial segments. | |
| Jun 10, 2011 at 10:51 | comment | added | Emil Jeřábek | (The point being that there exists a non-determined game whenever $2^\omega$ is well ordered.) | |
| Jun 10, 2011 at 10:28 | comment | added | Emil Jeřábek | Take a model of ZFC, then take its generic extension that preserves the reals, and then take its nontrivial symmetric submodel. | |
| Jun 10, 2011 at 10:00 | vote | accept | George Lazou | ||
| Jun 10, 2011 at 9:48 | vote | accept | George Lazou | ||
| Jun 10, 2011 at 10:00 | |||||
| Jun 10, 2011 at 2:16 | comment | added | Joel David Hamkins | But I would encourage you to ask that as a question, since perhaps someone knows a good model, and I would be interested to see it. | |
| Jun 10, 2011 at 0:29 | comment | added | Joel David Hamkins | As for your final question, I think you mean to ask whether the existence of a non-determined game is weaker than AC, rather than stronger, since ZFC proves the existence of such non-determined games for Cantor space in exactly the same way that it does for Baire space. I don't expect the existence of such a non-determined set to imply full AC, but I'd have to think a bit more to give a precise model showing this. If this is right, then the existence of non-determined sets would be a weak choice principle. | |
| Jun 10, 2011 at 0:24 | comment | added | Joel David Hamkins | I am not using the Cantor set as a subset of $\mathbb{R}$ and the Lebesgue measure on that, but rather, using the natural probability measure on $2^\omega$, for which $2^\omega$ is the whole space and has measure $1$; the measure of the basic open set determined by a finite binary sequence of length $n$ has measure $\frac{1}{2^n}$, like flipping a coin. A subset of $2^\omega$ which is $0$ in every fourth digit has measure $0$ (an easy calculation). The same idea works in Baire space, but you seem to be mapping spaces by homeomorphisms that may not be measure-preserving... | |
| Jun 9, 2011 at 23:51 | comment | added | George Lazou | I'm not convinced... Cantor Space is homeomorphic to the Cantor Set which has measure zero... Thus any payoff set in it will have measure zero when considered as a subset of $\mathbb{R}$ by virtue of being a subset of the Cantor Set... I'm not convinced that your argument about halving the measure carries through to non-measurable sets... further my guess (I should probably ask this as a question) is that the existence of a non-determined game in Cantor Space is stronger than AC... | |
| Jun 9, 2011 at 21:12 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Jun 9, 2011 at 19:35 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Jun 9, 2011 at 19:22 | history | undeleted | Joel David Hamkins | ||
| Jun 9, 2011 at 19:22 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Jun 9, 2011 at 19:10 | history | deleted | Joel David Hamkins | ||
| Jun 9, 2011 at 19:09 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |