Timeline for Real-valued measurability vs. Two-valued measurability in determining whether $CH$ holds or not
Current License: CC BY-SA 3.0
26 events
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| Sep 13, 2017 at 18:47 | comment | added | Noah Schweber | @ThomasBenjamin $L$ is a submodel of $M[G]$. It's not an elementary submodel (since it satisfies $V=L$ but $M[G]$ doesn't), but that's different.. | |
| Sep 13, 2017 at 16:34 | comment | added | Thomas Benjamin | @NoahSchweber: Since the forcing extension $\mathfrak M$[$G$] of $L$ was constructed to satisfy $\lnot$$CH$, $\mathfrak M$[$G$] $\vDash$ $ZFC$ + $\lnot$$CH$, so though $L$ $\vDash$ $ZFC$ (or, like $HF$ $\vDash$ $ZFC$ + $\lnot$_Infinity_, $L$ $vDash$ $ZFC$ + $V$=$L$), $ZFC$ is a fragment of $ZFC$ + $\lnot$$CH$ (but since $L$ $\vDash$ $ZFC$ + $V$=$L$, $L$ in that sense cannot be a submodel of $\mathfrak M$[$G$] (but that interpretation would correspond to $L$ not being an elementary substructure of $\mathfrak M$[$G$], correct?). | |
| Sep 13, 2017 at 15:09 | comment | added | Noah Schweber | Re: your second comment, adding random reals preserves cardinalities - since $\kappa>\omega_1$ in $M$, it remains $>\omega_1$ in $M[G]$, and since $(2^{\aleph_0})^{M[G]}=\kappa$ we have $M[G]\models\neg CH$. But you already know it has to satisfy $\neg$CH, since ZFC + CH contradicts "the continuum is real-valued measurable" and we've made $\kappa$ be the continuum and stay real-valued measurable. | |
| Sep 13, 2017 at 15:06 | comment | added | Noah Schweber | @ThomasBenjamin No, $L$ (or rather, $L^M=L^{M[G]}$) is an inner model of $M[G]$ - I don't know why you think it isn't. Whenever $N$ is a model of ZFC, $L^N$ is an inner model of $N$: it is a transitive (as far as $N$ is concerned) subclass of $N$ satisfying the ZFC axioms and containing all the ordinals in $N$. It's not generally an elementary substructure - if $N\not=L^N$, then $L^N\not\preccurlyeq N$ since $L^N\models V=L$ but $N\not\models V=L$ - but nobody said it had to be. | |
| Sep 13, 2017 at 15:01 | comment | added | Thomas Benjamin | @NoahSchweber: So in Andres' comment to you, the inner model in which $\kappa$ becomes two-valued measurable also satisfies $\lnot$$CH$? | |
| Sep 13, 2017 at 14:52 | comment | added | Thomas Benjamin | So basically, given a forcing extension $\mathfrak M$[$G$] of $L$ in which $\lnot$$CH$ holds, $L$ is not an inner model of $\mathfrak M$[$G$] (since $\mathfrak M$[$G$] $\vDash$ $ZFC$ + $\lnot$$CH$)? | |
| Sep 13, 2017 at 14:44 | comment | added | Noah Schweber | @AndrésE.Caicedo Oh yeah, d'oy. | |
| Sep 13, 2017 at 14:43 | comment | added | Andrés E. Caicedo | @Noah Just add random reals. | |
| Sep 13, 2017 at 14:37 | comment | added | Noah Schweber | @ThomasBenjamin "it seems reasonable (possibly) to (informally) infer that there is a [model satisfying] "The cardinality of the continuum is a real-valued measurable cardinal" in which CH holds." That can't happen - CH contradicts the real-valued measurability of the continuum. | |
| Sep 13, 2017 at 14:36 | comment | added | Thomas Benjamin | @NoahSchweber: I agree. That is an interesting question! | |
| Sep 13, 2017 at 14:33 | comment | added | Noah Schweber | An interesting question is whether, for $\kappa$ a measurable cardinal, there's a forcing which will kill the strong inaccessibility (when I said "inaccessible" above I meant "strongly inaccessible") of $\kappa$ (in particular, push the continuum above $\kappa$) but preserve the real-valued measurability of $\kappa$; I suspect the answer is yes, but I don't immediately see how to do it. @AndrésE.Caicedo Do you have any thoughts on this? | |
| Sep 13, 2017 at 14:32 | comment | added | Thomas Benjamin | (cont.) cardinal implies $\lnot$$CH$, it seems reasonable (possibly) to (informally) infer that there is a submodel $\mathfrak N$ of $\mathfrak M$$\vDash$ $ZFC$ + "The cardinality of the continuum is a real-valued measurable cardinal" in which $CH$ holds (and $\kappa$ is two-valued measurable)(I hope I have made no errors in what I wrote here). | |
| Sep 13, 2017 at 14:27 | comment | added | Noah Schweber | @ThomasBenjamin Measurable cardinals are real-valued measurable already. More precisely: if memory serves, a cardinal is measurable iff it is real-valued measurable and inaccessible (this should be in Jech's big book, but I don't have it on hand). | |
| Sep 13, 2017 at 14:20 | comment | added | Thomas Benjamin | @NoahSchweber: Thanks for clearing up that misconception of mine. I find the following comment of Andres intriguing: "For question 2, your recollection of what Solovay proved is correct: If there is a real-valued measurable cardinal $\kappa$, then in an inner model [of some forcing extension?--my comment] $\kappa$ is genuinely measurable". Question: If there is a two-valued measurable cardinal $\kappa$, then is there an inner model (of a forcing extension or otherwise) in which $\kappa$ is real-valued measurable? Since the assumption of the existence of a real-valued measurable | |
| Sep 13, 2017 at 13:22 | comment | added | Noah Schweber | @ThomasBenjamin That's not what equiconsistency means; $T_1$ and $T_2$ are equiconsistent if PRA (say) proves "$Con(T_1)\iff Con(T_2)$," not if $T_1$ proves $Con(T_2)$ and vice versa. (In fact, we can never have them prove each others' consistency if they're reasonable theories - see this MSE question.) Completeness says that we can replace "is consistent" with "has a model," of course, but I don't see how that's relevant here. Regardless, this doesn't seem to have anything to do with the question you're asking here. | |
| Sep 13, 2017 at 11:35 | comment | added | Andrés E. Caicedo | @Thomas No, that is not a correct restatement, because of the incompleteness theorem (among other reasons). (And yes, equiconsistency is an equivalence relation.) | |
| Sep 13, 2017 at 5:04 | comment | added | Thomas Benjamin | @NoahSchweber: Since by one version of G$\ddot o$del's completeness theorem ("A (first-order) theory $T$ is consistent iff it has a model"), equiconsistency merely states that two first-order theories $T_{\alpha}$ and $T_{\beta}$ are equiconsistent iff each proves the other has a model, correct? And equiconsistency is an equivalence relation on first-order theories $T_{\alpha}$, $T_{\beta}$ (correct also)? | |
| Sep 13, 2017 at 4:13 | vote | accept | Thomas Benjamin | ||
| Sep 13, 2017 at 3:42 | history | edited | Noah Schweber | CC BY-SA 3.0 |
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| Sep 13, 2017 at 3:40 | comment | added | Andrés E. Caicedo | It seems to be the fastest argument: If $I$ is the witnessing ideal, then $\kappa$ is measurable in $L[I]$. I of course like to argue instead as in my paper: If $\kappa$ is real-valued measurable, then there is a generic embedding of the universe into a transitive model, which in particular gives us $0^\sharp$. | |
| Sep 13, 2017 at 3:40 | history | edited | Noah Schweber | CC BY-SA 3.0 |
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| Sep 13, 2017 at 3:37 | comment | added | Noah Schweber | @AndrésE.Caicedo Sweet, I was paying attention! Is there an easier way to get countability of $\mathbb{R}^L$ from a real-valued measurable than this? | |
| Sep 13, 2017 at 3:35 | comment | added | Andrés E. Caicedo | For question 2, your recollection of what Solovay proved is correct: If there is a real-vaued measurable cardinal $\kappa$, then in an inner model $\kappa$ is genuinely measurable. So $0^\sharp$ exists, which is more than an overkill to see that $\mathbb R\cap L$ is countable. | |
| Sep 13, 2017 at 3:33 | comment | added | Noah Schweber | @AndrésE.Caicedo No, no, it is provable in ZFC since ZFC is clearly inconsistent :P. | |
| Sep 13, 2017 at 3:31 | comment | added | Andrés E. Caicedo | Re your first paragraph. Well, yes, it is the case. But it is not provable in $\mathsf{ZFC}$. :-) (Thomas, this comment is sort-of an inside joke, I am not actually contradicting anything Noah wrote.) | |
| Sep 13, 2017 at 3:28 | history | answered | Noah Schweber | CC BY-SA 3.0 |