Timeline for Real-valued measurability vs. Two-valued measurability in determining whether $CH$ holds or not
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16 events
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| Sep 18, 2017 at 23:39 | comment | added | Noah Schweber | @ThomasBenjamin If you add $\kappa$-many reals, then $\kappa\le2^{\aleph_0}$, so that will kill strong inaccessibility. Weak inaccessibility, though, can't be killed just by adding "enough reals:" forcing with finite partial maps from $\lambda\times\omega$ to $2$ adds $\lambda$-many reals but is c.c.c., so preserves cardinals and cofinalities. | |
| Sep 18, 2017 at 10:43 | comment | added | Thomas Benjamin | @NoahSchweber: Regarding real-valued measurables becoming 'small', Cantor's attic (upper attic) says of weakly inaccessible cardinals,"...but forcing arguments show that any [strongly?] inaccessible cardinal can become a non-inaccessible weakly inaccessible cardinal in a forcing extension, such as after adding an enormous number of Cohen reals ..." (e.g. Jech-Solovay's Theorem 22.1(ii) from Jech, Set Theory (2003), pg. 410). What "enormous number" would that be, exactly (it couldn't be a class-size forcing, because that would negate the Powerset axiom....)? | |
| Sep 17, 2017 at 1:37 | comment | added | Noah Schweber | @ThomasBenjamin You ask "How different can these principles be" - well, there's no precise definition of that, but the following facts paint a pretty complete picture: the measurables are exactly the inaccessible real-valued measurables, while real-valued measurables can be "small" (e.g. size continuum). That said, particular values being real-valued measurable can have arithmetic implications - namely, if the continuum is real-valued measurable then CH fails - while this is moot for actual measurables (since they can't be small). If this isn't satisfying, I'll need more specific questions. | |
| Sep 17, 2017 at 1:30 | comment | added | Noah Schweber | @ThomasBenjamin "can one construct forcing extensions of that inner model where CH, $\neg$CH hold respectively and $\kappa$ remains two-valued measurable?" Yes, and this has nothing to do with the inner model in question: measurability (unlike real-valued measurability) is "preserved by small forcing," i.e. if $\kappa$ is measurable then it stays measurable in any forcing extension by a forcing of cardinality $<\kappa$ (this is due to Levy and Solovay). Since CH and $\neg$CH can each be forced with small forcings, this means that we can always control CH while preserving a measurable. | |
| Sep 17, 2017 at 0:42 | comment | added | Thomas Benjamin | @NoahSchweber: Correct. No tension at all but regarding that inner model (where $\kappa$ is two-valued measurable), can one construct forcing extensions of that inner model where $CH$, $\lnot$$CH$ hold respectively and $\kappa$ remains two-valued measurable? What has to happen for that to occur (if in fact it does happen)? | |
| Sep 17, 2017 at 0:24 | comment | added | Noah Schweber | @ThomasBenjamin "the real-valued measurable cardinal $\kappa$ in the forcing extension can be a two-valued measurable cardinal in an inner model of that same forcing extension." Yes, $\kappa$ is inaccessible in that inner model, but not in the whole forcing extension. Which is why $\kappa$ is measurable in the inner model, but merely real-valued measurable in the whole forcing extension. There is no tension here at all. | |
| Sep 17, 2017 at 0:18 | comment | added | Thomas Benjamin | (cont.) How different can these principles be (and in what sense can these principles be different) if one prove a theorem like Theorem 22.1(i, ii) (and one of course can....)? | |
| Sep 17, 2017 at 0:14 | comment | added | Thomas Benjamin | @NoahSchweber: and yet a two-valued measurable cardinal $\kappa$ can become real-valued measurable in a forcing extension (and $\kappa$=$2^{\aleph_0}$ in that same forcing extension--Theorem 22.1(ii) in Jech 2003) , and the real-valued measurable cardinal $\kappa$ in the forcing extension can be a two-valued measurable cardinal in an inner model of that same forcing extension. You yourself said that " a cardinal is measurable iff it is real-valued measurable (to quote Jech's Corollary 10.15: "Every real-valued measurable cardinal is weakly inaccesible") and strongly inaccessible". | |
| Sep 16, 2017 at 18:02 | comment | added | Noah Schweber | Re: your edited last paragraph, like I say in my answer there's no tension here: "the continuum is real-valued measurable" and "there is a measurable cardinal" are very different principles, there's no reason to expect them to have the same consequences for cardinal arithmetic. | |
| Sep 16, 2017 at 15:03 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
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| Sep 13, 2017 at 4:25 | comment | added | bof | How is it possible that ZFC + "$2^{\aleph_0}=\aleph_2$" implies $\neg CH$ while ZFC is consistent with CH since the two theories are equiconsistent? | |
| Sep 13, 2017 at 4:13 | vote | accept | Thomas Benjamin | ||
| Sep 13, 2017 at 3:39 | review | Close votes | |||
| Sep 13, 2017 at 12:23 | |||||
| Sep 13, 2017 at 3:28 | answer | added | Noah Schweber | timeline score: 4 | |
| Sep 13, 2017 at 3:26 | comment | added | Andrés E. Caicedo | Yes, much weaker assumptions (consistencywise) imply that $\mathbb R\cap L$ is countable. On the other hand, any (atomless) real-valued measurable cardinal has size at most the continuum. Any such cardinal is weakly inaccessible (and much more), so $\mathsf{CH}$ fails badly under such an assumption. I suggest you study some basic model theory (completeness and compactness for first-order logic, for instance), even before studying axiomatic set theory, because your questions have been revealing serious gaps regarding basic concepts (in this case, consistency is not the same as provability). | |
| Sep 13, 2017 at 3:09 | history | asked | Thomas Benjamin | CC BY-SA 3.0 |