# Real-valued measurability vs. Two-valued measurability in determining whether $CH$ holds or not

The following fact is known:

If there is a measurable cardinal, then there are only countably many constructible reals.

It is also known that if $ZFC$ + "There is a (two-valued) mesurable cardinal" is consistent, then $ZFC$ + "There exists a (two-valued) measurable cardinal" + $CH$ is also consistent.

If one replaces "There exists a (two-valued) measurable cardinal" with "The cardinality of the continuum is a real-valued measurable cardinal", is

i)$ZFC$ + "The cardinality of the continuum is a real-valued measurable cardinal" + $CH$ is consistent if $ZFC$ + "The cardinality of the continuum is a real-valued measurable cardinal" is consistent (it is not, by a result of Banach and Kuratowski)? Also,

ii) Does $ZFC$ + "The cardinality of the continuum is a real-valued measurable cardinal" imply that there are only countably many constructible reals?

Since i) is inconsistent, how is it possible that $ZFC$ + "The cardinality of the continuum is a real-valued measurable" implies $\lnot$$CH while ZFC+ "There exists a (two-valued) measurable cardinal" is consistent with CH or \lnot$$CH$?

• 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). – Andrés E. Caicedo Sep 13 '17 at 3:26
• 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? – bof Sep 13 '17 at 4:25
• 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. – Noah Schweber Sep 16 '17 at 18:02
• @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". – Thomas Benjamin Sep 17 '17 at 0:14
• (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....)? – Thomas Benjamin Sep 17 '17 at 0:18

First of all, it is not the case that if ZFC is consistent, then ZFC + CH + "There is a measurable" is consistent; a measurable cardinal has much greater consistency strength than ZFC alone. What is true is that if ZFC + "There is a measurable" is consistent, then so is ZFC + CH + "There is a measurable;" I think this is what you mean.

A bit of clarification, just to pin things down: when we say "If $A$ is consistent, then $B$ is consistent," this is usually shorthand for "$T$ proves $Con(A)\implies Con(B)$" where $T$ is some reasonable weak theory (PRA is almost always enough); such an interpretation is needed to avoid stupidity (if ZFC is consistent, then the sentence "PA is consistent implies ZFC is consistent" is technically true, even though what we mean by it clearly isn't). So, elaborating a bit on the previous paragraph, here are a couple things PRA proves:

• "If ZFC + "There is a measurable" is consistent, then ZFC + CH + "There is a measurable" is consistent."

• "If PRA proves "If ZFC is consistent, then ZFC + "There is a measurable cardinal" is consistent," then ZFC is inconsistent."

No, that second one wasn't a typo.

Now re: question 1, you ask:

how is it possible that ZFC + "The cardinality of the continuum is a real-valued measurable" implies $\neg$CH while ZFC+ "There exists a (two-valued) measurable cardinal" is consistent with CH since the two theories are equiconsistent?

There's no tension here - equiconsistent theories can disagree. Equiconsistency doesn't mean that they're consistent together, but rather that each is consistent if and only if the other is consistent. For example, ZFC+CH and ZFC+$\neg$CH are equiconsistent, but clearly inconsistent with each other.

So there's no issue.

Re: question 2, off the top of my head I'm not sure, but if memory serves Solovay showed that if there is a real-valued measurable, then there is an inner model with a measurable (it is certainly true that he showed that "there is a real-valued measurable" and "there is a measurable" are equiconsistent over ZFC; but it's possible he build an inner model of a forcing extension, instead of an actual inner model, which would be a problem here). If my memory is correct, then the answer to 2 is "yes:" letting $M$ be that inner model, we have $M\models \vert\mathbb{R}^L\vert=\aleph_0$. But $L$ (hence $\mathbb{R}^L$) and $\aleph_0$ are absolute, so $V\models\vert\mathbb{R}^L\vert=\aleph_0$.

Note that this is stronger than what you ask for in (2): the conclusion is that if there is a real-valued measurable at all, regardless of its comparison with the continuum, then $\mathbb{R}^L$ is countable.

• 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. – Andrés E. Caicedo Sep 13 '17 at 3:35
• 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 $0^\sharp$. – Andrés E. Caicedo Sep 13 '17 at 3:40
• @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. – Noah Schweber Sep 13 '17 at 13:22
• @Noah Just add random reals. – Andrés E. Caicedo Sep 13 '17 at 14:43
• @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.. – Noah Schweber Sep 13 '17 at 18:47