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Although I know that "ZFC & there exists a measurable cardinal" and "ZFC & there exists a real-valued measurable cardinal" are equiconsistent with one another, I am not sure whether "ZFC & there exists a measurable cardinal k & there exists a real-valued measurable cardinal b" is equiconsistent with ZFC. (Obviously k is not equal to b.) I would be grateful for an answer.

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  • $\begingroup$ The question you're asking is equivalent to asking whether ZFC is inconsistent. $\endgroup$
    – Amit Kumar Gupta
    Commented Dec 26, 2012 at 10:19
  • $\begingroup$ A measurable cardinal is also real-valued measurable by definition. The consistency strength of the existence of (real-valued) measurable cardinals is much higher than that ZFC. $\endgroup$
    – François G. Dorais
    Commented Dec 26, 2012 at 11:26
  • $\begingroup$ First, a technicality: one doesn't compare a large cardinal axiom with ZFC, one compares "the axiom plus ZFC" to ZFC. So yes, your second sentence is correct, but I am not sure what the second sentence has to do with the question. Answer #1 to this question gives the link, but this is not something that is immediately evident from the definitions of these two large cardinals. So your answer seems to be rather dogmatic. $\endgroup$
    – David Reid
    Commented Dec 27, 2012 at 17:44

1 Answer 1

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Let me interpret the question as asking for a real-valued measurable cardinal that is not measurable, plus another (two-valued) measurable cardinal, which must be above it. For example, in a more extreme form, your question would ask: can the continuum be real-valued measurable while there is also another measurable cardinal?

The answer is yes.

Theorem. If there are two measurable cardinals, then there is a forcing extension in which the smaller one becomes the continuum and real-valued measurable, and the larger one remains measurable.

Proof. If $\kappa\lt\lambda$ are both measurable cardinals in $V$, then the forcing to add $\kappa$ many random reals will make $\kappa$ into the continuum and still real-valued measurable (by a result of Solovay), and the larger measurable cardinal $\lambda$ remains measurable, by the Levy-Solovay theorem, because the forcing was much smaller than $\lambda$. QED

Corollary. The following are equiconsistent.

  1. There are two measurable cardinals.
  2. There are two real-valued measurable cardinals.
  3. There is a non-measurable real-valued measurable cardinal and a measurable cardinal.
  4. The continuum is real-valued measurable and there is another measurable cardinal.

Proof. Statement 1 implies 4 in a forcing extension, by the argument above. Statement 4 implies statement 3 directly. Statement 3 implies statement 2 directly. Statement 2 implies statement 1 in an inner model. QED

Of course, one can generalize the arguments to handle any number of real-valued measurable cardinals--three instead of two, or any cardinal number---with measurable cardinals above them.

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