Questions about real-valued measurable cardinals and the continuum

[nomadreid]

Putting the following three statements together:

(a) Assuming that the continuum hypothesis is false, the power of the continuum 2^$\aleph₀$ is real-valued measurable.

(b) The existence of a real-valued measurable and the existence of a measurable (= real-valued measurable & inaccessible) cardinal are equiconsistent.

(c) If there exists a measurable cardinal, the continuum hypothesis is false.

it sounds like this would imply the following absurd statement:

(d) Assuming that the continuum hypothesis is false, the existence of 2^$\aleph₀$ and the existence of a measurable cardinal are equiconsistent.

What is wrong? Is (a) incorrect, or am I putting these together wrong? If (a) is incorrect, is there a clear example to show why?

 

[micromass]

I don't believe that $(a)$ is true. It is true that every real-valued measurable cardinal $\kappa$ is either measurable or $\kappa\leq 2^{\aleph_0}$.

But this does not imply that $2^{\aleph_0}$ is always real-valued measurable, even if CH is false. Indeed, it is perfectly possible that $2^{\aleph_0} = \aleph_2$ and this cannot be real-valued measurable since it is not weakly inaccessible (since it is not a limit cardinal).

As for (c), didn't you mean to say real-valued measurable. I don't think the existence of a measurable cardinal itself violates CH.

 

[nomadreid]

Thanks, micromass. Very informative answer. So, as you say, (a) is not true. As for (c) I was thinking of the violation of V=L, not the CH. Oops. So (c) is also not true. Also it would not have been true if I had put in "real valued measurable" where I put "measurable": at most I could have said that if the power of the continuum were real valued measurable, then CH would be false.