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Questions about real-valued measurable cardinals and the continuum

  1. Jun 20, 2014 #1
    Putting the following three statements together:

    (a) Assuming that the continuum hypothesis is false, the power of the continuum 2[itex]\aleph0[/itex] 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[itex]\aleph0[/itex] 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?
     
  2. jcsd
  3. Jun 20, 2014 #2

    micromass

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    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.
     
  4. Jun 20, 2014 #3
    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.
     
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