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What is known about the consistency strength of

ZFC + the continuum is real valued measurable + Martin's maximum?

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2 Answers 2

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Martin's maximum implies the continuum is $\aleph_2$, and therefore it is not real-valued measurable, since real-valued measurable cardinals must be limit cardinals and indeed weakly inaccessible and weakly Mahlo and more.

So unfortunately, what is known about your theory is that it is inconsistent.

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  • $\begingroup$ What would be the consistency strength of $ZFC$+ "The continuum is real-valued measurable"? $\endgroup$ Sep 9, 2017 at 7:59
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    $\begingroup$ This is equiconsistent with a measurable cardinal. (A classical result of Solovay, I think.) For one direction, add $\kappa$ many random reals. For the other direction, use $L[N]$, where $N$ is the ideal of measure zero sets. $\endgroup$
    – Goldstern
    Sep 9, 2017 at 9:36
  • $\begingroup$ @Goldstern: Thanks. Regarding $ZFC$ + "There exists a measurable cardinal", would this imply the existence of many real-valued measurable cardinals (my guess is yes)? I ask because I suppose (perhaps badly) that one could force the continuum to be any one of these (real-valued measurable) cardinals.... $\endgroup$ Sep 9, 2017 at 10:16
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    $\begingroup$ @ThomasBenjamin The theory ZFC + "there exists a measurable cardinal" does not imply the existence of even one real-valued measurable cardinal, let alone many (unless you define "real-valued measurable" to include the possibility of being 2-valued measurable). It is merely equiconsistent with ZFC + "there exists a real-valued measurable cardinal" (as Goldstern wrote). $\endgroup$ Sep 9, 2017 at 15:03
  • $\begingroup$ @AndreasBlass: Thanks for correcting me. Very helpful. $\endgroup$ Sep 9, 2017 at 15:05
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This is a supplement to Joel's answer:

In fact, even very weak forcing axioms are incompatible with the continuum being real-valued measurable. For example, if the continuum $\mathfrak{c}$ is real-valued measurable, then $\mathfrak{b}<\mathfrak{c}$, where $\mathfrak{b}$ is the minimum cardinality of an unbounded family in $(\vphantom{b}^\omega\omega, <^*)$, and $f<^*g$ means $f(n)<g(n)$ for all sufficiently large $n$.

Martin's Axiom (and much weaker forcing axioms) implies $\mathfrak{b}=\mathfrak{c}$, and hence implies that $\mathfrak{c}$ is not real-valued measurable.

Lemma 27.9 on page 304 Jech's Set Theory gives a slightly more general result, along with a proof. (One needs to note that if $\mathfrak{b}=\mathfrak{c}$, there is a $\mathfrak{c}$-scale.)

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    $\begingroup$ If one adopts some Gematria to these cardinal characteristics, then $\frak b=c$ is the same as $2=3$, which by subtracting $2$ from both sides gives us $0=1$. Ergo, Martin's Axiom and its large cardinal powered family are all inconsistent. $\endgroup$
    – Asaf Karagila
    Sep 9, 2017 at 16:26
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    $\begingroup$ Give that man a 6-9-5-12-4-19 medal! $\endgroup$ Sep 9, 2017 at 16:29

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