What is known about the consistency strength of
ZFC + the continuum is real valued measurable + Martin's maximum?
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
5
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.
answered Sep 9, 2017 at 3:14
-
$\begingroup$ What would be the consistency strength of $ZFC$+ "The continuum is real-valued measurable"? $\endgroup$ Commented Sep 9, 2017 at 7:59
-
2$\begingroup$ 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. $\endgroup$ Commented 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.... $\endgroup$ Commented Sep 9, 2017 at 10:16
-
3$\begingroup$ @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). $\endgroup$ Commented Sep 9, 2017 at 15:03
-
$\begingroup$ @AndreasBlass: Thanks for correcting me. Very helpful. $\endgroup$ Commented Sep 9, 2017 at 15:05
$\begingroup$
$\endgroup$
2
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.)
answered Sep 9, 2017 at 16:19
-
3$\begingroup$ 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 ♦Commented Sep 9, 2017 at 16:26
-
1$\begingroup$ Give that man a 6-9-5-12-4-19 medal! $\endgroup$ Commented Sep 9, 2017 at 16:29