3
$\begingroup$

We know that if ‎$‎‎\kappa‎‎$ is a‎ ‎measurable ‎cardinal ‎and ‎‎$‎‎\mu$ be a‎ ‎two-valued ‎non-trivial‎‎$‎‎‎\kappa‎$-additive ‎measure ‎on ‎it ‎the‎n the corresponding inner model produced by Mostowski collapse of Scott's ultraproduct (‎$‎‎M_{‎\kappa‎,\mu}$‎) is dependent on both ‎$‎‎‎\kappa‎$ ‎and ‎‎$‎‎\mu$. ‎So by ‎changing ‎measures ‎‎on a certain measurable cardinal we ‎can ‎produce ‎many different ‎inner ‎models. Even by transitivity if two such inner models be isomorphic then they are equal. Now a question is:

"Is it possible to have a measurable cardinal which its inner models be independent from its measures?"

Precisely:

Definition (1): A ‎"unique ‎measurable ‎cardinal" ‎is a‎ ‎measurable ‎cardinal which for all two-valued ‎non-trivial ‎‎$‎‎‎\kappa‎$-additive ‎measures like $\mu$ and $\mu'$ ‎on ‎it we have $M_{\kappa , \mu}=M_{\kappa , \mu'}$. We show this unique inner model by $M_{\kappa}$.

Question (1): Is the following statement true?

$Con(ZFC+ \text{there exists a measurable cardinal})\Longrightarrow$ $Con(ZFC+ \text{there exists a unique measurable cardianl})$ ‎

‎Now ‎the ‎main ‎question ‎is ‎about ‎the ‎behavio‎r of the "well defined" function ‎$‎\kappa ‎‎\mapsto ‎M_{‎\kappa‎}‎‎‎‎$:‎ ‎

Definition (2): Define:‎ ‎

The collection of all unique measurable cardinals: ‎ ‎

$‎‎‎UM:=‎\lbrace ‎‎‎\kappa‎~|~‎\kappa~‎\text{is a unique measurable cardinal‎}‎\rbrace‎$ ‎‎ ‎

The ‎(informal) ‎collection ‎of ‎all ‎inner ‎models ‎of ‎unique ‎measurable ‎cardinals: ‎ ‎

‎‎$‎‎IUM:=‎\lbrace ‎‎‎M_{\kappa}‎~|~‎\kappa ‎\in UM ‎\rbrace‎‎‎$‎ ‎

Question (2): ‎Is ‎the ‎function‎ ‎$‎\kappa ‎‎\mapsto ‎M_{‎\kappa‎}‎$ ‎from ‎‎$‎‎UM$ ‎to ‎‎$‎‎IUM$ (strictly) ‎‎increasing? ‎In ‎the ‎other ‎words which one of the following statements are true?‎ ‎

‎‎$‎‎(1)~‎\forall ‎‎\kappa‎,‎\lambda ‎\in UM~~~~~(‎\kappa < ‎‎‎\lambda ‎‎\longrightarrow ‎M_{‎\kappa‎}\subseteq M_{‎\lambda‎}‎‎‎‎‎)‎‎$‎ ‎

$‎‎(2)~‎\forall ‎‎\kappa‎,‎\lambda ‎\in UM~~~~~(‎\kappa < ‎‎‎\lambda ‎‎\longrightarrow ‎M_{‎\kappa‎}\subsetneq M_{‎\lambda‎}‎‎‎‎‎)‎‎$‎ ‎‎

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ It is not true that all (non-normal) measures on $\kappa$ are Rudin-Keisler equivalent to each other. (Unless right before definition 2, you decide that "measure" means "normal measure", but then there is no need to talk about RK reducibility at all.) $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 21, 2013 at 14:50
  • $\begingroup$ Dear Andres, Thanks. I edited the error. $\endgroup$
    – user36136
    Commented Sep 24, 2013 at 3:40
  • $\begingroup$ I'm not sure how to answer the second question, but the first question has a positive answer. If $D$ is a normal measure on $\kappa$ then $L[D]$ is a model of $\sf ZFC$ where $\kappa$ is the only measurable and $D$ is the unique normal measure on $\kappa$ (see Jech's book p. 348 for more information). $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 24, 2013 at 6:09
  • $\begingroup$ Dear Asaf, Thanks for your useful answer. Do you have any idea about a special application or property of these special measurable cardinals? What about question (2)? $\endgroup$
    – user36136
    Commented Sep 24, 2013 at 6:16
  • $\begingroup$ Well, these are all "small" measurable cardinals in some sense. Note that if $D$ is a normal measure on $\kappa$ then $D\notin M_{\kappa,D}$. Therefore all these measurable cardinals stop being measurable in the inner models, so in some sense they seem (to me, at least) as less interesting because we cannot iterate constructions with these cardinals. The second answer, I think is that $M_\kappa$ and $M_\lambda$ are not subsets of each other at all. In $M_\kappa$, $\lambda$ is still measurable and in $M_\lambda$, $\kappa$ is measurable. I'm not 100% behind this argument, though. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 24, 2013 at 11:34

1 Answer 1

Reset to default
4
$\begingroup$

Because Question 1 asked about arbitrary two-valued, $\kappa$-additive, non-trivial measures on $\kappa$, not only normal ones, the answer is negative. (If one considers only normal measures, then the answer becomes positive, as explained in a comment by Asaf Karagila.) If $U$ is a measure on $\kappa$ (by which I mean two-valued, $\kappa$-additive, and non-trivial throughout this answer), then there is a measure $U\otimes U$ on $\kappa\times\kappa$ defined as $$ \{X\subseteq\kappa\times\kappa:\{\alpha<\kappa:\{\beta<\kappa:(\alpha,\beta)\in X\}\in U\}\in U\}. $$ Applying a bijection from $\kappa\times\kappa$ to $\kappa$, we get a measure $U^2$ on $\kappa$. The ultrapower (by which I always mean its transitive collapse in this answer) of the universe with respect to this $U^1$ (or with respect to the isomorphic $U\otimes U$) is the two-step iterated ultrapower of the universe with respect to $U$, so it is a proper submodel of the (one-step) ultrapower of the universe with respect to $U$.

Improve this answer
$\endgroup$

You must log in to answer this question.