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For a measurable cardinal $\kappa$, we define an ordering over $\kappa$-complete ultrafilters as follows:

Suppose $W,U$ are both $\kappa$-complete free ultrafilters over $\kappa$, we say that $U\lhd W$ if and only if $U\in M_W$ (where $M_W$ is the Mostowski collapse of $Ult_W(V)$).

I am trying to give an alternative definition which will ease the proofs I have to write regarding this order (it is a well-founded, irreflexive and transitive ordering, for example). However I am getting somewhat confused about the minor details.

We know that $U\lhd W \iff U\in M_W\iff \exists g\in V^\kappa\colon\pi([g]_W)=U$ (where $\pi$ is the collapse of $V^\kappa/W$ to $M_W$, and $[g]_W$ is the equivalence class of $g$ in $V^\kappa/W$).

In Mitchell's original paper he defined the order for normal measures as follows:

$U\lhd W$ if and only if there is some $A\in W$ and a sequence $U_i,\ i\in A$ such that $x\in\kappa$ (I guess it supposed to be $x\subseteq\kappa$) then $x\in U$ if and only if $\{i\in A\mid x\cap i\in U_i\}\in W$.

This seems like a definition which relies to some extent on the normality of $W$, and I wonder if this is really the case and if so, whether or not my intuition is correct and there is a similar extension of the definition to all the measures on $\kappa$.

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1 Answer 1

up vote 5 down vote accepted

The $U_i$ sequence in Mitchell's definition is playing the role of your function $g$, and his formulation relies on normality in that it assumes that $U_i$ concentrates on $i$, which would be true when $W$ is normal.

Without any normality assumption on $W$, you can say $U$ is Mitchell below $W$ if and only if there is $A\in W$ and measures $U_i$ for $i\in A$, concentrating on $k(i)$, for which $[k]_W=\kappa$ and

$$x\in U\quad\text{ if and only if }\quad\{\ i\in A\mid x\cap k(i)\in U_i\ \}\in W,$$

which seems to be the style of statement that you desire.

The right hand side is equivalent to $x\in [\langle U_i\rangle_i]_W$ by Los, since $x=j(x)\cap \kappa$.

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I'm not sure what would the sequence of measures is going to be if $o(W)=2$, is it just going to be $W$ itself or $U$ itself? –  Asaf Karagila Jul 24 '11 at 14:56
    
For $o(W)=2$, you could just insist that each $U_i$ have $o(U_i)=1$. Alternatively, $W$ has order $2$ if there is a normal $U$ with $U_i$ as indicated, and each $U_i$ concentrates on measurable cardinals. –  JDH Jul 24 '11 at 22:20
    
I think that I got confused about the role of $U_i$ along the way. I will sleep over that and give it a fresh try in the morning. (I was told that in the lack of amphetamines one should opt for REM sleep as a source of energy :-)) –  Asaf Karagila Jul 24 '11 at 23:17
    
Okay, I think I understood the whole idea further (that sleep thing really works!), $U\lhd W$ if there is a sequence of measures over $k(i)$ (where $k$ represents $\kappa$) such that $U$ is coherent with respect to the sequence and to $W$. Now to see why the definitions are equivalent... –  Asaf Karagila Jul 25 '11 at 7:06
    
It took some time to chew and swallow, but I think I got it! Many thanks! –  Asaf Karagila Jul 26 '11 at 9:54

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