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Let $\kappa$ be a measurable cardinal, we say that $\mathcal{D}$ is a normal ultrafilter iff whenever $g\in\kappa^\kappa$ such that $g<_\mathcal{D} Id$, we have some $\alpha<\kappa$ such that $\{\gamma<\kappa | g(\gamma) = \alpha\}\in\mathcal{D}$.

Given $\kappa$ a measurable cardinal and some $M$ which is a transitive collapse of an ultrapower by some $\kappa$-complete ultrafilter, $\mathcal{U}$, on $\kappa$, and $j_\mathcal{U}$ the canonical embedding of $V$ into $M$, we can reconstruct a normal ultrafilter by: $\mathcal{D} = \{X\subseteq\kappa | \kappa\in j_\mathcal{U}(X)\}$.

Now, given a set $A\in\mathcal{U}$ then $\kappa\in j_\mathcal{U}(A)$ iff $[Id]_\mathcal{U} E [C_A]_\mathcal{U}$ iff $\{\alpha<\kappa | \alpha\in A\}\in\mathcal{U}$. (where $E$ is the relation on the ultraproduct, and $C_x$ is the constant function giving $x$ for every $\alpha$)

Now, clearly if $A\in\mathcal{U}$ then $\{\alpha<\kappa | \alpha\in A\}\in\mathcal{U}$ and therefore $\kappa\in j_\mathcal{U}(A)$ and therefore $A\in\mathcal{D}$.

But that means the ultraproduct was given by a normal ultrafilter, even though we took some general ultrafilter, and I was told by my prof. that not every ultrafilter can be normalized through a permutation on $\kappa$.

Where and what am I doing wrong here?

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up vote 5 down vote accepted

the problem with your argument is when you say that $\kappa\in j_{\mathcal U}(A)$ iff the class of the identity is a "member" of the class of the constantly-equal-to-$A$ function. This assumes that the identity represents $\kappa$, which is equivalent to the normality of ${\mathcal U}$.

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(It is not a bad idea to think how $\kappa$ is represented if ${\mathcal U}$ is not normal.) –  Andres Caicedo Dec 24 '10 at 1:20
    
Of course! I made the same mistake in class :) –  Asaf Karagila Dec 24 '10 at 7:11

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