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I apologize in advance for how basic this question is...

Let $j:V\rightarrow V/U$ be the ultrapower map where U is an ultrafilter on a set S, and $j(x)=[c_x]$. Now, let $f\in j(0)$. Then $f$ is undefined almost everywhere on S, but we're assuming $f$ is a function defined on all of S, so we conclude no such $f$ exists, and so $j(0)=0$. Okay, that I get. Now look at $j(1)$. By the same argument, $f\in j(1)$ iff $f=\emptyset$ a.e. So now my issue; this isn't unique! As long as it's 0 almost everywhere, I can give $f$ random values elsewhere. And in any case, it's a function on S, not $\{\emptyset\}$.

Thank you!

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Suppose $j$ is an elementary embedding, then it is 1-1, this is the definition of an embedding.

The critical point is the first ordinal such that $j(\kappa)\neq\kappa$. Suppose $\alpha=j(\kappa)<\kappa$ then what would $j(\alpha)$ be? It cannot be $\alpha$ because $j$ is 1-1. So you have to have $j(\kappa)>\kappa$.

So if $\kappa$ is the smallest one that moves, $j(\alpha)=\alpha$ for every ordinal below $\kappa$ and thus it is the identity map.

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    If $[f]\in_U j(1)$, then $f(i)\in 1$ for $U$-a.e. $i$, which means that $f(i)=0$ for $U$-a.e. $i$, and so $f=_U c_0$ which means $[f]_U=[c_0]=j(0)$, and so $j(1)=\{j(0)\}=\{0\}=1$.2011-03-05