If $U,D$ are $\kappa$-complete nonprincipal ultrafilters on $\kappa$ and $j_U = j_D$, is $U=D$?

129 Views Asked by At

Here, $j_U$, $j_D$ are the canonical elementary embeddings induced by the measures $U,D$ on $\kappa$.

I actually wish to ask three related questions.

Let Ult$_U(V)$, Ult$_D(V)$ be the transitive collapses of the ultrapowers of the universe $V$ mod $U,D$ respectively.

1) If $j_U = j_D$, is $U=D$?

2) If Ult$_U(V)$ = Ult$_D(V)$, is $U = D$?

3) If $j_U(U) = j_D(D)$, is $U=D$?

As you can see, these all revolve about the soft question, "To what extent can we reconstruct a measure given information about the elementary embedding it induces?"