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Stefan Kohl
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Here, $\, j_U, \, j_D$ are the canonical elementary embeddings induced by $U,D$ respectively.

I note that it is consistent with the existence of a measurable that the answer be yesyes: it is true in the model $L[D]$ for $D$ a measure on $\kappa$.

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

I note that it is consistent with the existence of a measurable that the answer be yes: it is true in the model $L[D]$ for $D$ a measure on $\kappa$.

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

I note that it is consistent with the existence of a measurable that the answer be yes: it is true in the model $L[D]$ for $D$ a measure on $\kappa$.

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

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

I note that it is consistent with the existence of a measurable that the answer be yes: it is true in the model $L[D]$ for $D$ a measure on $\kappa$.