Let M and N be transitive models of ZFC such that for every set x of ordinals, $x \in M \iff x \in N$. I want to show M=N. It will suffice to show $M \subset N$ by symmetry. I was thinking of assuming the contrary, and letting $x \in M$, $x \notin N$ be of least rank. So, we know that for all $y \in x$, $y \in N$. So now I just need to show that these $y$'s form a set in N. If there was a function in N enumerating these $y$'s whose domain is an ordinal, I could apply replacement and be done. I still haven't used the hypothesis of the models having the same sets of ordinals, and I'm not sure how to proceed.
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
HINT: Take any set $x\in M$, show that by knowing $\operatorname{tc}(\{x\})$ we know what $x$ is; then use the axiom of choice to find a set of ordinals $A$ such that by decoding $A$ into a set of ordered pairs of ordinals, we obtain a structure isomorphic to $\operatorname{tc}(\{x\})$, finally use Mostowski's collapse lemma.
answered Feb 11, 2015 at 7:21
-
$\begingroup$ Thanks Asaf - Just checking: For the first part, it's easy to see that tc({x})=tc({y}) implies x=y. We know tc({x}) is in M, and by AC there is a bijection f from some cardinal $\kappa$ into tc({x}). So, we can let A = {$(\alpha, \beta): \alpha , \beta < \kappa , f(\alpha) \in f(\beta)$}. If we could show A is in N, we could apply the Mostowski collapse to get tc({x}) in N, then by transitivity, x would be in N. How do we know A is in N? We know N has the same ordered pairs of ordinals as M by Axioms of Pairing and Union. How do we know N has the same sets of ordered pairs of ordinals as M? $\endgroup$– TaroCommented Feb 11, 2015 at 15:59
-
1$\begingroup$ Use the fact that there is a definable encoding of ordered pairs of ordinals as ordinals. $\endgroup$– Asaf Karagila ♦Commented Feb 11, 2015 at 17:00
-