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Prove that if $\kappa$ is an inaccessible cardinal, then $V_{\kappa}$ satisfies all the axioms of ZFC.

How is this done for the axiom of choice and for regularity?

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    For Choice, show that the (usual) product of a $\{A_{\lambda}\}_{\lambda\in\Lambda}$, where $A_{\lambda},\Lambda\in V_{\kappa}$ and $A_{\lambda}\neq\emptyset$ is also in $V_{\kappa}$. As for regularity, it is trivial!2010-12-15
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    Depending on how choice is defined, there is an easier argument: If we use the statement "For any $X$ there is an $f:{\mathcal P}(X)\setminus\{\emptyset\}\to X$ such that $f(A)\in A$ for all $A$", then it is trivial that this holds in $V_\kappa$: If $X\in V_\kappa$, then $X\in V_\alpha$ for some smaller $\alpha<\kappa$, and (since choice holds in $V$), there is such an $f\in V_{\alpha+7}\subset V_\kappa$ (where 7 can be reduced but I felt lazy).2010-12-15
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    @Andres: This is very good! @kittykat687: I really think a better title is needed for this question.2010-12-15

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