I read somewhere about the hyperuniverse of countable transitive models of ZFC (<http://www1.maths.leeds.ac.uk/maloa/lecturenotes/RW3%20Munster/friedman.pdf>). It states an assumption, namely that every real is a member of some countable transitive model of ZFC.


Is there some way to prove this assumption (assuming the hyperuniverse is not empty), or rather the (seemingly stronger) one: Every finite set of reals is a member of a ctm? 



Is every real a member of some CTM?
It is (relatively) consistent with ZFC that this is false. For example, suppose that there are $\alpha<\beta$ with $L\_\alpha$ and $L\_\beta$ both being models of ZFC. (This is a very weak large cardinal axiom). Let us take $\alpha$ and $\beta$ to be least. So $L\_\beta$ thinks that there is only one height $\alpha$ of a model of ZFC. It follows that $\alpha$ is countable inside $L\_\beta$, and so inside $L\_\beta$, the hyperverse is not empty---it contains $L\_\alpha$ and many forcing extensions of $L\_\alpha$ and so on. Let $x$ be a real in $L\_\beta$ coding a relation on $\omega$ of order-type $\alpha$. This real cannot be in any CTM of ZFC inside $L\_\beta$, since that model would have to have height at least $\beta$ by the minimality of $\beta$, and so it could not be an element of $L\_\beta$.


Another way to argue: let $M\_0$ be any countable transitive model of ZFC, and let $M$ be any $\in$-minimal countable transitive model of ZFC with $M\_0\in M$. So $M\_0$ is countable in $M$, but by minimality, $M$ can have no CTM containing a real coding $M\_0$. 


Meanwhile, if there is an inaccessible cardinal $\kappa$, or merely a worldly cardinal, then every real is contained in the corresponding $V\_\kappa$, and by Löwenheim-Skolem, we may collapse a countable elementary substructure to place it into a countable transitive model of ZFC.


