Is it possible that every set can be specified?

Is it possible for there to be a model of ZFC with the property that, for every set $S$ in the model, there is a unary predicate in the language of ZFC such that $S$ is the is the only set satisfying the predicate?

I'm pretty sure I've been told the answer is "yes", but I am never able to find a reference when I want one. So this question (assuming it is answered in the affirmative) is to be that reference!

• I'm reminded of Joel David Hamkins's explanation of the subtleties of undefinable numbers here ... mathoverflow.net/questions/44102/… – user4894 Dec 29 '14 at 3:00
• Was this actually a reference request? (maybe I should have asked first) – user2345215 Dec 29 '14 at 3:07
• @user2345215: An external reference is fine. So is a self-contained answer. I mainly just want something I can point to to remind myself that yes, the answer really is "yes". And sometimes, to point other people to. – Hurkyl Dec 29 '14 at 3:11
• Interesting question! Can one even find a model in which any class (not just the singletons) is definable? – Hanno Dec 29 '14 at 19:17
• @Hanno No, since any model of ZFC is infinite, and thus has at least continuum-many subsets, while there are only countably many formulas in the language of set theory. On the other hand, countable models have countably many elements, so it's possible that every element could be definable (which is what happens in pointwise definable models). – Alex Kruckman Apr 1 '16 at 4:24

The minimal model $M$ is such a model. $M$ is defined to be $L_{\delta}$ (the $\delta$-th level of constructible hierarchy) where $\delta$ is least such that $L_{\delta}$ models ZFC. Assuming that there are standard models of ZFC, the minimal model exists.
• Use the definable well ordering of $M$ to conclude that the set $D$ of definable members of $M$ forms an elementary substructure of $M$. Now take the transitive collapse $D'$ of $D$ and argue, using minimality of $M$, that $D = D' = M$. – user203787 Dec 29 '14 at 19:32