# What is the consistency strength of “Every set is a member of a transitive model”?

Recall that $\kappa$ is a worldly cardinal if $V_\kappa$ is a model of $\sf ZFC$. While every worldly cardinal is a strong limit cardinal, it is not necessarily regular. The point being that the short cofinal sequence is not first-order definable, so Replacement is not violated.

In particular, the first worldly cardinal has countable cofinality, and in fact the first worldly cardinal which is a limit of worldly cardinal has countable cofinality (as do the worldly cardinals below it).

Consider the following statement "For every $x$ there is a transitive model $M$ such that $x\in M$". Clearly this statement follows from "There is a proper class of worldly cardinals". Does it also imply it, or at least is it equiconsistent with it?

• If your transitive model $M$ is a transitive $ZFC$ model, then you may find your answer in the "Transitive $ZFC$ Model" entry of Cantor's Attic (it's in the first paragraph of that entry). Does the argument contained therein answer your question? – Thomas Benjamin Apr 28 '17 at 8:00
• Yes. Transitive model means a model of ZFC. The Cantor's Attic entry does have a section about the "Transitive model universe axiom". Nothing about its exact consistency strength, though. – Asaf Karagila Apr 28 '17 at 8:41

The answer is no, because I claim that if $\kappa$ is worldly, then $V_\kappa$ thinks that every set is a member of a transitive model of ZFC.
To see this, note first that every worldly cardinal $\kappa$ is a beth-fixed point $\beth_\kappa=\kappa$ and furthermore $V_\kappa=H_\kappa$, the set of sets whose transitive closures have size less than $\kappa$. Now consider any $x\in V_\kappa$. By the Löwenheim-Skolem theorem, we can find an elementary substructure $X\prec V_\kappa$ with $x\subseteq X$ and $x\in X$, with $|X|=|\text{TC}(x)|<\kappa$. The transitive collapse $M$ of $X$ will be a model of ZFC containing $x$ as an element. And even though $\kappa$ is singular, and so perhaps $X$ is unbounded in $V_\kappa$, nevertheless we will have $M\in V_\kappa$ since it is small enough. For example, $M$ will have fewer than $\kappa$ many ordinals, and so $M\subseteq V_\beta$ for some $\beta<\kappa$ and hence $M\in V_{\beta+1}\subset V_\kappa$.