Worldly vs Inaccessible Cardinals, why different?

Question

A worldly cardinal $\kappa $ is defined by $V_\kappa \vDash ZFC $ . An inaccessible cardinal $\iota $ is defined in such a way that $V_\iota \ $ is a Grothendieck universe and so provides a model of ZFC. Therefore inaccessible cardinals are worldly. But if they exist the smallest worldly cardinal is singular so not inaccessible.

My question is how can it be that $V_\kappa \vDash ZFC $ and yet not be Grothendieck? For example $V_\kappa \vDash \forall x(set x \implies set \wp x) $ so $V_\kappa $ is closed under $\wp $, and similarly for all the other Grothendieck properties (transitive, infinity, pairs, unions, powers, substitutions). If $V_\kappa $ is worldly shouldn't it therefore be Grothendieck, what am i missing?

Answer 1

I don't know the exact definition of Grothendieck universe you are using, but the following is the heart of the matter and can be adjusted to address some axiom or another of whatever your definition is.

One property that a Grothendieck universe $U$ must satisfy is that if $x\in U$ and $f:x\to U$ is an function, then the image of $f$ is an element of $U$. However, you cannot prove this statement from merely knowing that $U$ is a model of ZFC. The problem is that the only axiom you might be able to use to prove this is Replacement, but Replacement only applies to functions that are defined by some formula with parameters. It is possible that there is no formula in the language of set theory which defines the function $f$ when interpreted in $U$, even if you allow elements of $U$ as parameters in the formula.

To connect this with Asaf's answer, the least worldly cardinal $\kappa$ has countable cofinality, and is the limit of some sequence $(\alpha_n)_{n\in\omega}$, which is a function $\omega\to V_\kappa$. However, this function cannot be defined in the structure $V_\kappa$, and so Replacement in $V_\kappa$ does not require that this function actually is an element of $V_\kappa$ (and indeed it is not). This is no different from how a countable model of ZFC does not know it is countable, since the function from $\omega$ that would witness its countability is not an element of the model.

Answer 2

The least $\kappa$ such that $V_\kappa$ is a model of $\sf ZFC$—if such $\kappa$ exists that is—has cofinality $\omega$. It is certainly not inaccessible, since it is singular.

But a Grothendieck universe is closed under arbitrary functions with domains inside the universe. This means that if $\alpha_n$ is the cofinal sequence below $\kappa$, since $\omega\in V_\kappa$, the sequence itself $\langle\alpha_n\mid n<\omega\rangle$ has to be inside $V_\kappa$ as well. But it cannot be, since its supremum is $\kappa$.

Note that the Tarski–Grothendieck set theory also does not include Foundation as an axiom, but it follows for the same reason as above. If Foundation fails, there is a witness of this in the form of a decreasing $\omega$-sequence, but that would witness the failure of Foundation in the universe. And of course, this is impossible.

Answer 3

In addition to the other answers, I think it's useful to look at a proof for the

Fact. The least worldy cardinal $\kappa$, if it exists, is of cofinality $\omega$.

Proof (Sketch). Let $\lambda$ be a worldy cardinal of uncountable cofinality and fix a well-order $\prec$ of $V_{\lambda}$. Let $(\phi_n \mid n \in \omega)$ be an enumeration of all axioms of $\mathrm{ZFC}$ (or, if you want to, of the theory of $(V_{\lambda}; \in)$) which is closed under subformulae. Let $\lambda_0 = \omega$ and given $\lambda_i$ let $\lambda_{i} < \lambda_{i+1}$ be minimal such that $V_{\lambda_{i+1}}$ contains all the evaluations of the $\prec$-Skolem terms for $(\phi_n \mid n < \omega)$ with parameters in $V_{\lambda_{i}}$, i.e. whenever $\vec{p} \in [V_{\lambda_i}]^{<\omega}$ and $$ (V_\lambda; \in) \models \exists x \phi_{n}(x, \vec{p}), $$ then the $\prec$-least such $x$ is in $V_{\lambda_{i+1}}$. An easy calculation shows that $\kappa := \sup_{i < \omega} \lambda_i$ is wordly. Since $\mathrm{cof}(\lambda) > \omega$ we furthermore have that $\kappa < \lambda$ - hence $\lambda$ is not the least worldly cardinal. Q.E.D.

Answer 4

Although $V_{\kappa}$ is a model of $\mathrm{ZFC}$ for all worldly cardinals $\kappa$, you can't prove within $V_{\kappa}$ that $V_{\kappa}$ is (externally) a Grothendieck universe, since that requires you to quantify over classes of (external) size $\kappa$, which you can't do within $V_{\kappa}$ since $\kappa \not \in V_{\kappa}$.