# Worldly vs Inaccessible Cardinals, why different?

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?

• Can you say exactly what definition of Grothendieck universe you're using? The list of properties you named is not quite the one I am familiar with. – Eric Wofsey Dec 12 '17 at 22:51
• Thank you Eric, Asaf and Clive, great answers. My definition of Grothendiek universe is as per Wikipedia but with indexed union $\bigcup_{i \in I} x_i$ replaced by general union $\bigcup x$ and substitution $f:x\to U$ gives $range(f)\in U$. – Mark Kortink Dec 12 '17 at 23:45

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.

• Well, it's not as much as "missing" as it is just not entirely relevant to the discussion (the OP even knows that the least worldly is singular)... – Asaf Karagila Dec 13 '17 at 11:53
• @Asaf I might have misinterpreted OP's post. To me its seemed that they had heard about this fact but didn't know how this was possible. (If they had known about the proof, this surely would've answered their own question, wouldn't it?) – Stefan Mesken Dec 13 '17 at 11:56
• I think that the issue is a bit orthogonal. It's about the fact that first-order Replacement is only about definable functions, and the cofinal sequence is very much not definable internally to $V_\kappa$; whereas universes require what is essentially second-order Replacement, where you don't care about definability. I don't see how this problem is clarified by proving that the least worldly cardinal has countable cofinality. – Asaf Karagila Dec 13 '17 at 11:58
• You're probably right. But, still, I think looking at the proof clarifies the situation quite a bit and adds value to the discussion. Hence I'll just leave it here. – Stefan Mesken Dec 13 '17 at 12:02
• I didn't say it's not adding value to this question and its answer. Just that it's not "missing from the other answers", since it's not directly going to clarify the inherent difference between first- and second-order Replacement. :) – Asaf Karagila Dec 13 '17 at 12:03

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.

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.

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 $\kappa \not \in V_{\kappa}$.

• Well, to be accurate, the term "class" is sometimes referring to definable and sometimes to arbitrary collections. In this case, we are talking about arbitrary collections. Since the cofinal sequence is sure as hell not definable in any reasonable sense in $V_\kappa$. – Asaf Karagila Dec 12 '17 at 22:53