# Can there be elementary embedding between a universe and a universe inside it?

[EDIT] the prior question was trivially false, however the intention is to arrange a possible world of such universes, in other words the question is about if it is possible to have a proper class $$\mathcal {G}$$ of Grothenderick universes that contains $$V$$ in them, such that every object that is an element of some class (i.e. a set) then it is in $$\bigcup(\mathcal{G})$$, here $$\mathcal{G}$$ is a primitive constant symbol. So the formulation would be:

if $$\varphi(a_1,..,a_n)$$ is a formula in $$L(\in,=)$$, in which all and only $$a_1,...,a_n$$ occur free, then:

$$\forall W \big{(} W \in \mathcal {G} \to \\\forall a_1,..,a_n \in V [\text{if } \langle W,\in \rangle \models \varphi(a_1,..,a_n) \text{ then } \langle V,\in \rangle \models \varphi(a_1,..,a_n)] \big{)}$$

Add a constant primitive symbol $$V$$ to the language of $$ZF$$, standing for some "sub-world" of sets.Now axiomatize that $$V$$ is a Grothendieck universe. Add Tarki's axiom of every object is an element of some Grothendieck universe. Now is it possible to arrange that all Grothendieck universes containing $$V$$ do not add additional theorems about elements of $$V$$ in the pure set theoretic language (i.e. in $$L(\in,=)$$)?

Formally: if $$\varphi(a_1,..,a_n)$$ is a formula in $$L(\in,=)$$, in which all and only $$a_1,...,a_n$$ occur free, then:

$$\forall W \big{(}W \text { is a Grothendieck Universe } \wedge V \in W \to \\\forall a_1,..,a_n \in V [ \langle W,\in \rangle \models \varphi(a_1,..,a_n) \implies \\\langle V,\in \rangle \models \varphi(a_1,..,a_n)] \big{)}$$

The second question can we upgrade that to be from free variables in $$W$$? i.e. can we have the following?

if $$\varphi(a_1,..,a_n)$$ is a formula in $$L(\in,=)$$, in which all and only $$a_1,...,a_n$$ occur free, then:

$$\forall W \big{(}W \text { is a Grothendieck Universe } \wedge V \in W \to \\\forall a_1,..,a_n \in W [ \langle W,\in \rangle \models \varphi(a_1,..,a_n) \implies \\\langle V,\in \rangle \models \varphi(a_1,..,a_n)] \big{)}$$

• How much of your own time are you spending on these questions before bringing them here? Noah was able to see large flaws in your argument (and write them up) within 11 minutes (16 minutes with edits) -- this timeframe indicates that you should have caught them after any serious consideration if you're at a research level in this area. – Alec Rhea Dec 30 '18 at 3:38
• For the edited version, yes- this is just asserting that there is a proper class of inaccessible $\kappa$ such that $V_\kappa\prec V$. The consistency is below a Mahlo cardinal. This is trivial. – Monroe Eskew Dec 30 '18 at 18:34

No, even the "weak" version (indeed, a weakening of that) is impossible.

Consider the sentence (no parameters needed!) $$\mbox{\varphi\equiv "There is a largest inaccessible cardinal."}$$ If there is a proper class of inaccessible cardinals, then for proper class many inaccessible $$\kappa$$ we have $$V_\kappa\models\varphi$$ and for proper class many inaccessible $$\kappa$$ we have $$V_\kappa\models\neg\varphi$$. So the "theory of the inaccessibles" can never stabilize.

Remember that the Grothendieck universes are exactly the $$V_\kappa$$s for $$\kappa$$ inaccessible, so we're really just talking about the behavior of inaccessible cardinals. In particular, "Every set is contained in a Grothendieck universe" is equivalent to "There is a proper class of inaccessible cardinals."

EDIT: A better counterexample sentence might be $$\psi\equiv\mbox{"The class of inaccessibles has ordertype \alpha\cdot 2 for some ordinal \alpha."}$$

(In particular, note that $$\psi$$ implies that there are only set-many inaccessibles.) The reason this is preferable is that it works in a broader context: if we weaken our background theory to allow models with a proper class of inaccessibles of ordertype $$\omega$$ (that is, kill replacement), the $$\varphi$$ above doesn't work, but this $$\psi$$ still does. Indeed, this $$\psi$$ kills every version of the original question that makes sense, as far as I can tell.

Note that your title talks about elementary embeddings, but the above argument kills even the possibility of mere elementary equivalence between all sufficiently-large universes. Also, note that we could replace "inaccessible" with any definable large cardinal notion and get the same phenomenon. However, the word "definable" there is important. For example, if $$0^\sharp$$ exists then we have $$L_\alpha\prec L_\beta$$ whenever $$\alpha<\beta$$ are uncountable cardinals in the sense of $$V$$.

Re: your edit, the answer is now yes. Essentially - again, shifting from universes to cardinals (I really think this makes everything much easier to think about) - what you're basically asking is whether we can have a cofinal family of indiscernible inaccessible cardinals. As per my comment above, one way to get such a model is the following: suppose $$M\models$$ "There is a measurable cardinal," and consider the structure $$N=(L^M; \in, UncCard^M)$$ where $$UncCard^M$$ is the class of uncountable cardinals in $$M$$. We can now take $$W_0$$ (what you call "$$V$$," but I think that leads to overloading) to be $$(V_\kappa)^M$$ for any element $$\kappa$$ of $$UncCard^M$$ in this context (fine, after "throwing out" all elements of $$UncCard^M$$ which are $$<\kappa$$).

EDIT: And as Monroe Eskew observes we can do massively better than a measurable: if $$\kappa$$ is Mahlo, then $$V_\kappa$$ gives a model of the type you want.

And I believe even that is overkill: I suspect that, if we take a countable model $$M$$ of ZFC+"There is a proper class of inaccessbile cardinals" then the ultrapower of $$M$$ along any nonprincipal ultrafilter on $$\omega$$ has a cofinal indiscernible family of inaccessibles. However, I haven't checked the details, so I could be wrong. Also, note that even if this works it gives an illfounded model, whereas Monroe's observation gives a wellfounded model, admittedly from a significantly stronger assumption.

• just curious! for the original posting. if we work in "Ext. + Separation+ V is a Grothendieck universe + every object is a member of a Grothendieck universe", then we can have a proper class of inaccessible $\kappa$ such that in every $V_{\kappa}$ that contains $V$ as an member, there is a largest inaccessible cardinal. – Zuhair Al-Johar Dec 31 '18 at 16:23
• @ZuhairAl-Johar In the weaker theory you describe, the statement $(*)$ = "There are proper class many inaccessibles but only set-many which think there is no largest inaccessible" is indeed consistent, but it doesn't help the original formulation of your question: let $\varphi$ be the sentence "Either there is a proper class of inaccessibles, or the ordertype of the set of inaccessibles is an even ordinal." Then even in your weaker theory, this gives a counterexample. (cont'd) – Noah Schweber Dec 31 '18 at 17:45
• E.g. the simplest model of your theory looks like $V_\lambda$, where $\lambda$ is the (non-inaccessible!) limit of the first $\omega$-many inaccessibles in a model $V$ of ZFC containing at least $\omega$-many inaccessibles. (Again, I'm using $V$ in the usual way, not as you use it - I strongly recommend you use something like "$W_0$" instead, using "$V$" only makes things less clear.) Let $\kappa_0<\kappa_1<\kappa_2<...$ enumerate the inaccessibles of $V_\lambda$ in order; then $V_{\kappa_{2k}}\models\varphi$ but $V_{\kappa_{2k+1}}\models\neg\varphi$, for each $k\in\omega$. – Noah Schweber Dec 31 '18 at 17:48
• So while the specific example I gave doesn't work necessarily if you weaken the theory, a different example does - and indeed it works in any theory where the question makes sense, as far as I can tell. – Noah Schweber Dec 31 '18 at 17:48
• @ZuhairAl-Johar I've edited to include this. – Noah Schweber Dec 31 '18 at 17:51