# Belief in consistency of extremely large cardinals

One of the most common justifications for the consistency of large cardinals is the development of a coherent inner model theory for many large cardinal axioms. While the strength of this argument can be debated (has any axiom ever been shown to be inconsistent via the attempt to develop an inner model theory?), it appears to be pretty popular, e.g. it appears in Maddy's Believing the Axioms.

However, it is my understanding that assuming Woodin's HOD Conjecture, his Universality Theorem implies that given an extendible cardinal, there exists a weak extender model that reflects essentially all of the large cardinals that are currently studied.

If inner model theory can be extended so far in a single leap, can it still be used to justify the consistency of any particular large cardinal axiom stronger than a supercompact? If not, what is the replacement?

It's also my understanding that assuming the $\Omega$ Conjecture and a proper class of Woodin cardinals, the same class of large cardinal axioms can be ordered in consistency strength by the Borel degree of the initial segment of universally Baire sets $A$ for which they establish determinacy properties of $L(A, \mathbb{R})$. Would calibrating the place of a large cardinal axiom on this hierarchy replace other arguments for consistency?

• @მამუკაჯიბლაძე: We don't know any proof of inconsistency of "post-choice cardinals" (e.g. Berkeley cardinals and the likes of them). If one assumes the axiom of choice, then just taking a Reinhardt cardinal is already inconsistent. If one assumes the HOD Conjecture, then one can extrapolate some inconsistencies from ZFC to ZF again. – Asaf Karagila Nov 29 '16 at 10:09
• @მამუკაჯიბლაძე Kunen's inconsistency theorem established the inconsistency of several otherwise plausible large cardinals relative to ZFC. I believe there is no other argument for establishing inconsistency of an otherwise plausible large cardinal axiom. As Asaf mentions, this is not known to hold relative to ZF alone, but conjectures of Woodin would establish some of the consequence's of Kunen's theorem relative to ZF in a much deeper way than Kunen's original result for ZFC. – Cameron Zwarich Nov 29 '16 at 10:43
• [...] A famous result along those lines (due to Woodin) is the equiconsitency of $\operatorname{ZF} + \operatorname{AD}$ and $\operatorname{ZFC} + \text{ there are infinitely many Woodin cardinals}$. It's important to note that from a certain assumption $\dagger$ we - in general - don't get large cardinals in our universe, but that $\dagger$ allows us to prove the existence of certain large cardinals in a (canonical) inner model. Typically these large cardinals are in fact countable in our background universe - and hence far from being 'large' in that sense. – Stefan Mesken Nov 29 '16 at 13:58
• @Stefan Thanks! Fascinating and confusing at the same time :D When you say "we - in general - don't get large cardinals in our universe", do you mean that this universe represents a model for some axioms rigorously forbidding certain large cardinals, or you mean rather that the latter are absent in the background model "by accident"? Or this distinction does not matter here? – მამუკა ჯიბლაძე Nov 29 '16 at 15:36
• @მამუკა ჯიბლაძე Yes, this is possible. Let $\phi \equiv M_1^{\sharp} \text{ exists } + \text{ there is no Woodin cardinal}$. (If $\delta$ is the least Woodin cardinal in our universe, then $V_{\delta}$ satisfies this theory, so it is not outright inconsistent.) Now there is no Woodin cardinal in our universe, but we can linearly iterate the top extender of $M_1^\sharp$ out of the universe, thereby obtaining the inner model $M_1$ which has a Woodin cardinal (from its point of view - the Woodin cardinal of $M_1$ is countable in $V$). – Stefan Mesken Nov 29 '16 at 18:45

There is another path that can be used to justify the existence of very large cardinals. Consider, for example, the abstract to Magidor's paper, "On the Role of Supercompact and Extendible Cardinals in Logic" (Israel Journal of Mathematics, Vol. 10, 1971, pp. 147-157) :

It is proved that the existence of supercompact cardinals is equivalent to a certain Lowenheim-Skolem Theorem for second order logic, whereas the existence of [an] extendible cardinal is equivalent to a certain compactness theorem for that logic. It is also proved that a certain axiom schema related to model theory implies the existence of many extendible cardinals.

Here the theorems:

Theorem 1. There exists a supercompact cardinal iff there is a $\mu_0$ such that for all $R({\beta})$, $\beta$$\ge$$\mu_o$there is an $\alpha$$\lt$$\beta$ such that $<$$R({\alpha}), \epsilon$$>$ can be elementarily embedded in $<$$R({\beta}), \epsilon$$>$. The least such $\mu_0$ is the first supercompact cardinal.

Theorem 2. The first supercompact cardinal is the first $\mu_0$ such that for every structure $A$ = $<$$M, R_1,...,R_n > |$$M$$|$$\ge$$\mu_0 and every \Pi^{1}_1 sentence \varphi, such that A$$\vDash$$\varphi, there exists a substructure A^{'} = < M^{'}, R_1$$|$$M,..., R_n$$|$$M$$>$ of $A$ with $|$$M^{'}$$|$$\lt$$|$$M$$|$ and $A^{'}$$\vDash$$\varphi$ [this is the "Lowenheim-Skolem Theorem" Magidor refers to in his abstract--my comment].

Definition. Logic is called $\kappa$-compact iff for every set of formulae $A$ in this logic, if every subset of $A$ of cardinality $\lt$$\kappa has a model, then A has a model. The \mathrm L^{n}_{\kappa} logic is like the n-th order logic, except that we allow conjunction and disjunction of less than \kappa formulae. The usual second order logic is of course L^{2}_{\omega}. Theorem 4. \kappa is extendible iff L^{2}_{\kappa} is \kappa-compact. \kappa is the first extendible iff it is the first \alpha such that second-order logic [L^{2}_{\alpha}--my comment] is \alpha-compact. ...We shall now show that a certain axiom schema is a very strong axiom of infinity, namely, it implies the existence of many supercompact cardinals. The axiom schema is: (V) If \varphi(x) defines a proper class of structures in the same language, then there exist two members of the class that one can be elementary embedded in the second...This axiom schema is called Vopenka's principle. Let V^{'} be the following axiom schema: (V^{'}) if the formula \tau(x) defines a closed unbounded set of ordinals, then there is an extendible cardinal in this class (i.e. "the class of all extendable cardinals is 'stationary' "). Theorem 3. (V) implies (V^{'}) but is not equivalent to it. Note that the forward implication '\Rightarrow' of Theorems 1, 2, and 4 show that if the required large cardinal exists, the logic in question will have the required property. This suggests to me the following 'formalist' justification of the existence of large cardinals: If you want second-order logic to have certain model-theoretic properties, hypothesize the appropriate large cardinals. This also suggests the following research program: Discover what model-theoretic properties of n'th-order logic the existence of large cardinals imply (from I0 on down the large cardinal hierarchy), if any. Similarly, one can hypothesize the existence of certain Lowenheim-Skolem-Tarski numbers, as V\ddot a$$\ddot a$n$\ddot a$nen shows in his paper, "Sort Logic and Foundations of Mathematics" ("Sort Logic...is a many-sorted extension of second-order logic."):

The canonical hierarchy $\Delta_n$ ($n$$\lt$$\omega$) inside sort logic climbs up the large cardinal hierarchy by reference to Hanf-, Lowenheim-, and [Lowenheim-Skolem-Tarski]-numbers, reaching all the way up to Vopenka's Principle [pg. 185].

• That's an interesting approach, and it certainly seems more sensible to describe large cardinals via model-theoretic properties than the usual definitions. In the case of extendible cardinals, it's interesting that the property asserted for $\kappa$ is even stronger than that of $\omega$, because compactness fails quite badly for $L^2_\omega$. Can it provide an answer by which to judge the consistency of these new model-theoretic axioms? I assume you can come up with some that are obviously inconsistent, but is there a heuristic that tells you where to draw the line? – Cameron Zwarich Dec 3 '16 at 4:50
• @CameronZwarich: Regarding extendible cardinals, note the forward direction ($\Rightarrow$) of Magidor's Thm. 4: If $\kappa$ is extendible then $L^{2}_{\kappa}$ is $\kappa$-compact. This suggests (to me at least) the following: if one can show the relative consistency of some theory $T$+ "There exists an extendible cardinal" in which one can interpret $L^{2}_{\kappa}$ , one can prove $T$+"There exists an extendible cardinal"$\vDash$$L^{2}_{\kappa}$ is $\kappa$-compact (Question: can $\omega$ be deemed an extendible cardinal? If not, then one can use Thm. 4 to prove the non-compactness – Thomas Benjamin Dec 3 '16 at 8:40
• (cont.) of $L^{2}_{\omega}$). One might use Henkin models of $L^{2}_{\kappa}$ as the means to study such consistency results. Regarding the heuristic that "tells you where to draw the line"--finding such a heuristic would be an integral part of developing the "research program". – Thomas Benjamin Dec 3 '16 at 8:53