# Is this weak form of $V=L$ (in)consistent with large cardinals?

I have been considering a (definability-free) weak form of the constructibility axiom, which is intended to capture the coarse structure of the constructible hierarchy. This means that this weak form is intended to capture the ordinal pattern abstracted from the constructible hierarchy if the fine behavior of the ordinal values of the constructible rank is covered up. It turns out that basic consequences of the constructibility axiom remain valid in the resulting theory:

Let $ZF_{\rho}^-$ be the following theory: Its language has, in addition to $\in$, an unary function symbol $\rho$, and the axioms include all axioms of $ZF^-$ ($ZF$ minus the axiom of foundation) along with replacement and separation axioms for formulas containing $\rho$.

Now, add to $ZF_{\rho}^-$ the following axiom:

The function symbol $\rho$ is such that

1. $\forall x, \rho(x)$ is an ordinal.
2. $\forall \alpha, (\rho(\alpha)=\alpha)$.
3. $\forall x, y, (x\in y\rightarrow \rho(x)<\rho(y))$. (i.e., $\rho$ preserves membership.)
4. $\forall \alpha\: \exists f;(f:\alpha \cup \omega\rightarrow\left\{x:\rho(x)<\alpha\right\}$ is surjective). (i.e., $|\left\{x:\rho(x)<\alpha\right\}|\leq|\alpha\cup \omega|.)$
5. For every set $x$, $(i)$ if $x\in V_{\omega}$, then $\rho(x)=rk(x)$, and $(ii)$ if $x\notin V_{\omega}$, then given a transitive set $T$ containing $x$ and $r:T\rightarrow T$ satisfying 1-4 above, $\rho(x)<r(x)^+$.

A function symbol satisfying 1-5 is called a minimal ordinal connection in $V$. It is minimal with respect to cardinality. The constructible rank is a minimal ordinal connection in $L$ (the minimality clause 5 is a consequence of Godel's condensation lemma). In this sense, the above axiom is a weak form of the constructibility axiom. (If $a\subseteq\omega$, then the $L[a]$-rank is also a minimal ordinal connection in $L[a]$.)

The axiom of foundation, the axiom of choice and $GCH$ are theorems of the above theory. Also, if $\kappa$ is inaccessible, then $\left\{x:\rho(x)<\kappa\right\}=V_{\kappa}$. These are basic consequences of $ZF^- + V=L$ (if $\rho$ is interpreted as the constructible rank).

Now, it seems that the usual proofs of $V\neq L$ from, for example, the existence of a measurable cardinal will not work here, because they involve definability and absoluteness aspects of $L$ which are not present in this case. I have two questions:

• Is this weak form of $V=L$ consistent with measurable cardinals?

• Is this weak form of $V=L$ inconsistent with large cardinal notions assumed to be consistent with $ZFC$?

• This is consistent as far as we know. The natural stratification of the fine structural inner models $L[\cal E]$ shows that these models satisfy this theory by setting $\rho(x)$ to be the natural rank, the least $\alpha$ such that $x\in L_{\alpha+1}[\cal E]$. – Andrés E. Caicedo May 24 '16 at 15:46
• Thanks. How do you prove minimality? I suspected this to be the case, but I am ignorant on finestructural hierarchies and the version of condensation available in that setting seemed to me to be insufficient. – Rodrigo Freire May 24 '16 at 16:29
• This is a consequence of what we call the acceptability condition. (Take a look at section 2.2 of Steel's handbook article.) The point is that $\mathcal E$ is "formatted" in such a way that we have that whenever $J^{\mathcal E}_{\beta+1}\setminus J^{\mathcal E}_\beta$ adds a subset of an ordinal $\kappa$, and $\kappa$ is least for which this happens, then in fact $J^{\mathcal E}_{\beta+1}$ sees a surjection from $\kappa$ onto $J^{\mathcal E}_\beta$. The idea for this goes back to Dodd and Jensen, where it is explained below a measurable. – Andrés E. Caicedo May 24 '16 at 19:37
• Thanks, I think this is very interesting. I will take a look at this article. – Rodrigo Freire May 25 '16 at 2:18

• Thanks, very helpful. I can see clearly that, for sets of ordinals, the minimality (in my sense) of the $L[A]$-rank follows from strong acceptability. However, I have not been able to see how the general case follows from this. Of course, this is due to my lack of knowledge in the subject, and it is not a problem with your answer. – Rodrigo Freire May 25 '16 at 21:59