# Almost universal class

I so stuck with a problem of set theory. But first a recursive definition:

• Define $$R_0=\emptyset$$

• If $$R_\alpha$$ is defined, then $$R_{\alpha+1}=\mathcal{P}(R_\alpha)$$ (the power set).

• For a limit ordinal $$\gamma$$, if $$R_\alpha$$ is defined for all $$\alpha<\gamma$$, then define $$R_\gamma=\displaystyle\bigcup_{\alpha<\gamma} R_\alpha$$

We define $$\text{BF}=\displaystyle\bigcup_{\alpha\in\text{OR}} R_\alpha$$ (the class of well founded sets).

Next, my problem

Take $$A\subseteq \text{BF}$$ a proper transitive class such that $$(A,\in)\models\text{ZF}$$. Prove that $$A$$ is almost universal.

I think that the exercise is false because if it was true, then we could conclude that the strongly inaccessible cardinals doesn't exists. This because if $$\kappa$$ is strongly inaccessible then $$R_\kappa$$ satisfies the hypothesis but $$R_\kappa$$ isn't almost universal. I really appreciate any hint or/and suggestion.

Edit: my counterexample is wrong. But, then, how can I solve the exercise?

• $R_\kappa$ isn't a proper class. – Kenny Lau Oct 21 '18 at 3:33
• Let $B$ be a set with $B \subseteq A$. For each $x \in B$, define $\alpha(x)$ to be the smallest ordinal such that $x \in R_\alpha$. Let $\alpha_0 := \sup\limits_{x \in B} ~ \alpha(x)$. Then $B \in R_{\alpha_0+1}$. Now I don't know how to prove that $B \in A$... – Kenny Lau Oct 21 '18 at 3:39
• @KennyLau You don't need $B\in A$ (and generally won't have it). You need $B$ to be a subset of some set in $A.$ ($R_{\alpha_0+1}\cap A$ works, and is the thing that needs to be shown to be in $A.$) – spaceisdarkgreen Oct 21 '18 at 4:53

Since $$A$$ is a transitive proper class model of ZF, then for any ordinal $$\alpha,$$ we have $$R_\alpha\cap A \in A.$$ This follows from the absoluteness of the rank function. $$R_\alpha \cap A$$ is just $$A$$'s version of $$R_\alpha,$$ the sets of rank less than $$\alpha.$$
Now to see that $$A$$ is almost universal, consider any set $$B\subseteq A.$$ Then, since $$B$$ is a set, for some sufficiently large $$\alpha$$ we have $$B\subseteq R_\alpha,$$ and hence $$B\subseteq R_\alpha\cap A\in A.$$
• Can you explain me in detail why for any ordinal $\alpha$ we have that $R_\alpha\cap A\in A$? I can't see how it follows from the absoluteness of the rank function. – Carlos Jiménez Oct 21 '18 at 20:38
• I think that the proof that you wrote follows from the fact that $\text{OR}\subseteq A$. Am I correct? – Carlos Jiménez Oct 21 '18 at 20:50
• @CarlosJiménez $R_\alpha^A = R_\alpha\cap A$ holds for any transitive model of ZF, proper class or not (when $\alpha\in A,$ of course). This is just using enough of ZF to define the rank function and prove that$\{x:\operatorname{rank}(x) < \alpha\}$ is a set. Then absoluteness identifies this set (as defined in $A$) with $R_\alpha \cap A = \{x\in A :\operatorname{rank}(x) < \alpha\}.$ Then, as you suggest, the fact that $A$ is a proper class means it has arbitrarily large ordinals (hence by transitivity, all ordinals), so we have $R_\alpha\cap A\in A$ for all $\alpha$ . – spaceisdarkgreen Oct 21 '18 at 22:14
• Ok I got it. Only one question more: how can I prove that all the ordinals are in $A$? – Carlos Jiménez Oct 21 '18 at 22:37
• @CarlosJiménez my last sentence was meant to contain a sketch of how that goes. Although I guess you could just ask why the fact that it's a proper class means it has arbitrarily large ordinals (which after all can't be true just because it's a proper class). Actually, the absoluteness of rank can help us here again. Cause it is true that any proper class (of well-founded sets) must contain sets of arbitrarily large rank. So since by absoluteness, the rank of a set in $A$ is in $A,$ $A$ contains arbitrarily large ordinals. – spaceisdarkgreen Oct 21 '18 at 22:44