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?

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    $\begingroup$ $R_\kappa$ isn't a proper class. $\endgroup$ – Kenny Lau Oct 21 '18 at 3:33
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    $\begingroup$ 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 $B \in A$... $\endgroup$ – Kenny Lau Oct 21 '18 at 3:39
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    $\begingroup$ @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 $A.$) $\endgroup$ – 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 $B\subseteq R_\alpha\cap A\in A.$

  • $\begingroup$ 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. $\endgroup$ – Carlos Jiménez Oct 21 '18 at 20:38
  • $\begingroup$ I think that the proof that you wrote follows from the fact that $\text{OR}\subseteq A$. Am I correct? $\endgroup$ – Carlos Jiménez Oct 21 '18 at 20:50
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    $\begingroup$ @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 $\alpha$ . $\endgroup$ – spaceisdarkgreen Oct 21 '18 at 22:14
  • $\begingroup$ Ok I got it. Only one question more: how can I prove that all the ordinals are in $A$? $\endgroup$ – Carlos Jiménez Oct 21 '18 at 22:37
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    $\begingroup$ @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. $\endgroup$ – spaceisdarkgreen Oct 21 '18 at 22:44

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