# Beth fixed points and transitive models of ZFC minus Replacement

Let ZC denote ZFC without the axiom schema of replacement. Is there a minimal subset of the replacement schema that can be added to ZC to guarantee that for all models of the resulting theory of the form $$V_\alpha$$ for some ordinal $$\alpha$$, the ordinal $$\alpha$$ must be a beth fixed point (or, equivalently, $$\alpha = |V_\alpha|$$)?

For instance, for any model of ZFC of the form $$V_\alpha$$, the ordinal $$\alpha$$ must be a worldly cardinal, a beth fixed point, and more. (See If there is a "worldly ordinal," then must there be a worldly cardinal?)) On the other hand, $$V_{\alpha}$$ is a model of ZC for any limit ordinal $$\alpha > \omega$$. I'm looking for the weakest subschema of replacement that guarantees that $$\alpha$$ must be a beth fixed point for any model $$V_\alpha$$ of ZC plus that schema. Maybe such a subschema would have to include all of replacement?

Note that adding to ZC the axiom of "transfinite recursion on ordinals" is not enough, since $$V_{\aleph_1}$$ is a model for that theory, by a theorem in http://jdh.hamkins.org/transfinite-recursion-as-a-fundamental-principle-in-set-theory/

Claim. The following are equivalent for a limit ordinal $$\alpha$$ :

1. $$\alpha$$ is a beth-fixed point.

2. $$V_\alpha$$ thinks every set is equipotent with an ordinal.

For $$1\to 2$$, it suffices to show that for every $$\beta<\alpha$$, $$V_\alpha$$ thinks $$V_\beta$$ is equipotent with an ordinal. (This is because every set in $$V_\alpha$$ is a subset of some $$V_\beta$$, $$\beta<\alpha$$.) Let $$f:V_\beta\to\beth_\beta$$ be a bijection. Then $$f$$ is a subset of $$V_\beta\times \beth_\beta$$, which is a member of $$V_\alpha$$. Since $$V_\alpha$$ is closed under Cartesian products and power sets, $$f\in V_\alpha$$.

For $$2\to 1$$, observe that the assumption implies $$|V_\beta|\in V_\alpha$$ for every $$\beta<\alpha$$, so $$\beth_\beta<\alpha$$ for all $$\beta<\alpha$$, which means $$\alpha$$ is a $$\beth$$-fixed point.

I finally prove that the above characterization is equivalent to the validity of $$\Sigma_1$$-replacement over $$V_\alpha$$:

Claim. If $$\alpha$$ is a beth-fixed point, then $$V_\alpha$$ satisfies $$\Sigma_1$$-Replacement.

The main ingredient is the following version of Levy reflection principle (which is provable by the same proof of usual Levy reflection principle $$H_\kappa\prec_{\Sigma_1} V$$)

Theorem. Let $$\lambda<\kappa$$ be cardinals and $$\lambda$$ be regular. Then $$H_\lambda\prec_{\Sigma_1} V_\kappa$$.

Moreover, it is known that $$H_\lambda$$ is a model of ZFC without Power set if $$\lambda$$ is regular. Now let $$F$$ be a $$\Sigma_1$$-class function over $$V_\alpha$$ with a parameter $$p$$. Take $$x\in V_\alpha$$. Choose $$\xi<\alpha$$ such that $$p,x\in V_\xi$$. Since $$\alpha$$ is a beth-fixed point, $$\lambda:=|V_\xi|^+<\alpha$$. We can see that $$V_\xi\subseteq H_\lambda\subseteq V_\alpha$$.

Observe that $$F$$ is absolute between $$V_\alpha$$ and $$H_\lambda$$. Moreover, $$H_\lambda$$ satisfies Replacement for $$F$$. Let $$H_\lambda\models F^"[x]=y$$ for $$y\in H_\lambda$$. Since the formula $$[\forall v\in y\exists u\in x (F(u)=v)]\land [\forall u\in x\exists v\in y (F(u)=v)]$$ is $$\Sigma_1$$-formula, it also holds over $$V_\alpha$$. This shows $$y$$ witnesses the instance of replacement for $$F$$, $$x$$ and $$p$$.

• @JesseElliott I do not know, but the statement "every set is equipotent with an ordinal" is a consequence of replacement for $\Sigma_1$-formulas over ZC. Aug 31, 2020 at 10:26
• Inspecting the proof of Theorem 2.12 of Jech gives a more specific instance of Replacement that implies the assertion "every set is equipotent with an ordinal," but I do not know it is actually equivalent to that. Aug 31, 2020 at 10:31
• @JesseElliott I added the proof that $V_\alpha$ satisfies $\Sigma_1$-replacement if $\alpha$ is a beth fixed point. I hope it is satisfactory to you. Sep 5, 2020 at 7:24
• Oh, then KP plus Power Set doesn't prove that $V_{\omega+\omega}$ exists because $x \in V_{\omega+\omega}$ isn't $\Sigma_1$, not because it doesn't prove replacement for $\Sigma_1$-formulas. I think I understand now. Sep 7, 2020 at 6:12
• @JesseElliott I think $\Sigma_2$-replacement suffices, and it proves $V_\alpha$ exists for all $\alpha$, in fact. Sep 7, 2020 at 6:20