Beth fixed points and transitive models of ZFC minus Replacement

Question

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/

Answer 1

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.

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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 $p$.