Constructible subsets appear before successor cardinals

Choose a nonzero limit $\theta$ large enough that $x,\kappa ,L_{\kappa }$ and the relevant finite seed belong to $L_{\theta }$; this is possible because $x$ is constructible and the $L$-levels are nested and exhaustive for $L$. For the first clause set $A=\kappa \cup \{\kappa ,x\}$, and for the second set $A=L_{\kappa }\cup \{L_{\kappa },\kappa ,x\}$. Each seed is a subset of $L_{\theta }$. In the first case $\mathrm{\mid }A\mathrm{\mid }=\kappa$ because $\kappa$ is infinite; in the second F4 gives the same conclusion. Both seeds are well-orderable.

Let $H={\mathrm{Hull}}^{L_{\theta }}(A)$. By F1, $H\prec L_{\theta }$ and $H$ is well-orderable with $\mathrm{\mid }H\mathrm{\mid }=\kappa$. Collapse $H$ to $M$. By F2 there is an ordinal $\beta$ with $M=L_{\beta }$. Since $\beta =\mathrm{Ord}\cap M\subseteq M$ and the collapse bijects $H$ with $M$, there is an injection $\beta \to \kappa$; F5 therefore gives $\beta <{\kappa }^{+}$.

In the first case, $\kappa \subseteq A\subseteq H$, so F3 fixes every ordinal below $\kappa$. Since $x\in H$ and $x\subseteq \kappa$, the recursive collapse equation gives $\pi (x)=\{\pi (\xi ):\xi \in x\}=x$. Thus $x\in M=L_{\beta }$. In the second case, $L_{\kappa }\subseteq H$ is transitive by F6, so F3 fixes it pointwise; again $x\in H$ and $x\subseteq L_{\kappa }$ imply $\pi (x)=x$, whence $x\in L_{\beta }$.

We have proved the general clause with $\beta <{\kappa }^{+}$. For the first clause, nesting gives $L_{\beta }\subseteq L_{{\kappa }^{+}}$, so $x\in L_{{\kappa }^{+}}$. The case $x=\mathrm{\varnothing }$ is included, and no selection from a family of sets occurred: the hull is canonical and the cardinal bound uses only the Hartogs successor.