Symmetric and hereditarily symmetric sets

A ZFA atom is a hereditarily $\mathcal{F}$-symmetric base case: its singleton stabilizer belongs to the normal filter by Permutation groups, stabilizers, supports, and normal filters. For an atom $a$ put ${\mathrm{TC}}_{\mathrm{Z}\mathrm{F}\mathrm{A}}(a)=\mathrm{\varnothing }$. For a set $x$ define the membership-descendant closure by rank recursion,

$${\mathrm{TC}}_{\mathrm{Z}\mathrm{F}\mathrm{A}}(x)=\bigcup _{y\in x}(\{y\}\cup {\mathrm{TC}}_{\mathrm{Z}\mathrm{F}\mathrm{A}}(y))\mathrm{.}$$

A set $x$ is hereditarily $\mathcal{F}$-symmetric when every object in $\{x\}\cup {\mathrm{TC}}_{\mathrm{Z}\mathrm{F}\mathrm{A}}(x)$ is $\mathcal{F}$-symmetric. Rank induction on the displayed recursion proves equivalently that $x$ is symmetric and every $y\in x$ is hereditarily symmetric, using the atom base case when $y$ is an atom. On pure sets this closure agrees with Transitive closure of a set. Write ${\mathrm{H}\mathrm{S}}_{\mathcal{F}}$ for all hereditarily symmetric objects, with inherited membership and the original atoms. If $y\in x\in {\mathrm{H}\mathrm{S}}_{\mathcal{F}}$, the recursive clause gives $y\in {\mathrm{H}\mathrm{S}}_{\mathcal{F}}$, so this permutation subuniverse is transitive. Symmetry alone is insufficient: a symmetric set can contain a nonsymmetric member.