We say that a set $X$ is transitive whether for any $x$ in $X$ the implication $$ (x\in X)\to(x\subseteq X) $$ holds; moreover, we say that the set $S(y)$ is the successive of any set $y$ when the equality $$ S(y)=y\cup\{y\} $$ holds.
Now if $\mathfrak X$ is a collection of transitive sets then it is not complicated to show that even $\bigcup\mathfrak X$ is transitive so that if $\emptyset$ is transitive then for any $U$ in $\mathscr P(X)$ we can find the biggest transitive set $A_U$ (I call it transitive interior) there contained taking the union of all that transitive sets. So I am trying to calculate (provided it is possibile) explicitly $A_U$ so that I initially proved that the set $$ V:=\big\{x\in U\mid \text{ $y$ is transitive and $y\subseteq U$ for any $y \in S(x)$}\big\} $$ is transitive but unfortunately it seems to me that it is note equal to $A_U$.
Anyway, it is not complicated to show that $\bigcap\mathfrak X$ is transitive -provided any $X$ in $\mathfrak X$ is it: so if $X$ is a transitive set for any $U$ in $\mathscr P(X)$ there exists the littlest transitive set $B_U$ (I call it transitive clousure) of $\mathscr P(X)$ containing $U$ taking the intersection of all that sets. So I tried to calculate (provided it is possibile) explicitly $B_U$ but unfortunately I have not idea how procede here: to be honestly I tried to put $$ B_U:=U\cup\left(\bigcup_{x\in U}x\right) $$ but then $B_U$ it would not be transitive apparently.
Finally, I observed (I think it could be useful) that if $X$ is transitive then the collection $$ \mathcal T_X:=\{U\in\mathscr P(X)\mid \text{ $U$ is transitive}\} $$ is a topology of open or closed set so that if we consider it as "open" topology then $A_U$ would be (apparently) the (topological) interior of $U$ whereas if we consider it as "closed" topology then $B_U$ would be (apparently) the (topological) closure of $U$.
So I ask to calculate $A_U$ and $B_U$, hoping it is possible: could someone help me, please?