Let f : B $->$ B be a function, and assume that A $$ B$.$ We have two possible methods for

constructing the closure C of A under f$.$ First, define C$^$

to be the intersection of the closed supersets of A:

C$^=$the intersection of $(\{$X : A $$ X $$ B and f$[$X$]$ X$\}).$

Alternatively, we could apply the recursion theorem to obtain the function h for which

h$(0)=$ A

h$($n$^+)=$ h$($n$)\backslash$cup f$[$h$($n$)].$

Clearly h$(0)$ h$(1).$ Define C$\_$ to be union$($ranh$).$ In other words

C$\_=$union$($h$($i$)),$ where i belongs in omega.

$($a$)$ What is the map F : P$($A$)->$ P$($A$)$ in the language of the recursion theorem $($given in class$)$ that justifies the construction of h$?$ That is$,$ F must satisfy h$($n$^+)=$ F$($h$($n$)).$