Least Fixed Point of Enumeration Operator

Theorem
Let $\psi : \powerset \N \to \powerset \N$ be an enumeration operator.

Let $A_i$ be defined recursively as:
 * $A_0 = \empty$
 * $A_{n + 1} = \map \psi {A_n}$

Let $A = \ds \bigcup_{i \mathop \in \N} A_i$.

Then:
 * $A$ is a fixed point of $\psi$
 * Every fixed point of $\psi$ is a superset of $A$

Lemma
By definition of enumeration operator, let $\phi \subseteq \N$ be a recursively enumerable set that yields $\psi$.

$A$ is a fixed point
By definition of fixed point, we want to show that $\map \psi A = A$.

By definition of set equality, it suffices to show that:
 * $\map \psi A \subseteq A$
 * $A \subseteq \map \psi A$

Let $x \in \map \psi A$ be arbitrary.

Then, by definition of enumeration operator, there is some finite subset $B$ of $A$ such that:
 * $\map \pi {x, b} \in \phi$

where $b$ is the code number for $B$.

Suppose $B = \empty$.

Then, $x \in A_1 \subseteq A$ by the definition of enumeration operator and Union is Smallest Superset.

Otherwise, suppose $B \ne \empty$.

By definition of Union of Family, for each $y \in A$, there is some $i \in \N$ such that:
 * $y \in A_i$

By Principle of Finite Choice, choose such an $i_y$ for each $y \in B$.

Let $\ds j = \max_{y \mathop \in B} i_y$, which exists by Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements.

By definition of greatest element, every $i_y \le j$.

Thus, by the lemma, every $A_{i_y} \subseteq A_j$.

Therefore, for every $y \in B$:
 * $y \in A_j$

It follows from the definition of enumeration operator that $x \in A_{j + 1}$.

As $x$, was arbitrary, it follows from the definition that $\map \psi A \subseteq A$.

Now, let $x \in A$ be arbitrary.

Then, by definition of Union of Family, there is some $i \in \N$ such that:
 * $x \in A_i$

As $A_0 = \empty$, by definition $x \notin A_0$.

Therefore, $i > 0$.

By definition of enumeration operator, there is a finite subset $B \subseteq A_{i - 1}$ such that:
 * $\map \pi {x, b} \in \phi$

But, by definition of Union of Family, $A_{i - 1} \subseteq A$.

Thus, by Subset Relation is Transitive:
 * $B \subseteq A$

Therefore, it follows from the definition of enumeration operator that $x \in \map \psi A$.

As $x \in A$ was arbitrary, by the definition, $A \subseteq \map \psi A$.

$A$ is the least fixed point
Let $F \subseteq \N$ be a fixed point of $\psi$.

By Union of Subsets is Subset:
 * $\paren {\forall i \in \N: A_i \subseteq F} \implies A \subseteq F$

Thus, it suffices to show that $A_i \subseteq F$ for every $i \in \N$.

We will proceed by induction.

Basis for the Induction
Follows immediately from Empty Set is Subset of All Sets.

Induction Hypothesis
Suppose that, for $i \in \N$:
 * $A_i \subseteq F$

Induction Step
Let $x \in A_{i + 1}$ be arbitrary.

By definition of enumeration operator, there is a finite subset $B \subseteq A_i$ such that:
 * $\map \pi {x, b} \in \phi$

where $b$ is the code number for $B$.

By the induction hypothesis, $A_i \subseteq F$.

By Subset Relation is Transitive:
 * $B \subseteq F$

Thus, it follows from the definition of enumeration operator that:
 * $x \in \map \psi F$

But, as $F$ is a fixed point of $\psi$:
 * $\map \psi F = F$

Therefore:
 * $x \in F$

As $x$ was arbitrary, it follows from the definition that:
 * $A_{i + 1} \subseteq F$

Thus, by Principle of Mathematical Induction:
 * $\forall i \in \N: A_i \subseteq F$

completing our proof by the remarks above.