User:Dfeuer/Fixed Point of Subset-Inflationary Mapping is Greatest Element of Minimally Inductive Class

Theorem
Let $g$ be a $\subseteq$-inflationary mapping.

Let $A$ be minimally $g$-inductive.

Let $p$ be an element of $A$ which is a fixed point of $g$.

Then $p$ is the largest element of $A$.

Proof
Let $B$ be the set of all elements of $A$ that are subsets of $p$.

$\varnothing \subseteq B$.

Let $x \in B$.

Then by User:Dfeuer/Class Minimally Inductive under Inflationary Mapping forms Nest, $g(x) \subseteq p$ or $p \subseteq x$.

If $p \subseteq x$ then since $x \in B$, $x \subseteq p$, so $x = p$.

Thus since $p$ is a fixed point of $g$, $g(x) = g(p) = p \subseteq p$.

Thus is either case, $g(x) \subseteq p$ so $g(x) \in B$.

Therefore $B$ is inductive under $g$.

Since $A$ is minimally inductive under $g$, $B \subseteq A$.

That is, every element of $A$ is a subset of $p$, so $p$ is the greatest element of $A$.