Closed Set under Chain Unions with Choice Function is of Type M

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Theorem

Let $S$ be a set which is closed under chain unions.

Let there exist a choice function for $S$.


Then $S$ is of type $M$, that is:

every element of $S$ is a subset of a maximal element of $S$ under the subset relation.


Proof

Let $S$ be closed under chain unions.

Let $C$ be a choice function for $S$.

For $x \in S$, let $x^*$ be defined as the set of all elements $y$ of $S$ such that $x$ is a proper subset of $y$:

$\forall x \in S: x^* := \set {y \in S: x \subsetneqq y}$

We have that $x^*$ is empty if and only if $x$ is a maximal element of $S$ under the subset relation.


Let us define a progressing mapping $g$ on $S$ as follows:

$\forall x \in S: \map g x = \begin{cases} x & : \text {$x$ is maximal} \\ \map C {x^*} & : \text {otherwise} \end{cases}$

Thus:

if $x$ is a maximal element of $S$, then $\map g x = x$
if $x$ is not a maximal element of $S$, then $x$ is a proper subset of $\map g x$.

By Set which is Superinductive under Progressing Mapping has Fixed Point: Corollary:

for all $b \in S$, $b$ is a subset of some $x \in S$ such that $x = \map g x$.

This element $x$ is then a maximal element of $S$.

$\blacksquare$


Sources

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