Closure under Chain Unions with Choice Function implies Elements with no Immediate Extension

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Theorem

Let $S$ be a set of sets which:

is closed under chain unions
has a choice function $C$ for its union $\bigcup S$.

Let $b \in S$.

Then $b$ is the subset of an element of $S$ which has no immediate extension in $S$.


Proof

Let the hypothesis be assumed.

Let $A = \bigcup S$

Let $x \in S$ be arbitrary.

Let $\map E x$ be the set of all elements $a \in A$ such that $x \cup \set a$ is an immediate extension of $x$.

Note that:

$\map E x \subseteq \bigcup S$


Suppose $x$ has an immediate extension in $S$.

Then:

$\map E x$ is non-empty
because $\map E x \subseteq \bigcup S$, $\map E x$ is in the domain of the choice function $C$.


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

$\forall x \in S: \map g x = \begin {cases} x & : \text {$x$ has no immediate extension in $S$} \\ \\ x \cup \set {\map C {\map E x} } & : \text {$x$ does have an immediate extension in $S$} \end{cases}$


Thus $g$ is a progressing mapping whose fixed points are those elements of $S$ that have no immediate extensions in $S$.

Hence the criteria hold for Set which is Superinductive under Progressing Mapping has Fixed Point: Corollary to be applied.

That tells us that an arbitrary $b \in S$ is the subset of a fixed point $x$ of $g$.

Such an element, by definition of $g$, has no immediate extension in $S$.

$\blacksquare$


Sources

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