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| author | adelon <22380201+adelon@users.noreply.github.com> | 2024-02-10 02:22:14 +0100 |
|---|---|---|
| committer | adelon <22380201+adelon@users.noreply.github.com> | 2024-02-10 02:22:14 +0100 |
| commit | 442d732696ad431b84f6e5c72b6ee785be4fd968 (patch) | |
| tree | b476f395e7e91d67bacb6758bc84914b8711593f /library/set/fixpoint.tex | |
Initial commit
Diffstat (limited to 'library/set/fixpoint.tex')
| -rw-r--r-- | library/set/fixpoint.tex | 48 |
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diff --git a/library/set/fixpoint.tex b/library/set/fixpoint.tex new file mode 100644 index 0000000..700295b --- /dev/null +++ b/library/set/fixpoint.tex @@ -0,0 +1,48 @@ +\import{set/powerset.tex} +\import{function.tex} + +\subsection{Fixpoints} + +% NOTE: we need to explicitly require that a is an element of the domain, +% otherwise the emptyset becomes a fixpoint when it's not in the domain. +\begin{definition}\label{fixpoint} + $a$ is a fixpoint of $f$ iff $a\in\dom{f}$ and $f(a) = a$. +\end{definition} + +\begin{definition}\label{subseteqpreserving} + $f$ is \subseteq-preserving iff + for all $A, B\in\dom{f}$ such that $A\subseteq B$ we have $f(A)\subseteq f(B)$. +\end{definition} + +\begin{theorem}[Knaster--Tarski]\label{knastertarski} + Let $f$ be a \subseteq-preserving function from $\pow{A}$ to $\pow{A}$. + Then there exists a fixpoint of $f$. +\end{theorem} +\begin{proof} + $\dom{f} = \pow{A}$. + Let $P = \{a\in\pow{A}\mid a\subseteq f(a)\}$. + $P\subseteq\pow{A}$. + Thus $\unions{P}\in\pow{A}$. + Hence $f(\unions{P})\in\pow{A}$ by \cref{funs_type_apply}. + + Show $\unions{P}\subseteq f(\unions{P})$. + \begin{subproof} + It suffices to show that every element of $\unions{P}$ is an element of $f(\unions{P})$. + % + Fix $u\in\unions{P}$. + % + Take $p\in P$ such that $u\in p$. + Then $u\in f(p)$. + $p\subseteq\unions{P}$. + $f(p)\subseteq f(\unions{P})$ by \cref{subseteqpreserving}. + Thus $u\in f(\unions{P})$. + \end{subproof} + + Now $f(\unions{P})\subseteq f(f(\unions{P}))$ by \cref{subseteqpreserving}. + Thus $f(\unions{P})\in P$ by definition. + + Hence $f(\unions{P})\subseteq \unions{P}$. + + Thus $f(\unions{P}) = \unions{P}$ by \cref{subseteq_antisymmetric}. + Follows by \cref{fixpoint}. +\end{proof} |
