From 3845ab9020b3eb591ef999827503b483eb735bd7 Mon Sep 17 00:00:00 2001 From: adelon <22380201+adelon@users.noreply.github.com> Date: Tue, 21 May 2024 16:52:01 +0200 Subject: Add simple lemmas on filters --- library/set/filter.tex | 29 +++++++++++++++++++++++------ 1 file changed, 23 insertions(+), 6 deletions(-) (limited to 'library/set') diff --git a/library/set/filter.tex b/library/set/filter.tex index 2797d86..93309de 100644 --- a/library/set/filter.tex +++ b/library/set/filter.tex @@ -3,6 +3,8 @@ \section{Filters} +\subsection{Definition and basic properties of filters} + \begin{abbreviation}\label{upwardclosed} $F$ is upward-closed in $S$ iff for all $A, B$ such that $A\subseteq B\subseteq S$ and $A\in F$ we have $B\in F$. @@ -11,22 +13,37 @@ \begin{definition}\label{filter} $F$ is a filter on $S$ iff $F$ is a family of subsets of $S$ - and $S$ is inhabited and $S\in F$ and $\emptyset\notin F$ and $F$ is closed under binary intersections and $F$ is upward-closed in $S$. \end{definition} +\begin{proposition}\label{filter_ext_complement} + Let $F, G$ be filters on $S$. + Suppose for all $A\subseteq S$ we have $S\setminus A\in F$ iff $S\setminus A\in G$. + Then $F = G$. +\end{proposition} +\begin{proof} + Follows by set extensionality. +\end{proof} + +\begin{proposition}\label{filter_inter_in_iff} + Let $F$ be a filter on $S$. + Suppose $A, B\subseteq S$. + Then $A\inter B\in F$ iff $A, B\in F$. +\end{proposition} +\begin{proof} + We have $A\inter B\subseteq A, B$. + Follows by \cref{filter}. +\end{proof} + +\subsection{Principal filters over a set} + \begin{definition}\label{principalfilter} $\principalfilter{S}{A} = \{X\in\pow{S}\mid A\subseteq X\}$. \end{definition} -%\begin{proposition}\label{principalfilter_domain_inhabited} -% Suppose $F$ is a filter on $S$. -% Then $S$ is inhabited. -%\end{proposition} - \begin{proposition}\label{principalfilter_is_filter} Suppose $A\subseteq S$. Suppose $A$ is inhabited. -- cgit v1.2.3