summaryrefslogtreecommitdiff
path: root/library/set/filter.tex
diff options
context:
space:
mode:
authoradelon <22380201+adelon@users.noreply.github.com>2024-05-21 16:52:01 +0200
committeradelon <22380201+adelon@users.noreply.github.com>2024-05-21 16:52:01 +0200
commit3845ab9020b3eb591ef999827503b483eb735bd7 (patch)
tree492fa80da5901a02c30b3b4d6642890af3ed4ebd /library/set/filter.tex
parent59cec853a697a4dd793c216c9bd603bd775d6da2 (diff)
Add simple lemmas on filters
Diffstat (limited to 'library/set/filter.tex')
-rw-r--r--library/set/filter.tex29
1 files changed, 23 insertions, 6 deletions
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.