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| author | adelon <22380201+adelon@users.noreply.github.com> | 2024-05-22 16:58:06 +0200 |
|---|---|---|
| committer | adelon <22380201+adelon@users.noreply.github.com> | 2024-05-22 16:58:06 +0200 |
| commit | 3e4e7afc69bf43b3b45bde346c92f267e9b15c39 (patch) | |
| tree | 20cc299224860cb1e2db76841e2508c3441cdff5 /library | |
| parent | 311fff704f96869e0ea239f129a7cf846c206c4f (diff) | |
Add lemma `filter_setminus_in`
Diffstat (limited to 'library')
| -rw-r--r-- | library/set/filter.tex | 15 |
1 files changed, 13 insertions, 2 deletions
diff --git a/library/set/filter.tex b/library/set/filter.tex index 93309de..59b647f 100644 --- a/library/set/filter.tex +++ b/library/set/filter.tex @@ -38,6 +38,19 @@ Follows by \cref{filter}. \end{proof} +\begin{proposition}\label{filter_setminus_in} + Let $F$ be a filter on $S$. + Suppose $A\in F$. + Suppose $B\subseteq S$ and $S\setminus B\in F$. + Then $A\setminus B\in F$. +\end{proposition} +\begin{proof} + We have $A\subseteq S$. + Thus $A\setminus B = A\inter (S\setminus B)$ by \cref{setminus_eq_inter_complement}. + Now $S\setminus B\subseteq S$. + Follows by \cref{filter_inter_in_iff}. +\end{proof} + \subsection{Principal filters over a set} \begin{definition}\label{principalfilter} @@ -72,8 +85,6 @@ Suppose $X\notin\principalfilter{S}{A}$. Then $A\not\subseteq X$. \end{proposition} -\begin{proof} -\end{proof} \begin{definition}\label{maximalfilter} $F$ is a maximal filter on $S$ iff |
