From 27a23d004b264615760bf97b1867012e634653d6 Mon Sep 17 00:00:00 2001 From: saundersp Date: Fri, 21 Feb 2025 22:52:00 +0100 Subject: [PATCH] contents/topology.tex : fixed wrong proof closure of intersection --- contents/topology.tex | 6 ++---- 1 file changed, 2 insertions(+), 4 deletions(-) diff --git a/contents/topology.tex b/contents/topology.tex index 2f11609..b3cde9d 100644 --- a/contents/topology.tex +++ b/contents/topology.tex @@ -161,13 +161,11 @@ Source : \citeannexes{scholarpedia_topological_transitivity} \subseteqpart - Posons $A \union B \subseteq A$ par \ref{theorem:subset_implies_closure} $\implies \closure{A \union B} \subseteq \closure{A}$ et respectivement pour $B$, $\closure{A \union B} \subseteq \closure{B}$, et en faisant l'union de deux, cela donne $\closure{A \union B} \subseteq \closure{A} \union \closure{B}$. + Sachant que $A \subseteq \closure{A} \land B \subseteq \closure{B}$ par \ref{proposition:closure_is_smallest_closed} en faisait l'union des deux cela donne $A \union B \subseteq \closure{A} \union \closure{B}$, or $\closure{A} \union \closure{B} \equivalence E\setminus\closure{A} \intersection E\setminus\closure{B}$, il s'agit d'une intersection finie d'ouverts donc $\closure{A} \union \closure{B}$ est fermé donc par \ref{proposition:closure_is_smallest_closed} $\implies \closure{A \union B} \subseteq \closure{A} \union \closure{B}$. \Lsubseteqpart - Sachant que $A \subseteq \closure{A} \land B \subseteq \closure{B}$ par \ref{proposition:closure_is_smallest_closed} en faisait l'union des deux cela donne $A \union B \subseteq \closure{A} \union \closure{B}$, or $\closure{A} \union \closure{B} \equivalence E\setminus\closure{A} \intersection E\setminus\closure{B}$, il s'agit d'une intersection finie d'ouverts donc $\closure{A} \union \closure{B}$ est fermé donc par \ref{proposition:closure_is_smallest_closed} $\implies \closure{A \union B} \subseteq \closure{A} \union \closure{B}$. - - $(\closure{A \union B} \subseteq \closure{A} \union \closure{B}) \land (\closure{A \union B} \supseteq \closure{A} \union \closure{B}) \implies \closure{A \union B} = \closure{A} \union \closure{B}$ + Posons $A \union B \supseteq A$ par \ref{theorem:subset_implies_closure} $\implies \closure{A \union B} \supseteq \closure{A}$ et respectivement pour $B$, $\closure{A \union B} \supseteq \closure{B}$, et en faisant l'union de deux, cela donne $\closure{A \union B} \supseteq \closure{A} \union \closure{B}$. \end{proof} \langsection{Complétude}{Completeness}