saundersp/notebook
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

contents/topology.tex : fixed wrong proof closure of intersection

Browse Source
This commit is contained in:
saundersp 2025-02-21 22:52:00 +01:00
parent f97e3b72be
commit 27a23d004b
1 changed files with 2 additions and 4 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

6
contents/topology.tex
View File

@ -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}