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authorSimon-Kor <52245124+Simon-Kor@users.noreply.github.com>2024-09-23 03:14:06 +0200
committerGitHub <noreply@github.com>2024-09-23 03:14:06 +0200
commit8fd49ae84e8cc4524c19b20fa0aabb4e77a46cd5 (patch)
tree9848da3e57979a5a7e14ec99ee103cfa079e6fcb /library/topology/continuous.tex
parent18c79bcb98fb376f15b2b3e00972530df61b26a9 (diff)
parentf6b22fd533bd61e9dbcb6374295df321de99b1f2 (diff)
Abgabe
Submission of Formalisation
Diffstat (limited to 'library/topology/continuous.tex')
-rw-r--r--library/topology/continuous.tex16
1 files changed, 14 insertions, 2 deletions
diff --git a/library/topology/continuous.tex b/library/topology/continuous.tex
index 6a32353..95c4d0a 100644
--- a/library/topology/continuous.tex
+++ b/library/topology/continuous.tex
@@ -1,5 +1,9 @@
\import{topology/topological-space.tex}
\import{relation.tex}
+\import{function.tex}
+\import{set.tex}
+
+\subsection{Continuous}\label{form_sec_continuous}
\begin{definition}\label{continuous}
$f$ is continuous iff for all $U \in \opens[Y]$ we have $\preimg{f}{U} \in \opens[X]$.
@@ -8,18 +12,25 @@
\begin{proposition}\label{continuous_definition_by_closeds}
Let $X$ be a topological space.
Let $Y$ be a topological space.
+ Let $f \in \funs{X}{Y}$.
Then $f$ is continuous iff for all $U \in \closeds{Y}$ we have $\preimg{f}{U} \in \closeds{X}$.
\end{proposition}
\begin{proof}
Omitted.
- %We show that if $f$ is continuous then for all $U \in \closeds{Y}$ we have $\preimg{f}{U} \in \closeds{X}$.
+ %We show that if $f$ is continuous then for all $U \in \closeds{Y}$ such that $U \neq \emptyset$ we have $\preimg{f}{U} \in \closeds{X}$.
%\begin{subproof}
% Suppose $f$ is continuous.
% Fix $U \in \closeds{Y}$.
% $\carrier[Y] \setminus U$ is open in $Y$.
% Then $\preimg{f}{(\carrier[Y] \setminus U)}$ is open in $X$.
% Therefore $\carrier[X] \setminus \preimg{f}{(\carrier[Y] \setminus U)}$ is closed in $X$.
- % $\carrier[X] \setminus \preimg{f}{(\carrier[Y] \setminus U)} \subseteq \preimg{f}{U}$.
+ % We show that $\carrier[X] \setminus \preimg{f}{(\carrier[Y] \setminus U)} \subseteq \preimg{f}{U}$.
+ % \begin{subproof}
+ % It suffices to show that for all $x \in \carrier[X] \setminus \preimg{f}{(\carrier[Y] \setminus U)}$ we have $x \in \preimg{f}{U}$.
+ % Fix $x \in \carrier[X] \setminus \preimg{f}{(\carrier[Y] \setminus U)}$.
+ % Take $y \in \carrier[Y]$ such that $f(x)=y$.
+ % It suffices to show that $y \in U$.
+ % \end{subproof}
% $\preimg{f}{U} \subseteq \carrier[X] \setminus \preimg{f}{(\carrier[Y] \setminus U)}$.
%\end{subproof}
%We show that if for all $U \in \closeds{Y}$ we have $\preimg{f}{U} \in \closeds{X}$ then $f$ is continuous.
@@ -27,3 +38,4 @@
% Omitted.
%\end{subproof}
\end{proof}
+