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| author | Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> | 2024-09-23 03:14:06 +0200 |
|---|---|---|
| committer | GitHub <noreply@github.com> | 2024-09-23 03:14:06 +0200 |
| commit | 8fd49ae84e8cc4524c19b20fa0aabb4e77a46cd5 (patch) | |
| tree | 9848da3e57979a5a7e14ec99ee103cfa079e6fcb /library/topology/continuous.tex | |
| parent | 18c79bcb98fb376f15b2b3e00972530df61b26a9 (diff) | |
| parent | f6b22fd533bd61e9dbcb6374295df321de99b1f2 (diff) | |
Abgabe
Submission of Formalisation
Diffstat (limited to 'library/topology/continuous.tex')
| -rw-r--r-- | library/topology/continuous.tex | 16 |
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} + |
