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| author | adelon <22380201+adelon@users.noreply.github.com> | 2025-07-02 20:28:22 +0200 |
|---|---|---|
| committer | GitHub <noreply@github.com> | 2025-07-02 20:28:22 +0200 |
| commit | 793849dd0b20bc70ea0e0ecfd5008a3806eca0d9 (patch) | |
| tree | 280949f358a695c5471212cc5b22950399d8cd57 /library/topology/continuous.tex | |
| parent | 3caadfbe0fdb417b8edebc17002ddafe795a4bcc (diff) | |
| parent | 8fd49ae84e8cc4524c19b20fa0aabb4e77a46cd5 (diff) | |
Merge pull request #2 from Simon-Kor/main
Merge (finally)
Diffstat (limited to 'library/topology/continuous.tex')
| -rw-r--r-- | library/topology/continuous.tex | 41 |
1 files changed, 41 insertions, 0 deletions
diff --git a/library/topology/continuous.tex b/library/topology/continuous.tex new file mode 100644 index 0000000..95c4d0a --- /dev/null +++ b/library/topology/continuous.tex @@ -0,0 +1,41 @@ +\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 $\preimg{f}{U} \in \opens[X]$. +\end{definition} + +\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 $U \in \closeds{Y}$ 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 $X$. + % 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. + %\begin{subproof} + % Omitted. + %\end{subproof} +\end{proof} + |
