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\import{topology/topological-space.tex}
\import{relation.tex}
\import{function.tex}
\import{set.tex}
\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}
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