\import{topology/topological-space.tex}
\import{topology/separation.tex}
\import{topology/continuous.tex}
\import{topology/basis.tex}
\import{numbers.tex}
\import{function.tex}
\import{set.tex}
\import{cardinal.tex}
\import{relation.tex}
\import{relation/uniqueness.tex}
\import{set/cons.tex}
\import{set/powerset.tex}
\import{set/fixpoint.tex}
\import{set/product.tex}

\section{Urysohns Lemma}


\begin{definition}\label{one_to_n_set}
    $\seq{m}{n} = \{x \in \naturals \mid  m \leq x \leq n\}$.   
\end{definition}

\begin{struct}\label{sequence}
    A sequence $X$ is a onesorted structure equipped with
    \begin{enumerate}
        \item $\index$
        \item $\indexset$
    \end{enumerate}
    such that
    \begin{enumerate}
        \item\label{indexset_is_subset_naturals} $\indexset[X] \subseteq \naturals$.
        \item\label{index_is_bijection} $\index[X]$ is a bijection from $\indexset[X]$ to $\carrier[X]$.
    \end{enumerate}
\end{struct}

\begin{abbreviation}\label{urysohnspace}
    $X$ is a urysohn space iff
    $X$ is a topological space and
    for all $A,B \in \closeds{X}$ such that $A \inter B = \emptyset$
    we have there exist $A',B' \in \opens[X]$
    such that  $A \subseteq A'$ and $B \subseteq B'$ and $A' \inter B' = \emptyset$.    
\end{abbreviation}


\begin{definition}\label{intervalclosed}
    $\intervalclosed{a}{b} = \{x \in \reals \mid a \leq x \leq b\}$.
\end{definition}

\begin{theorem}\label{urysohnsetinbeetween}
    Let $X$ be a urysohn space.
    Suppose $A,B \in \closeds{X}$.
    Suppose $\closure{A}{X} \subseteq \interior{B}{X}$.
    Suppose $\carrier[X]$ is inhabited.
    There exist $U \subseteq \carrier[X]$ such that $U$ is closed in $X$ and $\closure{A}{X} \subseteq \interior{U}{X} \subseteq \closure{U}{X} \subseteq \interior{B}{X}$.
\end{theorem}



\begin{theorem}\label{urysohn}
    Let $X$ be a urysohn space.
    Suppose $A,B \in \closeds{X}$.
    Suppose $A \inter B$ is empty.
    Suppose $\carrier[X]$ is inhabited.
    There exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ 
    and $f(A) = \zero$ and $f(B)= 1$ and $f$ is continuous.
\end{theorem}
\begin{proof}

    Let $H = \carrier[X] \setminus B$.
    Let $P = \{x \in \pow{X} \mid x = A \lor x = H \lor (x \in \pow{X} \land (\closure{A}{X} \subseteq \interior{U}{X} \subseteq \closure{U}{X} \subseteq \interior{H}{X}))\}$.
    Let $\eta = \carrier[X]$.

    
    % Provable 
    % vvv
    Define $F : \eta \to \reals$ such that $F(x) =$
    \begin{cases}
        & \zero                 &\text{if} x \in  A\\
        & \rfrac{1}{1+1}    &\text{if} x \in  (\carrier[X] \setminus (A \union B))\\
        & 1                 &\text{if} x \in  B
    \end{cases}

    %Define $f : \naturals \to \pow{P}$ such that $f(x)=$
    %\begin{cases}
    %    & \emptyset             & \text{if} x = \zero \\
    %    & \{A, H\}             & \text{if} x = 1  \\
    %    & G                     & \text{if} x \in (\naturals \setminus \{1, \zero\}) \land  G = \{g \in \pow{P} \mid g \in f(n-1) \lor (g \notin f(n-1) \land g \in P) \}
    %\end{cases}

    Let $D = \{d \mid d \in \rationals \mid \zero \leq d \leq 1\}$.
    Take $R$ such that for all $q \in D$ we have for all $S \in P$ we have $q \mathrel{R} S$ iff 
        $q = \zero \land S = A$ or $q = 1 \land S = H$ or 
        for all $q_1, q_2, S_1, S_2$ 
        such that $q_1 \leq q \leq q_2$ and $q_1 \mathrel{R} S_1$ and $q_2 \mathrel{R} S_2$ 
        we have $\closure{S_1}{X} \subseteq \interior{S}{X} \subseteq \closure{S}{X} \subseteq \interior{S_2}{X}$ 
        and $q \mathrel{R} S$.
    
    Trivial.
    
    
\end{proof}

\begin{theorem}\label{safe}
    Contradiction.     
\end{theorem}