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


\section{The canonical topology on $\mathbb{R}$}

\begin{definition}\label{topological_basis_reals_eps_ball}
    $\topoBasisReals = \{ \epsBall{x}{\epsilon} \mid x \in \reals, \epsilon \in \realsplus\}$.
\end{definition}

\begin{axiom}\label{reals_carrier_reals}
    $\carrier[\reals] = \reals$.
\end{axiom}

\begin{lemma}\label{intervals_are_connected_in_reals}
    Suppose $a,b \in \reals$.
    Then for all $c \in \reals$ such that $a < c < b$ we have $c \in \intervalopen{a}{b}$.
\end{lemma}

\begin{lemma}\label{epsball_are_subset_reals_elem}
    Suppose $x \in \reals$.
    Suppose $\epsilon \in \realsplus$.
    Then for all $y \in \epsBall{x}{\epsilon}$ we have $y \in \reals$.
\end{lemma}

\begin{lemma}\label{intervalopen_iff}
    Suppose $a,b,c \in \reals$.
    Suppose $a < b$.
    $c \in \intervalopen{a}{b}$ iff $a < c < b$.
\end{lemma}

\begin{lemma}\label{epsball_are_subseteq_reals_set}
    Suppose $x \in \reals$.
    Suppose $\epsilon \in \realsplus$.
    Then $\epsBall{x}{\epsilon} \subseteq \reals$.
\end{lemma}

\begin{lemma}\label{epsball_are_subset_reals_set}
    Suppose $x \in \reals$.
    Suppose $\epsilon \in \realsplus$.
    Then $\epsBall{x}{\epsilon} \subset \reals$.
\end{lemma}

\begin{lemma}\label{reals_order_minus_positiv}
    Suppose $x,y \in \reals$.
    Suppose $\zero < y$.
    $x - y < x$.
\end{lemma}

\begin{lemma}\label{realsplus_bigger_zero}
    For all $x \in \realsplus$ we have $\zero < x$.
\end{lemma}

\begin{lemma}\label{realsplus_in_reals}
    For all $x \in \realsplus$ we have $x \in \reals$.
\end{lemma}

\begin{lemma}\label{epsball_are_inhabited}
    Suppose $x \in \reals$.
    Suppose $\epsilon \in \realsplus$.
    Then $\epsBall{x}{\epsilon}$ is inhabited.
\end{lemma}
\begin{proof}
    $x < x + \epsilon$.
    $x - \epsilon < x$.
    $x \in \epsBall{x}{\epsilon}$.
\end{proof}

\begin{lemma}\label{reals_elem_inbetween}
    For all $a,b \in \reals$ such that $a < b$ we have there exists $c \in \reals$ such that $a < c < b$.
\end{lemma}

\begin{lemma}\label{epsball_equal_openinterval}
    Suppose $x \in \reals$.
    Suppose $\epsilon \in \realsplus$.
    Then $\epsBall{x}{\epsilon} = \intervalopen{x - \epsilon}{x + \epsilon}$.
\end{lemma}

\begin{lemma}\label{minus_behavior1}
    For all $x \in \reals$ we have $x - x = \zero$.
\end{lemma}

\begin{lemma}\label{minus_behavior2}
    For all $x \in \reals$ we have $x + \neg{x} = \zero$.
\end{lemma}

\begin{lemma}\label{minus_behavior3}
    For all $x \in \reals$ we have $\neg{x} = \zero - x$.
\end{lemma}

\begin{lemma}\label{reals_order_is_addition_with_positiv_number}
    For all $x,y \in \reals$ such that $x < y$ we have there exists $z \in \realsplus$ such that $x + z = y$.
\end{lemma}
\begin{proof}
    Omitted.
\end{proof}

\begin{lemma}\label{reals_order_is_transitive}
    For all $x,y,z \in \reals$ such that $x < y$ and $y < z$ we have $x < z$.
\end{lemma}
\begin{proof}
    Omitted.
\end{proof}

\begin{lemma}\label{reals_order_plus_minus}
    Suppose $a,b \in \reals$.
    Suppose $\zero < b$.
    Then $(a-b) < (a+b)$.
\end{lemma}
\begin{proof}
    We show that $a < (a+b)$.
    \begin{subproof}
        Trivial.
    \end{subproof}
    We show that $(a-b) < a$.
    \begin{subproof}
        Trivial.
    \end{subproof}
\end{proof}

\begin{lemma}\label{epsball_are_connected_in_reals}
    Suppose $x \in \reals$.
    Suppose $\epsilon \in \realsplus$.
    Then for all $c \in \reals$ such that $(x - \epsilon) < c < (x + \epsilon)$ we have $c \in \epsBall{x}{\epsilon}$.
\end{lemma}
\begin{proof}
    $x - \epsilon \in \reals$.
    $x + \epsilon \in \reals$.


    It suffices to show that for all $c$ such that $c \in \reals \land (x - \epsilon) \rless c \rless (x + \epsilon)$ we have $c \in \epsBall{x}{\epsilon}$.
    Fix $c$ such that $(c \in \reals) \land (x - \epsilon) \rless c \rless (x + \epsilon)$.
    $(x - \epsilon) < c < (x + \epsilon)$.
\end{proof}

\begin{theorem}\label{topological_basis_reals_is_prebasis}
    $\topoBasisReals$ is a topological prebasis for $\reals$.
\end{theorem}
\begin{proof}
    We show that $\unions{\topoBasisReals} \subseteq \reals$.
    \begin{subproof}
        It suffices to show that for all $x \in \unions{\topoBasisReals}$ we have $x \in \reals$.
        Fix $x \in \unions{\topoBasisReals}$.
        \begin{byCase}
            \caseOf{$x = \emptyset$.}
                Trivial.
            \caseOf{$x \neq \emptyset$.}
                %There exists $U \in \topoBasisReals$ such that $x \in U$.
                Take $U \in \topoBasisReals$ such that $x \in U$.
                Follows by \cref{epsball_are_subseteq_reals_set,topological_basis_reals_eps_ball,epsilon_ball,minus,subseteq}.
        \end{byCase}
    \end{subproof}
    We show that $\reals \subseteq \unions{\topoBasisReals}$.
    \begin{subproof}
        It suffices to show that for all $x \in \reals$ we have $x \in \unions{\topoBasisReals}$.
        Fix $x \in \reals$.
        $\epsBall{x}{1} \in \topoBasisReals$.
        Therefore $x \in \unions{\topoBasisReals}$ by \cref{one_in_reals,reals_one_bigger_zero,unions_intro,realsplus,plus_one_order,reals_order_minus_positiv,epsball_are_connected_in_reals}.
    \end{subproof}
\end{proof}

%\begin{lemma}\label{intervl_intersection_is_interval}
%    Suppose $a,b,a',b' \in \reals$.
%    Suppose there exist $x \in \reals$ such that $x \in \intervalopen{a}{b} \inter \intervalopen{a'}{b'}$.
%    Then there exists $q,p \in \reals$ such that $q < p$ and $\intervalopen{q}{p} \subseteq \intervalopen{a}{b} \inter \intervalopen{a'}{b'}$.
%\end{lemma}
%

\begin{lemma}\label{reals_order_total}
    For all $x,y \in \reals$ we have either $x < y$ or $x \geq y$. 
\end{lemma}
\begin{proof}
    It suffices to show that for all $x \in \reals$ we have for all $y \in \reals$ we have either $x < y$ or $x \geq y$. 
    Fix $x \in \reals$.
    Fix $y \in \reals$.
    Omitted.
\end{proof}

\begin{lemma}\label{topo_basis_reals_eps_iff}
    $X \in \topoBasisReals$ iff there exists $x_0, \delta$ such that $x_0 \in \reals$ and $\delta \in \realsplus$ and $\epsBall{x_0}{\delta} = X$.
\end{lemma}

\begin{lemma}\label{topo_basis_reals_intro}
For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have $\epsBall{x}{\delta} \in \topoBasisReals$.
\end{lemma}

\begin{lemma}\label{realspuls_in_reals_plus}
    For all $x,y$ such that $x \in \reals$ and $y \in \realsplus$ we have $x + y \in \reals$.
\end{lemma}

\begin{lemma}\label{realspuls_in_reals_minus}
    For all $x,y$ such that $x \in \reals$ and $y \in \realsplus$ we have $x - y \in \reals$.
\end{lemma}

\begin{lemma}\label{eps_ball_implies_open_interval}
    Suppose $x \in \reals$.
    Suppose $\epsilon \in \realsplus$. 
    Then there exists $a,b \in \reals$ such that $a < b$ and $\intervalopen{a}{b} = \epsBall{x}{\epsilon}$.
\end{lemma}

\begin{lemma}\label{open_interval_eq_eps_ball}
    Suppose $a,b \in \reals$.
    Suppose $a < b$.
    Then there exist $x,\epsilon$ such that $x \in \reals$ and $\epsilon \in \realsplus$ and $\intervalopen{a}{b} = \epsBall{x}{\epsilon}$.
\end{lemma}
\begin{proof}
    Let $\delta = (b-a)$.
    $\delta$ is positiv by \cref{minus_}.
    There exists $\epsilon \in \reals$ such that $\epsilon + \epsilon = \delta$.

\end{proof}



\begin{theorem}\label{topological_basis_reals_is_basis}
    $\topoBasisReals$ is a topological basis for $\reals$.
\end{theorem}
\begin{proof}
    $\topoBasisReals$ is a topological prebasis for $\reals$ by \cref{topological_basis_reals_is_prebasis}.
    Let $B = \topoBasisReals$.
    It suffices to show that for all $U \in B$ we have for all $V \in B$ we have for all $x$ such that $x \in U, V$ there exists $W\in B$ such that $x\in W\subseteq U, V$.
    Fix $U \in B$.
    Fix $V \in B$.
    It suffices to show that for all $x \in U \inter V$ there exists $W\in B$ such that $x\in W\subseteq U, V$.
    Fix $x \in U \inter V$.
    \begin{byCase}
        \caseOf{$U \inter V = \emptyset$.}
            Trivial.
        \caseOf{$U \inter V \neq \emptyset$.}
            Then $U \inter V$ is inhabited.
            $x \in \reals$ by \cref{inter_lower_left,subseteq,topological_prebasis_iff_covering_family,omega_is_an_ordinal,naturals_subseteq_reals,subset_transitive,suc_subseteq_elim,ordinal_suc_subseteq}.
            There exists $x_1, \alpha$ such that $x_1 \in \reals$ and $\alpha \in \realsplus$ and $\epsBall{x_1}{\alpha} = U$.
            There exists $x_2, \beta$ such that $x_2 \in \reals$ and $\beta \in \realsplus$ and $\epsBall{x_2}{\beta} = V$.
            Then $ (x_1 - \alpha) < x < (x_1 + \alpha)$.
            Then $ (x_2 - \beta) < x < (x_2 + \beta)$.
            We show that if there exists $\delta \in \realsplus$ such that $\epsBall{x}{\delta} \subseteq U \inter V$ then there exists $W\in B$ such that $x\in W\subseteq U, V$.
            \begin{subproof}
                Suppose there exists $\delta \in \realsplus$ such that $\epsBall{x}{\delta} \subseteq U \inter V$.
                $x \in \epsBall{x}{\delta}$.
                $\epsBall{x}{\delta} \subseteq U$.
                $\epsBall{x}{\delta} \subseteq V$.
                $\epsBall{x}{\delta} \in B$.
            \end{subproof}
            It suffices to show that there exists $\delta \in \realsplus$ such that $\epsBall{x}{\delta} \subseteq U \inter V$.


            %It suffices to show that there exists $x_0, \delta$ such that $x_0 \in \reals$ and $\delta \in \realsplus$ and $\epsBall{x_0}{\delta} \subseteq u \inter V$.
            %There exists $x_1, \alpha$ such that $x_1 \in \reals$ and $\alpha \in \realsplus$ and $\epsBall{x_1}{\alpha} = U$.
            %There exists $x_2, \beta$ such that $x_2 \in \reals$ and $\beta \in \realsplus$ and $\epsBall{x_2}{\beta} = V$.
            %Then $ (x_1 - \alpha) < x < (x_1 + \alpha)$.
            %Then $ (x_2 - \beta) < x < (x_2 + \beta)$.
            %\begin{byCase}
            %    \caseOf{$x_1 = x_2$.}
            %        Take $\gamma \in \realsplus$ such that either $\gamma = \alpha \land \gamma \leq \beta$ or $\gamma \leq \alpha \land \gamma = \beta$.
            %    \caseOf{$x_1 < x_2$.}
            %    \caseOf{$x_1 > x_2$.}
            %\end{byCase}
            %%Take $m$ such that $m \in \min{\{(x_1 + \alpha), (x_2 + \beta)\}}$.
            %Take $n$ such that $n \in \max{\{(x_1 - \alpha), (x_2 - \beta)\}}$.
            %Then $m < x < n$.
            %We show that there exists $x_1 \in \reals$ such that $x_1 \in U \inter V$ and $x_1 < x$.
            %\begin{subproof}
            %    Suppose not.
            %    Then For all $y \in U \inter V$ we have $x \leq y$.
            %\end{subproof}
            %We show that there exists $x_2 \in \reals$ such that $x_2 \in U \inter V$ and $x_2 > x$.
            %\begin{subproof}
            %    Trivial.
            %\end{subproof}
    \end{byCase}
\end{proof}

\begin{axiom}\label{topological_space_reals}
    $\opens[\reals] = \genOpens{\topoBasisReals}{\reals}$.
\end{axiom}

\begin{theorem}\label{reals_is_topological_space}
    $\reals$ is a topological space.
\end{theorem}
\begin{proof}
    $\topoBasisReals$ is a topological basis for $\reals$.
    Let $B = \topoBasisReals$.
    We show that $\opens[\reals]$ is a family of subsets of $\carrier[\reals]$.
    \begin{subproof}
        It suffices to show that for all $A \in \opens[\reals]$ we have $A \subseteq \reals$.
        Fix $A \in \opens[\reals]$.
        Follows by \cref{powerset_elim,topological_space_reals,genopens}.
    \end{subproof}
    We show that $\reals \in\opens[\reals]$.
    \begin{subproof}
        $B$ covers $\reals$ by \cref{topological_prebasis_iff_covering_family,topological_basis}.
        $\unions{B} \in \genOpens{B}{\reals}$.
        $\reals \subseteq \unions{B}$.
    \end{subproof}
    We show that for all $A, B\in \opens[\reals]$ we have $A\inter B\in\opens[\reals]$.
    \begin{subproof}
        Follows by \cref{topological_space_reals,inters_in_genopens}.
    \end{subproof}
    We show that for all $F\subseteq \opens[\reals]$ we have $\unions{F}\in\opens[\reals]$.
    \begin{subproof}
        Follows by \cref{topological_space_reals,union_in_genopens}.
    \end{subproof}
    $\carrier[\reals] = \reals$.
    Follows by \cref{topological_space}.
\end{proof}

\begin{proposition}\label{open_interval_is_open}
    Suppose $a,b \in \reals$.
    Then $\intervalopen{a}{b} \in \opens[\reals]$.
\end{proposition}