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diff --git a/library/topology/real-topological-space.tex b/library/topology/real-topological-space.tex new file mode 100644 index 0000000..db7ee94 --- /dev/null +++ b/library/topology/real-topological-space.tex @@ -0,0 +1,786 @@ +\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{Topology Reals}\label{form_sec_toporeals} + +\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} + $x < x + \epsilon$ by \cref{reals_order_behavior_with_addition,realsplus,reals_axiom_zero_in_reals,reals_axiom_kommu,reals_axiom_zero}. + $x - \epsilon < x$. + $x \in \epsBall{x}{\epsilon}$. +\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_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{realsplus_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{one_in_realsplus} + $1 \in \realsplus$. +\end{lemma} + +\begin{lemma}\label{reals_existence_addition_reverse} + For all $\delta \in \reals$ there exists $\epsilon \in \reals$ such that $\epsilon + \epsilon = \delta$. +\end{lemma} +\begin{proof} + Fix $\delta \in \reals$. + Follows by \cref{one_in_realsplus,reals_disstro2,reals_axiom_disstro1,reals_rmul,suc_eq_plus_one,reals_axiom_mul_invers,suc,suc_neq_emptyset,realsplus_in_reals_plus,naturals_addition_axiom_2,naturals_1_kommu,reals_axiom_zero,naturals_inductive_set,one_is_suc_zero,realsplus,reals_one_bigger_zero,one_in_reals,reals_axiom_one,minus_in_reals}. +\end{proof} + +\begin{lemma}\label{reals_addition_minus_behavior1} + For all $a,b,c \in \reals$ such that $a = b + c$ we have $b = a - c$. +\end{lemma} +\begin{proof} + It suffices to show that for all $a \in \reals$ for all $b \in \reals$ for all $c \in \reals$ if $a = b + c$ then $b = a - c$. + Fix $a \in \reals$. + Fix $b \in \reals$. + Fix $c \in \reals$. + Suppose $a = b + c$. + Then $a + \neg{c} = b + c + \neg{c}$. + Therefore $a - c = b + c + \neg{c}$. + $a - c = (b + c) - c$. + $(b + c) - c = (b + c) + \neg{c}$. + $(b + c) + \neg{c} = b + (c + \neg{c})$. + $b + (c + \neg{c}) = b + (\zero)$. + $a - c = b$. +\end{proof} + +\begin{lemma}\label{reals_addition_minus_behavior2} + For all $a,b,c \in \reals$ such that $a = b - c$ we have $b = c + a$. +\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_in_reals,minus_behavior3,reals_axiom_zero_in_reals,reals_order_behavior_with_addition,minus_behavior1,minus}. + There exists $\epsilon \in \reals$ such that $\epsilon + \epsilon = \delta$. + Let $x = a + \epsilon$. + $a + \delta = b$. + $a + \epsilon + \epsilon = b$. + $x + \epsilon = b$. + $\epsilon \in \realsplus$ by \cref{reals_order_behavior_with_addition,reals_axiom_kommu,reals_axiom_zero,reals_order_is_transitive,reals_add,minus_behavior1,minus_behavior3,minus,reals_order_total,reals_axiom_zero_in_reals,realsplus}. + $a = x - \epsilon$. + $b = x + \epsilon$. + We show that $\intervalopen{a}{b} \subseteq \epsBall{x}{\epsilon}$. + \begin{subproof} + It suffices to show that for all $y \in \intervalopen{a}{b}$ we have $y \in \epsBall{x}{\epsilon}$. + Fix $y \in \intervalopen{a}{b}$. + \end{subproof} + We show that $\epsBall{x}{\epsilon} \subseteq \intervalopen{a}{b}$. + \begin{subproof} + It suffices to show that for all $y \in \epsBall{x}{\epsilon}$ we have $y \in \intervalopen{a}{b}$ by \cref{subseteq}. + Fix $y \in \epsBall{x}{\epsilon}$. + \end{subproof} + +\end{proof} + +\begin{lemma}\label{intersection_openinterval_inclusion_of_border} + Suppose $a,b,x,y \in \reals$. + Suppose $a < b$. + Suppose $x < y$. + Suppose $a \leq x < y \leq b$. + Then $\intervalopen{a}{b} \inter \intervalopen{x}{y} = \intervalopen{x}{y}$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{lemma}\label{intersection_openinterval_lower_border_eq} + Suppose $a,b,x,y \in \reals$. + Suppose $a < b$. + Suppose $x < y$. + Suppose $a = x$ and $b \leq y$. + Then $\intervalopen{a}{b} \inter \intervalopen{x}{y} = \intervalopen{a}{b}$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{lemma}\label{intersection_openinterval_upper_border_eq} + Suppose $a,b,x,y \in \reals$. + Suppose $a < b$. + Suppose $x < y$. + Suppose $a \leq x$ and $b = y$. + Then $\intervalopen{a}{b} \inter \intervalopen{x}{y} = \intervalopen{x}{y}$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{lemma}\label{intersection_openinterval_none_border_eq} + Suppose $a,b,x,y \in \reals$. + Suppose $a < b$. + Suppose $x < y$. + If $a \leq x < b \leq y$ then $\intervalopen{a}{b} \inter \intervalopen{x}{y} = \intervalopen{x}{b}$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{lemma}\label{reals_order_total2} + For all $a,b \in \reals$ we have $a < b \lor a > b \lor a = b$. +\end{lemma} + + +\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)$. + Let $a = (x_1 - \alpha)$. + Let $b = (x_1 + \alpha)$. + Let $c = (x_2 - \beta)$. + Let $d = (x_2 + \beta)$. + We have $a < b$ and $a < x$ and $x < b$. + We have $c < d$ and $c < x$ and $x < d$. + We have $a \in \reals$. + We have $b \in \reals$. + We have $c \in \reals$. + We have $d \in \reals$. + We show that there exist $a',b'\in \reals$ such that $\intervalopen{a}{b} \inter \intervalopen{c}{d} = \intervalopen{a'}{b'}$. + \begin{subproof} + \begin{byCase} + \caseOf{$a < c$.} + \begin{byCase} + \caseOf{$b < d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus}. + \caseOf{$b = d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus}. + \caseOf{$b > d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus}. + \end{byCase} + \caseOf{$a = c$.} + \begin{byCase} + \caseOf{$b < d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus}. + \caseOf{$b = d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus}. + \caseOf{$b > d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus}. + \end{byCase} + \caseOf{$a > c$.} + \begin{byCase} + \caseOf{$b < d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus,reals_add,minus_in_reals,realsplus,inter_comm,epsilon_ball}. + \caseOf{$b = d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus,reals_add,minus_in_reals,realsplus,inter_comm,epsilon_ball}. + \caseOf{$b > d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus,reals_add,minus_in_reals,realsplus,inter_comm,epsilon_ball}. + \end{byCase} + \end{byCase} + \end{subproof} + + Take $a',b'\in \reals$ such that $\intervalopen{a}{b} \inter \intervalopen{c}{d} = \intervalopen{a'}{b'}$. + We have $a',b' \in \reals$ by assumption. + We have $a' < b'$ by \cref{id_img,epsilon_ball,minus,intervalopen,reals_order_is_transitive}. + Then there exist $x',\epsilon'$ such that $x' \in \reals$ and $\epsilon' \in \realsplus$ and $\intervalopen{a'}{b'} = \epsBall{x'}{\epsilon'}$. + Then $x \in \epsBall{x'}{\epsilon'}$ by \cref{epsilon_ball}. + + Follows by \cref{inter_lower_left,inter_lower_right,epsilon_ball,topological_basis_reals_eps_ball}. + %Then $(x_1 - \alpha) < (x_2 + \beta)$. + + %Therefore $U \inter V = \intervalopen{}{(x_2 + \beta)}$. + + %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} +\begin{proof} + If $a > b$ then $\intervalopen{a}{b} = \emptyset$. + If $a = b$ then $\intervalopen{a}{b} = \emptyset$. + It suffices to show that if $a < b$ then $\intervalopen{a}{b} \in \opens[\reals]$. + Suppose $a \rless b$. + Take $x, \epsilon$ such that $x \in \reals$ and $\epsilon \in \realsplus$ and $\intervalopen{a}{b} = \epsBall{x}{\epsilon}$. + It suffices to show that $\epsBall{x}{\epsilon} \in \opens[\reals]$. + $\topoBasisReals$ is a topological basis for $\reals$. + $\epsBall{x}{\epsilon} \in \topoBasisReals$. + $\topoBasisReals \subseteq \opens[\reals]$ by \cref{basis_is_in_genopens,topological_space_reals,topological_basis_reals_is_basis}. +\end{proof} + +\begin{lemma}\label{reals_minus_to_realsplus} + Suppose $a,b \in \reals$. + Suppose $a < b$. + Then $(b - a) \in \realsplus$. +\end{lemma} + +\begin{lemma}\label{existence_of_epsilon_upper_border} + Suppose $a,b \in \reals$. + Suppose $a < b$. + Then there exists $\epsilon \in \realsplus$ such that $b \notin \epsBall{a}{\epsilon}$. +\end{lemma} +\begin{proof} + Let $\epsilon = b - a$. + Then $\epsilon \in \realsplus$. + It suffices to show that $b \notin \epsBall{a}{\epsilon}$ by \cref{epsilon_ball,reals_addition_minus_behavior2,realsplus,minus,intervalopen,order_reals_lemma0}. + Suppose not. + Then $ b \in \epsBall{a}{\epsilon}$. + Therefore $ (a - \epsilon) < b < (a + \epsilon)$. + $b = (a + \epsilon)$. + Contradiction. +\end{proof} + +\begin{lemma}\label{existence_of_epsilon_lower_border} + Suppose $a,b \in \reals$. + Suppose $a > b$. + Then there exists $\epsilon \in \realsplus$ such that $b \notin \epsBall{a}{\epsilon}$. +\end{lemma} +\begin{proof} + Let $\epsilon = a - b$. + Then $\epsilon \in \realsplus$. + It suffices to show that $b \notin \epsBall{a}{\epsilon}$ by \cref{epsilon_ball,reals_addition_minus_behavior2,realsplus,minus,intervalopen,order_reals_lemma0}. + Suppose not. + Then $ b \in \epsBall{a}{\epsilon}$. + Therefore $ (a - \epsilon) < b < (a + \epsilon)$. + $b = (a - \epsilon)$. + Contradiction. +\end{proof} + +\begin{proposition}\label{openinterval_infinite_left_in_opens} + Suppose $a \in \reals$. + Then $\intervalopenInfiniteLeft{a} \in \opens[\reals]$. +\end{proposition} +\begin{proof} + Let $E = \{ B \in \pow{\reals} \mid \exists x \in \intervalopenInfiniteLeft{a} . \exists \delta \in \realsplus . B = \epsBall{x}{\delta} \land a \notin \epsBall{x}{\delta} \}$. + We show that for all $x \in \intervalopenInfiniteLeft{a}$ we have there exists $e \in E$ such that $x \in e$. + \begin{subproof} + Fix $x \in \intervalopenInfiniteLeft{a}$. + Then $x < a$. + Take $\delta' \in \realsplus$ such that $a \notin \epsBall{x}{\delta'}$. + $x \in \epsBall{x}{\delta'}$ by \cref{intervalopen,epsilon_ball,reals_addition_minus_behavior1,reals_order_minus_positiv,minus,reals_add,realsplus,intervalopen_infinite_left}. + $a \notin \epsBall{x}{\delta'}$. + $\epsBall{x}{\delta'} \in E$. + \end{subproof} + $E \subseteq \topoBasisReals$. + We show that $\unions{E} = \intervalopenInfiniteLeft{a}$. + \begin{subproof} + We show that $\unions{E} \subseteq \intervalopenInfiniteLeft{a}$. + \begin{subproof} + It suffices to show that for all $x \in \unions{E}$ we have $x \in \intervalopenInfiniteLeft{a}$. + Fix $x \in \unions{E}$. + Take $e \in E$ such that $x \in e$ by \cref{unions_iff}. + $x \in \reals$. + Take $x',\delta'$ such that $x' \in \reals$ and $\delta' \in \realsplus$ and $e = \epsBall{x'}{\delta'}$ by \cref{epsilon_ball,minus,topo_basis_reals_eps_iff,setminus,setminus_emptyset,elem_subseteq}. + $\epsBall{x'}{\delta'} \in E$. + We show that for all $y \in e$ we have $y < a$. + \begin{subproof} + Fix $y \in e$. + Then $y \in \epsBall{x'}{\delta'}$. + $e = \epsBall{x'}{\delta'}$. + There exists $x'' \in \intervalopenInfiniteLeft{a}$ such that there exists $\delta'' \in \realsplus$ such that $e = \epsBall{x''}{\delta''}$ and $a \notin \epsBall{x''}{\delta''}$. + Take $x'',\delta''$ such that $x'' \in \intervalopenInfiniteLeft{a}$ and $\delta'' \in \realsplus$ and $e = \epsBall{x''}{\delta''}$ and $a \notin \epsBall{x''}{\delta''}$. + Suppose not. + Take $y' \in e$ such that $y' > a$. + $x'' < a$. + $(x'' - \delta'') < y' < (x'' + \delta'')$ by \cref{minus,intervalopen,epsilon_ball,realsplus,reals_add,reals_addition_minus_behavior1,reals_order_minus_positiv}. + $(x'' - \delta'') < x'' < (x'' + \delta'')$ by \cref{minus,intervalopen,epsilon_ball,realsplus,reals_add,reals_addition_minus_behavior1,reals_order_minus_positiv}. + Then $x'' < a < y'$. + Therefore $(x'' - \delta'') < a < (x'' + \delta'')$ by \cref{realspuls_in_reals_minus,intervalopen_infinite_left,reals_order_is_transitive,reals_add,realsplus_in_reals,powerset_elim,subseteq}. + Then $a \in e$ by \cref{epsball_are_connected_in_reals,intervalopen_infinite_left,neq_witness}. + Contradiction. + \end{subproof} + $x < a$. + Then $x \in \intervalopenInfiniteLeft{a}$. + \end{subproof} + We show that $\intervalopenInfiniteLeft{a} \subseteq \unions{E}$. + \begin{subproof} + Trivial. + \end{subproof} + \end{subproof} + $\unions{E} \in \opens[\reals]$ by \cref{opens_unions,reals_is_topological_space,basis_is_in_genopens,topological_space_reals,topological_basis_reals_is_basis,subset_transitive}. +\end{proof} + +\begin{lemma}\label{continuous_on_basis_implies_continuous_endo} + Suppose $X$ is a topological space. + Suppose $B$ is a topological basis for $X$. + Suppose $f$ is a function from $X$ to $X$. + $f$ is continuous iff for all $b \in B$ we have $\preimg{f}{b} \in \opens[X]$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{proposition}\label{openinterval_infinite_right_in_opens} + Suppose $a \in \reals$. + Then $\intervalopenInfiniteRight{a} \in \opens[\reals]$. +\end{proposition} +\begin{proof} + Let $E = \{ B \in \pow{\reals} \mid \exists x \in \intervalopenInfiniteRight{a} . \exists \delta \in \realsplus . B = \epsBall{x}{\delta} \land a \notin \epsBall{x}{\delta} \}$. + We show that for all $x \in \intervalopenInfiniteRight{a}$ we have there exists $e \in E$ such that $x \in e$. + \begin{subproof} + Fix $x \in \intervalopenInfiniteRight{a}$. + Then $a < x$. + Take $\delta' \in \realsplus$ such that $a \notin \epsBall{x}{\delta'}$. + $x \in \epsBall{x}{\delta'}$ by \cref{intervalopen,epsilon_ball,reals_addition_minus_behavior1,reals_order_minus_positiv,minus,reals_add,realsplus,intervalopen_infinite_right}. + $a \notin \epsBall{x}{\delta'}$. + $\epsBall{x}{\delta'} \in E$. + \end{subproof} + $E \subseteq \topoBasisReals$. + We show that $\unions{E} = \intervalopenInfiniteRight{a}$. + \begin{subproof} + We show that $\unions{E} \subseteq \intervalopenInfiniteRight{a}$. + \begin{subproof} + It suffices to show that for all $x \in \unions{E}$ we have $x \in \intervalopenInfiniteRight{a}$. + Fix $x \in \unions{E}$. + Take $e \in E$ such that $x \in e$ by \cref{unions_iff}. + $x \in \reals$. + Take $x',\delta'$ such that $x' \in \reals$ and $\delta' \in \realsplus$ and $e = \epsBall{x'}{\delta'}$ by \cref{epsilon_ball,minus,topo_basis_reals_eps_iff,setminus,setminus_emptyset,elem_subseteq}. + $\epsBall{x'}{\delta'} \in E$. + We show that for all $y \in e$ we have $y > a$. + \begin{subproof} + Fix $y \in e$. + Then $y \in \epsBall{x'}{\delta'}$. + $e = \epsBall{x'}{\delta'}$. + There exists $x'' \in \intervalopenInfiniteRight{a}$ such that there exists $\delta'' \in \realsplus$ such that $e = \epsBall{x''}{\delta''}$ and $a \notin \epsBall{x''}{\delta''}$. + Take $x'',\delta''$ such that $x'' \in \intervalopenInfiniteRight{a}$ and $\delta'' \in \realsplus$ and $e = \epsBall{x''}{\delta''}$ and $a \notin \epsBall{x''}{\delta''}$. + Suppose not. + Take $y' \in e$ such that $y' < a$ by \cref{reals_order_total,intervalopen,eps_ball_implies_open_interval}. + $x'' > a$. + $(x'' - \delta'') < y' < (x'' + \delta'')$ by \cref{minus,intervalopen,intervalclosed,epsilon_ball,realsplus,reals_add,reals_addition_minus_behavior1,reals_order_minus_positiv}. + $(x'' - \delta'') < x'' < (x'' + \delta'')$ by \cref{minus,intervalopen,intervalclosed,epsilon_ball,realsplus,reals_add,reals_addition_minus_behavior1,reals_order_minus_positiv}. + Then $x'' > a > y'$. + Therefore $(x'' - \delta'') > a > (x'' + \delta'')$ by \cref{realspuls_in_reals_minus,intervalopen_infinite_right,reals_order_is_transitive,reals_add,realsplus_in_reals,powerset_elim,subseteq,epsball_are_connected_in_reals,subseteq}. + Then $a \in e$ by \cref{reals_order_is_transitive,reals_order_total,reals_add,realsplus,epsball_are_connected_in_reals,intervalopen_infinite_right,neq_witness}. + Contradiction. + \end{subproof} + $x > a$. + Then $x \in \intervalopenInfiniteRight{a}$. + \end{subproof} + We show that $\intervalopenInfiniteRight{a} \subseteq \unions{E}$. + \begin{subproof} + Trivial. + \end{subproof} + \end{subproof} + $\unions{E} \in \opens[\reals]$. + + %Let $I = \{\neg{b} \mid b \in \intervalopenInfiniteRight{a} \}$. + %Let $f(x) = \neg{x}$ for $x \in \reals$. + %$f$ is a function from $\reals$ to $\reals$. + %We show that $f$ is continuous. + %\begin{subproof} + % It suffices to show that for all $b \in \topoBasisReals$ we have $\preimg{f}{b} \in \opens[\reals]$. + % Fix $b \in \topoBasisReals$. + % Take $x, \epsilon$ such that $x \in \reals$ and $\epsilon \in \realsplus$ and $b = \epsBall{x}{\epsilon}$. + % Let $y = \neg{x}$. + % It suffices to show that $\preimg{f}{b} \in \topoBasisReals$ by \cref{topological_space_reals,topological_basis_reals_is_basis,basis_is_in_genopens,cons_remove,cons_subseteq_iff}. + % It suffices to show that $\epsBall{y}{\epsilon} = \preimg{f}{\epsBall{x}{\epsilon}}$. + % $\preimg{f}{\epsBall{x}{\epsilon}} \subseteq \reals$. + % $\epsBall{y}{\epsilon} \subseteq \reals$ by \cref{intervalopen,subseteq,minus,epsilon_ball}. + % %It suffices to show that for all $x \in \reals$ we have $x \in \epsBall{y}{\epsilon}$ iff $x \in \preimg{f}{\epsBall{x}{\epsilon}}$. + % %Fix $x \in \reals$. + % Let $u = (y - \epsilon)$. + % Let $v = (y + \epsilon)$. + % $u = \neg{(x - \epsilon)}$. + % $v = \neg{(x + \epsilon)}$. + % %$v - u = \epsilon + \epsilon$. + % We show that for all $z \in \epsBall{y}{\epsilon}$ we have $z \in \preimg{f}{\epsBall{x}{\epsilon}}$. + % \begin{subproof} + % Fix $z \in \epsBall{y}{\epsilon}$. + % Then $u < z < v$. + % Let $z' = z - u$. + % Then $z = u + z'$. + % Suppose not. + % Let $h = \neg{z}$. + % $\neg{h} = \neg{\neg{z}}$. + % $\neg{h} = z$. + % Then $f(h) = \neg{h}$. + % $f(h) = z$. + % Then $z \in \preimg{f}{\{h\}}$. +% + % \end{subproof} + % We show that for all $z \in \preimg{f}{\epsBall{x}{\epsilon}}$ we have $z \in \epsBall{y}{\epsilon}$. + % \begin{subproof} + % Fix $z \in \preimg{f}{\epsBall{x}{\epsilon}}$. + % Take $h \in \epsBall{x}{\epsilon}$ such that $f(h) = z$. + % \end{subproof} + % Follows by set extensionality. + %\end{subproof} + %$\intervalopenInfiniteLeft{a} \in \opens[\reals]$. + %We show that $\preimg{f}{\intervalopenInfiniteLeft{a}} = \intervalopenInfiniteRight{a}$. + %\begin{subproof} + % Omitted. + %\end{subproof} + %Then $\intervalopenInfiniteRight{a} \in \opens[\reals]$ by \cref{continuous,preim_eq_img_of_converse,openinterval_infinite_left_in_opens}. +\end{proof} + +\begin{lemma}\label{reals_as_union_of_open_closed_intervals1} + Suppose $a,b \in \reals$. + Then $\reals = \intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b} \union \intervalclosed{a}{b}$. +\end{lemma} +\begin{proof} + We show that for all $x \in \reals$ we have $x \in (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b} \union \intervalclosed{a}{b})$. + \begin{subproof} + Fix $x \in \reals$. + Follows by \cref{union_intro_left,intervalopen_infinite_left,reals_order_total,reals_order_total2,union_iff,intervalopen_infinite_right,union_assoc,union_intro_right,intervalclosed}. + \end{subproof} +\end{proof} + +\begin{lemma}\label{reals_as_union_of_open_closed_intervals2} + Suppose $a \in \reals$. + Then $\reals = \intervalopenInfiniteLeft{a} \union \intervalclosedInfiniteRight{a}$. +\end{lemma} +\begin{proof} + It suffices to show that for all $x \in \reals$ we have either $x \in \intervalopenInfiniteLeft{a}$ or $x \in \intervalclosedInfiniteRight{a}$ by \cref{intervalopen_infinite_left,union_intro_left,neq_witness,intervalclosed_infinite_right,union_intro_right,union_iff}. + Trivial. +\end{proof} + +\begin{lemma}\label{reals_as_union_of_open_closed_intervals3} + Suppose $a \in \reals$. + Then $\reals = \intervalopenInfiniteRight{a} \union \intervalclosedInfiniteLeft{a}$. +\end{lemma} +\begin{proof} + It suffices to show that for all $x \in \reals$ we have either $x \in \intervalclosedInfiniteLeft{a}$ or $x \in \intervalopenInfiniteRight{a}$. + Trivial. +\end{proof} + +\begin{lemma}\label{intersection_of_open_closed__infinite_intervals_open_right} + Suppose $a \in \reals$. + Then $\intervalopenInfiniteRight{a} \inter \intervalclosedInfiniteLeft{a} = \emptyset$. +\end{lemma} +\begin{proof} + Follows by \cref{reals_order_total,inter_lower_left,intervalopen_infinite_right,order_reals_lemma6,inter_lower_right,foundation,subseteq,intervalclosed_infinite_left}. +\end{proof} + +\begin{lemma}\label{intersection_of_open_closed__infinite_intervals_open_left} + Suppose $a \in \reals$. + Then $\intervalopenInfiniteLeft{a} \inter \intervalclosedInfiniteRight{a} = \emptyset$. +\end{lemma} + +\begin{proposition}\label{closedinterval_infinite_right_in_closeds} + Suppose $a \in \reals$. + Then $\intervalclosedInfiniteRight{a} \in \closeds{\reals}$. +\end{proposition} +\begin{proof} + $\intervalclosedInfiniteRight{a} = \reals \setminus \intervalopenInfiniteLeft{a}$ by \cref{intersection_of_open_closed__infinite_intervals_open_left,reals_as_union_of_open_closed_intervals2,setminus_inter,double_relative_complement,subseteq_union_setminus,subseteq_setminus,setminus_union,setminus_disjoint,setminus_partition,setminus_subseteq,setminus_emptyset,setminus_self,setminus_setminus,double_complement_union}. +\end{proof} + +\begin{proposition}\label{closedinterval_infinite_left_in_closeds} + Suppose $a \in \reals$. + Then $\intervalclosedInfiniteLeft{a} \in \closeds{\reals}$. +\end{proposition} +\begin{proof} + $\intervalclosedInfiniteLeft{a} = \reals \setminus \intervalopenInfiniteRight{a}$. +\end{proof} + +\begin{proposition}\label{closedinterval_eq_openintervals_setminus_reals} + Suppose $a,b \in \reals$. + Then $\reals \setminus (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b}) = \intervalclosed{a}{b}$. +\end{proposition} +\begin{proof} + We have $\intervalclosed{a}{b} \subseteq \reals$. + We show that $\reals \setminus (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b}) = (\reals \setminus \intervalopenInfiniteLeft{a}) \inter (\reals \setminus \intervalopenInfiniteRight{b})$. + \begin{subproof} + We show that for all $x \in \reals \setminus (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b})$ we have $x \in (\reals \setminus \intervalopenInfiniteLeft{a}) \inter (\reals \setminus \intervalopenInfiniteRight{b})$. + \begin{subproof} + Fix $x \in \reals \setminus (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b})$. + Then $x \in \reals \setminus \intervalopenInfiniteLeft{a}$ by \cref{setminus,double_complement_union}. + Then $x \in \reals \setminus \intervalopenInfiniteRight{b}$ by \cref{union_upper_left,subseteq,union_comm,subseteq_implies_setminus_supseteq}. + Follows by \cref{inter_intro}. + \end{subproof} + We show that for all $x \in (\reals \setminus \intervalopenInfiniteLeft{a}) \inter (\reals \setminus \intervalopenInfiniteRight{b})$ we have $x \in \reals \setminus (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b})$. + \begin{subproof} + Fix $x \in (\reals \setminus \intervalopenInfiniteLeft{a}) \inter (\reals \setminus \intervalopenInfiniteRight{b})$. + Then $x \in (\reals \setminus \intervalopenInfiniteLeft{a})$ by \cref{setminus_setminus,setminus}. + Then $x \in (\reals \setminus \intervalopenInfiniteRight{b})$ by \cref{inter_lower_right,elem_subseteq,setminus_setminus}. + \end{subproof} + Follows by \cref{setminus_union}. + \end{subproof} + We show that $\reals \setminus \intervalopenInfiniteLeft{a} = \intervalclosedInfiniteRight{a}$. + \begin{subproof} + For all $x \in \intervalclosedInfiniteRight{a}$ we have $x \geq a$. + For all $x \in (\reals \setminus \intervalopenInfiniteLeft{a})$ we have $x \geq a$. + Follows by set extensionality. + \end{subproof} + $\reals \setminus \intervalopenInfiniteRight{b} = \intervalclosedInfiniteLeft{b}$. + It suffices to show that $\intervalclosedInfiniteLeft{b} \inter \intervalclosedInfiniteRight{a} = \intervalclosed{a}{b}$. + For all $x \in \intervalclosed{a}{b}$ we have $a \leq x \leq b$. + For all $x \in (\intervalclosedInfiniteLeft{b} \inter \intervalclosedInfiniteRight{a})$ we have $a \leq x \leq b$. + Follows by set extensionality. +\end{proof} + +\begin{proposition}\label{closedinterval_is_closed} + Suppose $a,b \in \reals$. + Then $\intervalclosed{a}{b} \in \closeds{\reals}$. +\end{proposition} +\begin{proof} + We have $\reals = \intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b} \union \intervalclosed{a}{b}$. + It suffices to show that $\reals \setminus (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b}) = \intervalclosed{a}{b}$ by \cref{closeds,setminus_subseteq,powerset_intro,closed_minus_open_is_closed,opens_type,subseteq_refl,union_open,is_closed_in,reals_carrier_reals,setminus_self,emptyset_open,reals_is_topological_space,openinterval_infinite_left_in_opens,openinterval_infinite_right_in_opens}. +\end{proof} |