diff options
Diffstat (limited to 'library/topology/real-topological-space.tex')
| -rw-r--r-- | library/topology/real-topological-space.tex | 36 |
1 files changed, 21 insertions, 15 deletions
diff --git a/library/topology/real-topological-space.tex b/library/topology/real-topological-space.tex index d9790aa..b2e5ea9 100644 --- a/library/topology/real-topological-space.tex +++ b/library/topology/real-topological-space.tex @@ -382,7 +382,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have 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 $x \in \epsBall{x'}{\epsilon'}$. + 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 $(x_1 - \alpha) < (x_2 + \beta)$. @@ -536,7 +536,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have \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 $x \in e$. + 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}. $\epsBall{x'}{\delta'} \in E$. @@ -550,8 +550,8 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have Suppose not. Take $y' \in e$ such that $y' > a$. $x'' < a$. - $(x'' - \delta'') < y' < (x'' + \delta'')$. - $(x'' - \delta'') < x'' < (x'' + \delta'')$. + $(x'' - \delta'') < y' < (x'' + \delta'')$ 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}. @@ -565,7 +565,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have Trivial. \end{subproof} \end{subproof} - $\unions{E} \in \opens[\reals]$. + $\unions{E} \in \opens[\reals]$ 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} @@ -600,7 +600,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have \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 $x \in e$. + 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}. $\epsBall{x'}{\delta'} \in E$. @@ -612,10 +612,10 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have 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 $y' < a$. + Take $y' \in e$ such that $y' < a$ by \cref{reals_order_total,intervalopen,eps_ball_implies_open_interval}. $x'' > a$. - $(x'' - \delta'') < y' < (x'' + \delta'')$. - $(x'' - \delta'') < x'' < (x'' + \delta'')$. + $(x'' - \delta'') < y' < (x'' + \delta'')$ 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}. @@ -689,6 +689,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have 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} @@ -697,7 +698,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have 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 $x \in \intervalclosedInfiniteRight{a}$. + 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} @@ -714,6 +715,9 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have 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 $a \in \reals$. @@ -725,7 +729,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have Then $\intervalclosedInfiniteRight{a} \in \closeds{\reals}$. \end{proposition} \begin{proof} - $\intervalclosedInfiniteRight{a} = \reals \setminus \intervalopenInfiniteLeft{a}$. + $\intervalclosedInfiniteRight{a} = \reals \setminus \intervalopenInfiniteLeft{a}$ 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} @@ -747,15 +751,17 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have 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 $x \in \reals \setminus \intervalopenInfiniteLeft{a}$. - Then $x \in \reals \setminus \intervalopenInfiniteRight{b}$. + 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 $x \in (\reals \setminus \intervalopenInfiniteLeft{a})$. - Then $x \in (\reals \setminus \intervalopenInfiniteRight{b})$. + 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} |
