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-rw-r--r--library/topology/real-topological-space.tex36
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}