Prove `emptyset_open` to replace structure axiom

2 files changed, 11 insertions, 5 deletions

diff --git a/library/set.tex b/library/set.tex
index e7e062f..fcd2642 100644
--- a/library/set.tex
+++ b/library/set.tex
@@ -131,8 +131,7 @@ which applies it to goals of the form “$A = B$” and  “$A \neq B$”.
     If $x$ and $y$ are empty, then $x = y$.
 \end{proposition}
 
-\begin{proposition}%
-\label{emptyset_subseteq}
+\begin{proposition}\label{emptyset_subseteq}
     For all $a$ we have $\emptyset \subseteq a$.
     % LATER $\emptyset$ is a subset of every set.
 \end{proposition}
@@ -266,8 +265,7 @@ The $\operatorname{\textsf{cons}}$ operation is determined by the following axio
     There exists $B\in C$ such that $A\in B$.
 \end{proof}
 
-\begin{proposition}%
-\label{unions_emptyset}
+\begin{proposition}\label{unions_emptyset}
     $\unions{\emptyset} = \emptyset$.
 \end{proposition}
 
diff --git a/library/topology/topological-space.tex b/library/topology/topological-space.tex
index e467d48..2bbdf09 100644
--- a/library/topology/topological-space.tex
+++ b/library/topology/topological-space.tex
@@ -11,7 +11,6 @@
     such that
     \begin{enumerate}
         \item\label{opens_type} $\opens[X]$ is a family of subsets of $\carrier[X]$.
-        \item\label{emptyset_open} $\emptyset\in\opens[X]$.
         \item\label{carrier_open} $\carrier[X]\in\opens[X]$.
         \item\label{opens_inter} For all $A, B\in \opens[X]$ we have $A\inter B\in\opens[X]$.
         \item\label{opens_unions} For all $F\subseteq \opens[X]$ we have $\unions{F}\in\opens[X]$.
@@ -26,6 +25,15 @@
     $U$ is open in $X$ iff $U\in\opens[X]$.
 \end{abbreviation}
 
+\begin{proposition}\label{emptyset_open}
+    Let $X$ be a topological space.
+    Then $\emptyset$ is open in $X$.
+\end{proposition}
+\begin{proof}
+    We have $\unions{\emptyset} = \emptyset\subseteq\opens[X]$ by \cref{unions_emptyset,emptyset_subseteq}.
+    Follows by \cref{opens_unions}.
+\end{proof}
+
 \begin{proposition}\label{union_open}
     Let $X$ be a topological space.
     Suppose $A$, $B$ are open.