Cleanup

src/topology/main/compact.tex

@@ -17,26 +17,30 @@
     \[
     U_J = \bigcup_{j \in J}E_j^c
     \]
+
     then $U_J \subset X$ is open. For any $J, J' \subset I$, $U_J \cup U_{J'} = U_{J \cup J'}$.
 
     Suppose for contradiction that $\bigcap_{i \in I}E_i = \emptyset$, then
     \[
     \mathbf{U} = \bracs{U_J|J \subset I \text{ finite}}
     \]
+
     is an open cover for $X$. By assumption, $U_J \subsetneq X$ for all $J \subset I$ finite. Thus $\mathbf{U}$ admits no finite subcover, contradiction.
 
     (2) $\Rightarrow$ (3): Let $\fF \subset 2^X$ be a filter, then $\bracsn{\overline{E}| E \in \fF}$ satisfies the hypothesis of (2).
 
-    (3) $\Leftrightarrow$ (4): By \ref{definition:accumulation-point}, the cluster points and the limit points of an ultrafilter coincide.
+    (3) $\Leftrightarrow$ (4): By \autoref{definition:accumulation-point}, the cluster points and the limit points of an ultrafilter coincide.
 
     (3) $\Rightarrow$ (1): For each $J \subset I$, let
     \[
     E_J = \bigcap_{j \in J}U_j^c
     \]
+
     then for each $J, J' \subset I$, $E_J \cap E_{J'} = E_{J \cup J'}$. Assume for contradiction that $\mathbf{U}$ admits no finite subcover. Let
     \[
     \fB = \bracs{E_H|J \subset I \text{ finite}}
     \]
+
     then $\fB$ is a filter base consisting of closed sets. By assumption, there exists $x \in \bigcap_{i \in I}U_j^c$, so $\mathbf{U}$ is not an open cover, contradiction.
 \end{proof}