Cleanup

src/topology/main/interiorclosureboundary.tex

@@ -18,7 +18,7 @@
 
     $(3) \Rightarrow (1)$: By (F1) of $\cn(x)$, $A \in \cn(x)$.
 
-    Let $U \subset A$ be open, then $U \in \cn(x)$ for all $x \in U$ by \ref{lemma:openneighbourhood}. By (3), $U \subset A^o$, so $A^o$ contains every open subset of $A$. On the other hand, for any $x \in A^o$, (2) implies that there exists $U \in \cn^o(x)$ with $U \subset A$. In which case, $U \subset A^o$, and $A^o \in \cn(x)$. Therefore $A^o$ itself is open by \ref{lemma:openneighbourhood}.
+    Let $U \subset A$ be open, then $U \in \cn(x)$ for all $x \in U$ by \autoref{lemma:openneighbourhood}. By (3), $U \subset A^o$, so $A^o$ contains every open subset of $A$. On the other hand, for any $x \in A^o$, (2) implies that there exists $U \in \cn^o(x)$ with $U \subset A$. In which case, $U \subset A^o$, and $A^o \in \cn(x)$. Therefore $A^o$ itself is open by \autoref{lemma:openneighbourhood}.
 \end{proof}
 
 \begin{definition}[Closure]
@@ -72,9 +72,9 @@
     If the above holds, then $A$ is a \textbf{dense} subset of $X$.
 \end{definition}
 \begin{proof}
-    $(1) \Rightarrow (2)$: Since $U \ne \emptyset$, there exists $x \in U$, so $U \in \cn(x)$. By (2) of \ref{definition:closure}, $U \cap A \ne \emptyset$.
+    $(1) \Rightarrow (2)$: Since $U \ne \emptyset$, there exists $x \in U$, so $U \in \cn(x)$. By (2) of \autoref{definition:closure}, $U \cap A \ne \emptyset$.
 
-    $(2) \Rightarrow (3)$: Let $\emptyset \ne U \subset X$ open. For each $x \in U$ and $V \in \cn(x)$, $U \cap V \in \cn(x)$ by (F2). Thus there exists $y \in A \cap U \cap V$. By (3) of \ref{definition:closure}, $x \in \overline{A \cap U}$.
+    $(2) \Rightarrow (3)$: Let $\emptyset \ne U \subset X$ open. For each $x \in U$ and $V \in \cn(x)$, $U \cap V \in \cn(x)$ by (F2). Thus there exists $y \in A \cap U \cap V$. By (3) of \autoref{definition:closure}, $x \in \overline{A \cap U}$.
 
     $(3) \Rightarrow (1)$: $X$ is open.
 \end{proof}
@@ -92,7 +92,7 @@
     Let $\seqf{X_j}$ be separable topological spaces, then $\prod_{j = 1}^n X_j$ is also separable.
 \end{proposition}
 \begin{proof}
-    Let $\seq{A_j}$ such that $A_j \subset X_j$ is countable and dense for each $1 \le j \le n$. By \ref{proposition:dense-product}, $\prod_{j = 1}^n A_j$ is countable and dense in $\prod_{j = 1}^n X_j$.
+    Let $\seq{A_j}$ such that $A_j \subset X_j$ is countable and dense for each $1 \le j \le n$. By \autoref{proposition:dense-product}, $\prod_{j = 1}^n A_j$ is countable and dense in $\prod_{j = 1}^n X_j$.
 \end{proof}
 
 
@@ -101,7 +101,7 @@
     Let $X$ be a topological space and $A \subset X$, then $(A^o)^c = \overline{A^c}$.
 \end{lemma}
 \begin{proof}
-    Let $x \in X$. By (3) of \ref{definition:interior}, $x \in (A^o)^c$ if and only if there exists no $U \in \cn(x)$ such that $U \subset A$. Thus $x \in (A^o)^c$ if and only if $U \cap A^c \ne \emptyset$ for all $U \in \cn(x)$. By (2) of \ref{definition:closure}, this is equivalent to $x \in \ol{A^c}$.
+    Let $x \in X$. By (3) of \autoref{definition:interior}, $x \in (A^o)^c$ if and only if there exists no $U \in \cn(x)$ such that $U \subset A$. Thus $x \in (A^o)^c$ if and only if $U \cap A^c \ne \emptyset$ for all $U \in \cn(x)$. By (2) of \autoref{definition:closure}, this is equivalent to $x \in \ol{A^c}$.
 \end{proof}
 
 
@@ -120,9 +120,9 @@
 \begin{proof}
     $(1) \Rightarrow (2)$: $\cn(x) \supset \fB$.
 
-    $(2) \Rightarrow (3)$: By (2) of \ref{definition:closure}, $x \in \overline{A}$. Since $V \cap A^c \ne \emptyset$ for all $V \in \fB$, there exists no open set $U \subset A$ with $x \in A$. By (2) of \ref{definition:interior}, $x \not\in A^o$.
+    $(2) \Rightarrow (3)$: By (2) of \autoref{definition:closure}, $x \in \overline{A}$. Since $V \cap A^c \ne \emptyset$ for all $V \in \fB$, there exists no open set $U \subset A$ with $x \in A$. By (2) of \autoref{definition:interior}, $x \not\in A^o$.
 
-    $(3) \Rightarrow (4)$: By \ref{lemma:closurecomplement}.
+    $(3) \Rightarrow (4)$: By \autoref{lemma:closurecomplement}.
 
-    $(4) \Rightarrow (1)$: By (2) of \ref{definition:closure}.
+    $(4) \Rightarrow (1)$: By (2) of \autoref{definition:closure}.
 \end{proof}