Cleanup

src/fa/tvs/continuous.tex

@@ -13,9 +13,9 @@
     If the above holds, then $T$ is a \textbf{continuous linear map}. The set $L(E; F)$ denotes the vector space of all continuous linear maps from $E$ to $F$.
 \end{definition}
 \begin{proof}
-    $(1) \Rightarrow (2) \Rightarrow (3)$: By \ref{proposition:uniform-continuous} and \ref{definition:continuity}.
+    $(1) \Rightarrow (2) \Rightarrow (3)$: By \autoref{proposition:uniform-continuous} and \autoref{definition:continuity}.
 
-    $(3) \Rightarrow (1)$: Let $U$ be an entourage of $F$, there exists an entourage $V$ of $E$ such that $T(V(0)) \subset U(0)$.  Using \ref{proposition:tvs-uniform} and \ref{lemma:translation-invariant-symmetric}, assume without loss of generality that $U$ and $V$ are symmetric and translation-invariant.
+    $(3) \Rightarrow (1)$: Let $U$ be an entourage of $F$, there exists an entourage $V$ of $E$ such that $T(V(0)) \subset U(0)$.  Using \autoref{proposition:tvs-uniform} and \autoref{lemma:translation-invariant-symmetric}, assume without loss of generality that $U$ and $V$ are symmetric and translation-invariant.
 
     For any $x, y \in V$, $x - y \in V(0)$, so $Tx - Ty \in U(0)$, $Ty \in U(Tx)$ by symmetry, and $(Tx, Ty) \in U$. Therefore $T$ is uniformly continuous.
 \end{proof}
@@ -47,6 +47,7 @@
         \prod_{i \in I}E_i \ar@{->}[r]_{\pi_i} & E_i
         }
         \]
+
     \end{enumerate}
     The uniformity $\fU$ and its induced topology are the \textbf{product uniformity/topology}, and $E$ equipped with $\fU$ is the \textbf{product TVS} of $\seqi{E}$.
 \end{definition}