Housekeeping.

src/dg/derivative/sets.tex

@@ -82,7 +82,6 @@
 \label{proposition:chain-rule-sets-conditions}
     Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated. If $\sigma \subset B(E)$ and $\tau \subset B(F)$ correspond to the following families of sets on $E$ and $F$:
     \begin{enumerate}
-        \item Finite sets.
         \item Compact sets.
         \item Bounded sets.
     \end{enumerate}
@@ -96,9 +95,7 @@
 \begin{proof}
     (1): Let $A \in \sigma$ and $U \in \cn_G(0)$. Since $T$ is continuous, there exists $V \in \cn_F(0)$ such that $T(V) \subset U$. Since $r \in \mathcal{R}_\sigma(E; G)$, there exists $t > 0$ such that $r(sA)/s \in V$ for all $s \in (0, t)$. In which case, $T \circ r(sA)/s \in U$ for all $s \in (0, t)$.
 
-    (2): Firstly, for any $A \subset E$ finite, $(T + r)(A)$ is also finite, so the claim holds directly in the case of finite sets.
-
-    To show that $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$, it is sufficient to show that for every $A \in \sigma$, $\seq{t_n} \subset \real_{> 0}$ with $t_n \downto 0$ as $n \to \infty$, and $\seq{a_n} \subset A$,
+    (2): To show that $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$, it is sufficient to show that for every $A \in \sigma$, $\seq{t_n} \subset \real_{> 0}$ with $t_n \downto 0$ as $n \to \infty$, and $\seq{a_n} \subset A$,
     \[
     \limv{n} \frac{1}{t_n}s \circ (T + r)(t_na_n) = \limv{n}\frac{1}{t_n}s\braks{t_n\paren{Ta_n + \frac{r(t_na_n)}{t_n}}} = 0
     \]
@@ -112,13 +109,14 @@
     \[
     \limv{n}\frac{1}{t_n}s\braks{t_n\paren{Ta_n + \frac{r(t_na_n)}{t_n}}} = 0
     \]
-
 \end{proof}
 
 
 
 
 \begin{remark}
+\label{remark:differentiation-bornology}
+
     In \autoref{definition:differentiation-small}, the system $\sigma$ can be chosen based on the bornology of $E$, and the definition of small-ness depends exclusively on $\sigma$. As such, there is an apparent disconnect between differentiation and the topology of the domain.
 
     Consider for example a Hilbert space equipped with its norm and weak topology. The norm itself is differentiable with respect to both topologies, because the bounded sets coincide. Moreover, the data for differentiability needs to only come from a neighbourhood of $0$ in the norm topology. As such, a function may be differentiable even if its domain is too small to have an interior.