Added uniform structures for completely regular spaces. Added calculus lemma.

src/dg/derivative/sets.tex

@@ -209,33 +209,3 @@
 
 
 
-\begin{proposition}
-\label{proposition:derivative-sets-real}
-    Let $E$ be a separated topological vector space and $\sigma \subset \mathfrak{B}(\real)$ be a covering ideal, then
-    \begin{enumerate}
-        \item $\mathcal{R}_{\sigma}(\real; E) = \mathcal{R}_{\mathfrak{B}(\real)}(\real; E)$. Hence, all forms of $\sigma$-differentiability on $\real$ are equivalent.
-        \item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, $f$ is differentiable at $x_0$ if and only if
-        \[
-        \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}
-        \]
-
-        exists. In which case, the above limit is identified with the derivative of $f$ at $0$.
-        \item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, if $f$ is differentiable at $x_0$, then $f$ is continuous at $x_0$.
-    \end{enumerate}
-\end{proposition}
-\begin{proof}
-    (1): Let $r \in \mathcal{R}_\sigma(\real; E)$. For any $R > 0$ and $U \in \cn_E(0)$, there exists $\delta > 0$ such that $t^{-1}r(tR), t^{-1}r(-tR) \in U$ for all $t \in (0, \delta)$. Thus $t^{-1}r(tB(0, R)) \subset U$, and $r \in \mathcal{R}_{\mathfrak{B}(\real)}(\real; E)$.
-
-    (2): Suppose that $f$ is differentiable at $x_0$, then there exists $r \in \mathcal{R}_\sigma$ such that for any $t \in \real$ with $x_0 + t \in U$,
-    \begin{align*}
-        f(x_0 + t) - f(x_0) &= Df(x_0)(t) + r(t) \\
-        \frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1) + t^{-1}r(t) \\
-        \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1)
-    \end{align*}
-    Now suppose that $v = \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}$ exists. Let $T: \real \to E$ be defined by $t \mapsto tv$, then
-    \[
-    \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0) - Tt}{t} = \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} - v = 0
-    \]
-
-    and $Df(x_0) = T$.
-\end{proof}