Added facts about vector measures.

src/topology/functions/uniform.tex

@@ -17,29 +17,6 @@
 \end{remark}
 
 
-\begin{definition}[Space of Bounded Functions]
-\label{definition:bounded-function-space}
-    Let $T$ be a set, $E$ be a TVS over $K \in \RC$, and $f: T \to E$ be a function, then $f$ is \textbf{bounded} if $f(T)$ is a bounded subset of $E$. The set $B(T; E) \subset E^T$ is the space of all bounded functions from $T$ to $E$, and:
-    \begin{enumerate}
-        \item $B(T; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
-        \item $B(T; E)$ is a closed subset of $E^T$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $B(T; E)$.
-    \end{enumerate}
-\end{definition}
-\begin{proof}
-    (1): For any $f, g \in B(T; E)$ and $\lambda \in K$, $(\lambda f + g)(T) \subset (\lambda f)(T) + g(T)$, which is bounded by \autoref{proposition:bounded-operations}, so $B(T; E)$ is closed under addition and scalar multiplication. 
-    
-    Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by \autoref{proposition:tvs-set-uniformity}.
-
-    (2): Let $f \in \ol{B(T; E)}$ and $U \in \cn_E(0)$ be circled, then there exists $g \in B(T; E)$ such that $(f - g)(E) \subset U$. Since $g \in B(T; E)$, there exists $\lambda \ge 0$ such that $g(E) \subset \lambda U$. In which case,
-    \[
-    f(E) \subset g(E) + (f - g)(E) \subset (\lambda + 1)U
-    \]
-    so $f \in B(T; E)$.
-
-    If $E$ is complete, then $E^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $B(T; E)$ is also complete by \autoref{proposition:complete-closed}.
-\end{proof}
-
-
 \begin{proposition}
 \label{proposition:uniform-limit-continuous}
     Let $X$ be a topological space and $Y$ be a uniform space, then:
@@ -72,20 +49,3 @@
     If $Y$ is complete, then $Y^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $C(T; X)$ and $UC(T; X)$ are both complete subspaces by \autoref{proposition:complete-closed}.
 \end{proof}
 
-
-\begin{definition}[Space of Bounded Continuous Functions]
-\label{definition:bounded-continuous-function-space}
-    Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then $BC(X; E) = B(X; E) \cap C(X; E)$ is the space of bounded and continuous functions from $X$ to $E$, and
-    \begin{enumerate}
-        \item $BC(X; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
-        \item $BC(X; E)$ is a closed subspace of $C(X; E)$ and $B(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $BC(X; E)$.
-    \end{enumerate}
-\end{definition}
-\begin{proof}
-    (1): Since addition and scalar multiplication are continuous, $BC(X; E)$ is a subspace of $B(X; E)$, and hence a TVS over $K \in \RC$ by \autoref{definition:bounded-function-space}.
-
-    (2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^X$ by \autoref{definition:bounded-function-space}, $BC(X; E)$ is a closed subspace. 
-    
-    If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by \autoref{definition:bounded-function-space}. Therefore $BC(X; E)$ is also complete. 
-\end{proof}
-

src/topology/uniform/cauchy.tex

@@ -24,6 +24,21 @@
     Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a filter on $X$, then $\fF$ is \textbf{Cauchy} if for every $V \in \fU$, there exists $E \in \fF$ such that $E$ is $V$-small.
 \end{definition}
 
+\begin{definition}[Cauchy Net]
+\label{definition:cauchy-net}
+    Let $(X, \fU)$ be a uniform space and $\net{x} \subset X$ be a net, then $\net{x}$ is \textbf{Cauchy} if for any $V \in \fU$, there exists $\alpha_0 \in A$ such that $(x_\alpha, x_\beta) \in V$ for all $\alpha, \beta \ge \alpha_0$.
+\end{definition}
+
+\begin{lemma}
+\label{lemma:cauchy-subnet}
+    Let $(X, \fU)$ be a uniform space and $\net{x} \subset X$ be a Cauchy net. If there exists a subnet $\angles{x_\beta}_B \subset \net{x}$ and $x \in X$ such that $x_\beta \to x$, then $x_\alpha \to x$.
+\end{lemma}
+\begin{proof}
+    Let $U \in \fU$ and $V \in \fU$ such that $V \circ V \subset U$, then there exists $\alpha_0 \in A$ such that $(x_\alpha, x_{\alpha'}) \in V$ for all $\alpha, \alpha' \ge \alpha_0$, and $\beta \in B$ with $\beta \ge \alpha_0$ such that $(x_\beta, x) \in V$. Thus, $(x_\alpha, x) \in V \circ V \subset U$ for all $\alpha \ge \alpha_0$, so $x_\alpha \to x$.
+\end{proof}
+
+
+
 \begin{proposition}[Cauchy Criterion]
 \label{proposition:cauchycriterion}
     Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a convergent filter, then $\fF$ is Cauchy.