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src/fa/tvs/bounded.tex

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+\section{Bounded Sets}
+\label{section:bounded}
+
+\subsection{Bounded Sets}
+\label{subsection:tvs-bounded}
+
+\begin{definition}[Bounded]
+\label{definition:bounded}
+    Let $E$ be a TVS over $K \in \RC$ and $B \subset E$, then $B$ is \textbf{bounded} if for every $U \in \cn(0)$, there exists $\lambda \in K$ such that $\lambda U \supset B$.
+\end{definition}
+
+\begin{proposition}[{{\cite[1.5.1]{SchaeferWolff}}}]
+\label{proposition:bounded-operations}
+    Let $E$ be a TVS over $K \in \RC$ and $A, B \subset E$ be bounded, then the following sets are bounded:
+    \begin{enumerate}
+        \item Any $C \subset B$.
+        \item The closure $\ol{B}$.
+        \item $\lambda B$ where $\lambda \in K$.
+        \item $A \cup B$.
+        \item $A + B$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    Let $U \in \cn(0)$.
+
+    (2): Using \ref{proposition:uniform-neighbourhoods}, assume without loss of generality that $U$ is closed. Let $0 \ne \lambda \in K$ with $\lambda U \supset B$, then since $\lambda U$ is closed, $\lambda U \supset \ol B$.
+
+    (4), (5): By \ref{proposition:tvs-good-neighbourhood-base}, there exists $V \in \cn(0)$ circled such that $V + V \subset U$, and $\lambda, \lambda' \in K$ such that $\lambda V \supset A$ and $\lambda' V \supset B$.
+
+    Let $\mu > \abs{\lambda}, \abs{\lambda'}$, then
+    \[
+    \mu U \supset \mu V \supset \lambda V \cup \lambda' V \supset A \cup B
+    \]
+    and
+    \[
+    \mu U \supset \mu(V + V) \supset \lambda V + \lambda' V \supset A + B
+    \]
+\end{proof}