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garden/src/fa/tvs/bounded.tex
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\section{Bounded Sets}
\label{section: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}