\section{Bounded Sets}
\label{section:bounded}

\begin{definition}[Bounded]
\label{definition:bounded}
    Let $(E, \topo)$ be a TVS over $K \in \RC$ and $B \subset E$, then the following are equivalent:
    \begin{enumerate}
        \item For every $U \in \cn(0)$, there exists $\lambda \in K$ such that $\lambda U \supset B$. 
        \item For every $\seq{x_n} \subset B$ and $\seq{\lambda_n} \subset K$ such that $\lambda_n \to 0$, $\lambda_n x_n \to 0$ as $n \to \infty$.
    \end{enumerate}

    If the above holds, then $B$ is \textbf{bounded}. The collection $B(E) = B(E, \topo)$ is the set of all bounded sets of $E$.
\end{definition}
\begin{proof}
    (1) $\Rightarrow$ (2): Let $U \in \cn_E(0)$ be circled, then there exists $k \in \natp$ such that $kU \supset B$. Since $\lambda_n \to 0$ as $n \to \infty$, there exists $N \in \natp$ such that $|\lambda_n| \le 1/k$ for all $n \ge N$. In which case, $\lambda_n x_n \in \lambda_n B \subset U$ for all $n \ge N$. 

    $\neg (1) \Rightarrow \neg (2)$: Let $U \in \cn_E(0)$ be circled such that $B \not\subset nU$ for all $n \in \natp$. For each $n \in \natp$, let $x_n \in B \setminus nU$, then $x_n/n \not\in U$ for all $n \in \natp$. 
\end{proof}

\begin{proposition}[{{\cite[I.5.1]{SchaeferWolff}}}]
\label{proposition:bounded-operations}
    Let $E$ be a TVS over $K \in \RC$ and $A, B \in B(E)$, 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 \autoref{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 \autoref{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}