Proposition 8.3.2. Let $E$ be a TVS over $K \in \RC$ and $A, B \in B(E)$, then the following sets are bounded:
Any $C \subset B$.
The closure $\ol{B}$.
$\lambda B$ where $\lambda \in K$.
$A \cup B$.
$A + B$.
Proof. Let $U \in \cn(0)$.
(2): Using Proposition 5.1.16, 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 Proposition 8.1.11, 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
\[\mu U \supset \mu(V + V) \supset \lambda V + \lambda' V \supset A + B\]
$\square$