10.4 Bounded Sets
Definition 10.4.1 (Bounded).label Let $(E, \topo)$ be a TVS over $K \in \RC$ and $B \subset E$, then the following are equivalent:
- (1)
For every $U \in \cn(0)$, there exists $\lambda \in K$ such that $\lambda U \supset B$.
- (2)
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 $n \to \infty$.
If the above holds, then $B$ is bounded. The collection $\mathfrak{B}(E) = \mathfrak{B}(E, \topo)$ is the set of all bounded sets of $E$.
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 $n \in \natp$.$\square$
Proposition 10.4.2 ([I.5.1, SW99]).label Let $E$ be a TVS over $K \in \RC$ and $A, B \in \mathfrak{B}(E)$, then the following sets are bounded:
- (1)
Any $C \subset B$.
- (2)
The closure $\ol{B}$.
- (3)
$\lambda B$ where $\lambda \in K$.
- (4)
$A \cup B$.
- (5)
$A + B$.
Proof. Let $U \in \cn(0)$.
(2): Using Proposition 6.1.16, 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 10.1.11, 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
and
$\square$
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