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 $n \in \natp$.$\square$
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