Q:

Equivalent definition of bounded set in norm linear space

Definition of Bounded set in norm linear space.

If $X$ is norm linear space and $B \subseteq X$, then $B$ is bounded if there exists $M>0$ such that $\sup_{n\in \mathbb N} \lvert \lvert x_n \rvert \rvert \leq M$.

Please help me to prove this statement.

Let $X$ be norm linear space and $B \subseteq X$. Then $B$ is bounded if and only if for every sequence $(x_n)$ contained in $B$ and for every sequence $(\alpha_n) \in c_0$, $\alpha_nx_n \to 0$ where $c_0$ is the set of all sequences in $\mathbb C$ that converge to $0$.

A:

First let us assume that $B$ is bounded. Let $(x_n)$ be any sequence in $B$ and $(a_n)\in c_0$. Then $\|x_n\|\leq M$ for all $n$ and given $\epsilon >0$ there exist a natural number $N$ such that $|a_n|<\frac{\epsilon}{M}$ for all $n\geq N$. So, for $n\geq N$ we have 
\begin{equation}
\|a_nx_n\| = |a_n|\|x_n\|<\frac{\epsilon}{M}M = \epsilon.
\end{equation}
Therefore, $a_nx_n \longrightarrow 0$.
Conversely, let us assume that for any sequence $(x_n)$ in $B$ and for any $(a_n)\in c_0$; $a_nx_n \longrightarrow 0$. If possible let us assume that $B$ is not bounded. Then given any natural number $n$ there exist will $x_n \in B$ such that $\|x_n\|> n$. Now, consider the sequence $(a_n) = (\frac{1}{n}) \in c_0$. Then $\|a_nx_n\| = |a_n|\|x_n\| > \frac{1}{n}n = 1$. So, $a_nx_n$ does not converge to $0$, which contradicts our hypothesis. Hence it follows that $B$ must be bounded.

