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Let $R^\infty$ be the vector space of all sequence $\{a_j\}$ of real numbers. Put $\|\{a_j\}\|_n:= \sum_{j=0}^n |a_j|$. This collection of semi norms make this as Frechet space.

A set $B$ is bounded if every continuous seminorm is bounded on $B$. That is bounded set will be as $\{a=\{a_i\}: \sum_{i=0}^k |a_i|<M_k\text{ for some } M_k>0 \text{ and } \forall k\in \mathbb N\}.$

Can I have an explicit example of one bounded set and one unbounded set.

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What is your question? –  Chris Eagle Jul 18 '12 at 20:52
    
sir, question edited.. i hope this make sense now? –  zapkm Jul 18 '12 at 20:57
    
Does "the set of the series that are absolutely convergent" answer your question? –  S4M Jul 18 '12 at 21:02
    
@S4M, Take (1,1,1,,,) this series is not absolutely convergent, still lies in bounded set.. –  zapkm Jul 18 '12 at 21:04
    
@zapkm then, $A_K=\{a=\{a_i\},\displaystyle\sum_{i=0}^{\infty}|a_i|<K\}$ is an explicit example of a bounded set. For an unbounded set, you can take $a=\{a_i\}$ such as $\displaystyle\sum_{i=0}^{\infty}|a_i|=1$. Then $\{\{a^n_i\}=\{n\times a_i\} \text{where } n\in\mathbb{N}\}$ is clearly unbounded. –  S4M Jul 18 '12 at 21:22

2 Answers 2

up vote 1 down vote accepted

We can check that a subset $B$ of $\Bbb R^{\infty}$ is bounded if and only if all the subsets of $\Bbb R$ defined by $$B_k:=\{x_k,(x_n)_{n\in\Bbb N}\in B\}$$ are bounded.

For example, the set $\{(x_n)_{n\geq 1}\},|x_n|\leq n\}$ is bounded.

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thanks for the answer.. –  zapkm Jul 18 '12 at 21:18

The set $\{(x_n)_{n\in N}\}$ with the sequence $x_n=0$ is bounded. Define the set $B=\{(a_n^1),(a_n^2),...,(a_n^m),...\}$ such that $a_n^m = n$, if $n\leq m$ and 0 otherwise. It is clear that $|(a_n^m)| = \sum\limits_{i=1}^{m} i = \frac{(m+1)m}{2}$. Since the value of the norms are unbounded the set is unbounded.

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