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 October 1st, 2009, 02:51 PM #1 Newbie   Joined: Oct 2009 Posts: 1 Thanks: 0 Prove S is not finitely generated. please help Let S={(a1,a2,a3,...)l ai c-(belongs to) R, forall i c- N} be the vector space of a infinite sequences of real numbers. (We know that S is a vector space over the feild R.) Prove carefully that S is not finitely generated. Can a spanning set equal all of the elements of S? Then the largest spanning set would be infinite because S has no limit to the elements in it. If I am wrong can someone please prove this or if I'm right and you would like to prove this as well, please be my guess. thank you
October 1st, 2009, 03:59 PM   #2
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Quote:
 Originally Posted by mattpi Suppose $T=\{t_1,t_2,t_3,\ldots\}$ is a finite set that is contained in S. Since S contains only finite sequences (I am taking this to mean that for each  there exists an $N\in \mathbb{N}$ such that $x_n=0$ for all $n\,>\,N$ ), each $t_i$ has a highest non-zero term. Let $K=\max(n\in\mathbb{N}t_i)_n\neq0\text{ for some }i)." /> Then the sequence $(0,0,0,\ldots,0,1,0,0,0,\ldots)$ where the '1' is in the (K+1)th position cannot be expressed as a linear combination of the elements of $T.$ Therefore T does not span S.
OK - given the original typo, that's all irrelevant now...

Suppose there exists a finite basis $V=\{v_1,v_2,\ldots,v_N\}.$ Consider the vectors $e_1=(1,0,0,\ldots),\ e_2=(0,1,0,0,\ldots),\ e_3=(0,0,1,0,0,\ldots)$ etc.

The set $\{e_1,e_2,e_3,\ldots,e_{N+1}\}$ is contained in $S,$ and is linearly independent (easy to show). It is of dimension larger than the basis $V.$ Can you see why $V$ cannot be a basis?

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