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I'm trying to understand sequence spaces as vector spaces. I know that various sequence spaces are vector spaces (over $\mathbb{R}$, say) with componentwise addition and scalar multiplication, such as $$\{ (x_n) \mid x_n \in \mathbb{R} \} = \{ \text{ real-valued sequences}\}$$ $$\{ (x_n) \mid x_n \in \mathbb{R},\; \lim x_n =0 \} = \{ \text{ real-valued sequences converging to zero}\}$$ $$\{ (x_n) \mid x_n \in \mathbb{R},\; \lim x_n \text{ exists} \} = \{ \text{convergent real-valued sequences}\}.$$

However, these spaces do not have bases, in the sense that there is no collection of elements such that every sequence can be expressed as a finite linear combination of basis elements. You need infinite combinations of the standard basis vectors $e_i$, (where $(e_i)_n = 1$ if $n =i$, and zero else).

I've been reading through standard proofs of the existence of bases, such as this one, but I can't see where they break down in these cases.

Can anyone point out what goes wrong, or where I may be misunderstanding the definitions? Thanks.

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Those spaces do have bases (any vector space does). These do just not consist simply of the standard basis vectors from the finite dimensional subspaces. – Tobias Kildetoft Feb 8 '13 at 16:48
And infinite linear combinations are not defined at all. (You are talking about vector spaces, not Banach spaces or alike) – Martin Brandenburg Feb 8 '13 at 17:37
up vote 3 down vote accepted

Why do you say that your vector spaces do not have bases? If you assume the Axiom of Choice the proof you are referring to, applies to your examples. Only the bases have a more than countable number of elements...

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