Consider the $\mathbb R-$ vector space of sequences $\{(x_1,x_2,...)\mid x_i\in\mathbb R\}$. A basis of this vector space is $\{e_i\mid i\in\mathbb N\}$ where $e_i=(\delta _{ij})_{j\in\mathbb N}$. So a sequence can be written as $$(x_i)_{i\in\mathbb N}=\sum_{i=1}^\infty x_i e_i,$$

but I'm not sure in which sense I have to think this sum... formally it makes sense, but the limit looks weird. What do you think ?

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    $\begingroup$ If you're permitting infinite sequences, then you're mistaken about the basis. Infinite sums aren't defined in a general vector space, so $(1, 1, 1, \ldots)$ is not a linear combination of the $e_i$. The space does have a basis, but you need the axiom of choice to prove that. $\endgroup$ – Robert Shore Apr 9 at 22:27
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    $\begingroup$ Making sense of this sum requires taking a limit. Taking a limit requires a choice in topology. We get a different answer to your question, for instance, with different choices of norm. $\endgroup$ – Omnomnomnom Apr 9 at 22:27
  • $\begingroup$ @RobertShore: How can this space doesn't have a basis ? Every vector space has a basis. $\endgroup$ – user657324 Apr 10 at 5:24
  • $\begingroup$ It does have a basis (assuming the Axiom of Choice). But the set that you're calling a basis is merely linearly independent. It doesn't span, because when you talk about linear combinations infinite sums aren't allowed. $\endgroup$ – Robert Shore Apr 10 at 7:07

For infinite-dimensional vector spaces there are two notions of "basis":

1) The notion of Hamel basis. This is the one from your basic linear algebra course. Only finite linear combinations are allowed. In particular your $(e_i)_{i\in\mathbb{N}}$ is not a Hamel basis for the space of sequences $V=\mathbb{R}^{\mathbb{N}}$. Also note that Hamel bases are essentially useless for infinite-dimensional spaces. In that case, the important bases are...

2) Schauder bases.
In the generation property you are allowed the use infinite linear combinations, i.e., limits of finite sums. Your $(e_i)_{i\in\mathbb{N}}$ is in fact a Schauder basis for $\mathbb{R}^{\mathbb{N}}$ seen as a topological vector space when equipped with the product topology. It is even an unconditional Schauder basis.


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