Basis of infinite Vector Space $\mathbb R^{\infty}$ I have a question about one example in Linear Algebra.
Let $\mathbb R^∞$ be the vector space of infinite sequences $(\alpha_1, \alpha_2, \alpha_3, \ldots )$ of real numbers.
Scalar multiplication are defined in the natural way: the sum of $(\alpha_1 , \alpha_2 , \alpha_3 , \ldots )$ and $(\beta_1,\beta_2,\beta_3,\ldots)$ is $(\alpha_1 +\beta_1, \alpha_2 + \beta_2, \alpha_3 + \beta_3,\ldots)$ the product of $(\alpha_1,\alpha_2,\alpha_3,\ldots)$ by a scalar $\lambda$ is the sequence $(\lambda \alpha_1, \lambda\alpha_2, \lambda\alpha_3, \ldots )$.
There exists infinite linear independent set of vectors $(e_1, e_2, e_3, \ldots)$
\begin{align}
e_1 &= (1, 0, 0, \ldots)\\
e_2 &= (0, 1, 0, \ldots)\\
& \,\,\,\vdots
\end{align}
The problem is that this set (lets call it $X$) is not a basis of this vector space. Because for example $v = (1, 1, 1, \ldots)$ cannot be written as a linear combination of set $X$ (Linear combination must be a finite sum).
My task is to add "some vectors" to the set $X$ to create a basis of that vector space. If I add $v$ its not basis ($\langle X, v\rangle \ne\mathbb R^∞$)
Is there any proof that the process of adding vectors to set $X$ is not finite? Or is it possible to create a basis with adding vectors to $X$?
Thanks for answers
 A: Take all non-zero sequences, and impose a well-order on them (which is possible assuming the Axiom of Choice). If you want to have the partial basis you identified as part of the final basis, arrange that the well-ordering starts with that partial basis.
Now take the set of those vectors that cannot be written as a linear combination of vectors preceding it in that well-order. Those vectors will form a basis of the vector space.
Linear independence of that set is quite obvious from the construction, and that it spans the complete vector space is seen from the fact that any non-zero vector that is not a linear combination of the vectors in that set would in particular not be a linear combination of those that precede it in the total order; but then by construction it should be in that set.
A: You have noticed the difference between $\prod^\infty R$ (the vector space of all sequences of elements in $R$) and $\bigoplus^\infty R$ (the vector space of sequences which are only finitely often nonzero). The latter has a basis of cardinality $\aleph_0$ (the basis you write down in your question), the former has a basis of cardinality $2^{\aleph_0}$, and some choice principle must be invoked to prove the existence of this basis (and there even exist models of set theory where this space provably does not have a basis). In particular, yes, you cannot get a basis for $\prod^\infty R$ by adding finitely many basis elements to the standard basis for $\bigoplus^\infty R$, since it has strictly larger cardinality.
How can we describe a Hamel basis for $\prod^\infty R$? We can't, not with standard set theoretic constructions. Since it is consistent with ZF set theory that no such basis exist, no ZF constructions (like tuples, powerset, set builder) will allow you to write down this basis.
How can we prove that such a basis exists then, using the axiom of choice or Zorn's lemma? Well linearly independent sets are ordered by inclusion. Each chain of such sets has a maximum (take the union). Therefore by Zorn's lemma there is a maximal linearly independent set, also known as a basis. Or see this answer by Michael Hardy.
