Basis for $\mathbb{R}^{\infty}$ It follows from Zorn's Lemma that every vector space $V$ has a basis (this means, a subset $B$ of $V$ that generate any $v \in V$ by means a finite linear combination, and such that $B$ is LI) . But, seems quite impossible to me to find a basis for $\mathbb{R}^{\infty}$; I have listened somewhere that is really impossible find an actual basis for this space, because of something that I don't remember. 
There exist a concrete basis? There exists some reason for we never shall find one?
 A: Finding a basis for $\Bbb R^\infty$ depends on how you choose to define $\Bbb R^\infty$. Let's expand on @MattSamuel's comment by introducing some terminology.
The support of a function $f:X\to\Bbb R$ is the set
$$
\DeclareMathOperator{supp}{supp}\supp(f)=\{x\in X:f(x)\neq0\}
$$
One reasonable definition of $\Bbb R^\infty$ is
$$
\Bbb R^\infty=\left\{\Bbb N\xrightarrow{f}\Bbb R:\lvert\supp(f)\rvert<\infty\right\}
$$
It is not difficult that this definition makes $\Bbb R^\infty$ an $\Bbb R$-vector space. To exhibit a basis, for $j\in\Bbb N$ let $\chi_j:\Bbb N\to\Bbb R$ be 
$$
\chi_j(n)=
\begin{cases}
1 & n=j \\
0 & n\neq j
\end{cases}
$$
One then shows that
$$
\beta=\{\chi_j:j\in\Bbb N\}
$$
is a basis for $\Bbb R^\infty$.
If, however, one chooses to define $\Bbb R^\infty$ as the collection of all maps $f:\Bbb N\to\Bbb R$, then $\Bbb R^\infty$ is still an $\Bbb R$-vector space and therefore Zorn's lemma implies that $\Bbb R^\infty$ has a basis. This time a basis is much more difficult (or maybe even impossible!) to write down.
