Find a basis of a vector space where F(R) is the set of all functions from R to R F(R) is the set of all functions from R to R.
(F(R),R,+,•) is a vector space.
What is a basis of given vector space?
What if we restrict the set of functions to functions with finite domain?
 A: The theorem that states that every vector space has a basis uses the axiom of choice (particularly for infinite dimensional vector spaces such as this one). The fact that $F(R)$ has a basis follows from the axiom of choice, but that doesn't really allow us to construct a basis for it (i.e, the proof of the existence of the basis is "non constructive").
So if you use the axiom of choice, set theory will bestow upon you a nice basis, but if asked to explicitly construct or describe a basis, you will be unable to do so. That is, you need axiom of choice in a strong sense, and without choice, there are models of set theory where there is no such basis.
An intuitive way of seeing why you will never be able to describe such a basis is: given some set of functions, we may always use those functions to obtain something "stranger" (for example, if given a bunch of continuous functions, we may bring up a non continuous). In that sense any infinite family of functions $B$ that we supply will be insufficient since we will be able to construct some function $f$ sufficiently pathalogical so that $f \not \in span(B)$.
Additional links: basis for space of continuous functions, basis for $\mathbb{R}^{\mathbb{R}}$, which is the space of functions $\mathbb{R} \rightarrow \mathbb{R}$

What if we restrict the set of functions to functions with finite domain?

If $A \subset \mathbb{R}$ is finite, and $F(A, \mathbb{R}) = \{f : A \rightarrow \mathbb{R}\}$, then this is in fact a finite dimensional vector space. Suppose $A = \{a_1, ... a_n\}$ for some $n \in \mathbb{N}$. Then the basis is of the functions $f_i : A \rightarrow \mathbb{R}$ where $f_i(a_i) = 1$ and $f_i(a_j) = 0$ if $j \not = i$.
