Let $V$ = {$f$ : $f$ is function $\mathbb{R} \rightarrow \mathbb{R}$} $-$ vector space over $\mathbb{R}$. Is it possible to explicitly present a subset $B\subset V$, such that
P.S. $B$ isn't necessary a basis.
Not a complete answer. I believe we can more or less reduce to the same question for $V_2 = \mathbb{F}_2^{\mathbb{R}}$ as a vector space over $\mathbb{F}_2$.
The first step of the reduction, which is heuristic, is that I don't believe we gain anything by allowing our functions to take infinitely many values; the large dimension of this vector space is "because of" $\mathbb{R}$ having many subsets, so that's where our size needs to "come from." That is, I believe that WLOG we can assume our functions are indicator functions $i_A$ of subsets $A \subseteq \mathbb{R}$.
The second step of the reduction is that if a family of indicator functions $i_A(x)$ is linearly independent over $\mathbb{F}_2$ then it is linearly independent over $\mathbb{R}$. First observe that if there is a nontrivial linear dependence $\sum c_A i_A(x) = 0$ between finitely many indicator functions, then the functions $i_A$ collectively only take finitely many distinct values (e.g. at most the size of the Boolean algebra generated by the sets $A$); so we can restrict our attention to thinking of the functions $i_A$ as living on this Boolean algebra WLOG, which is finite. Then we can argue as in this answer that the question of linear dependence or independence is captured by the nonvanishing of some minors, and since those minors are integers in this case, if they don't vanish over $\mathbb{F}_2$ then they don't vanish over $\mathbb{R}$. Moreover I also believe we don't gain anything by working over $\mathbb{R}$.
So, I believe this question reduces to the question of exhibiting $|2^{\mathbb{R}}|$ subsets $A \subseteq \mathbb{R}$ whose indicator functions $i_A$ are linearly independent over $\mathbb{F}_2$.
