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Let $V=\{f : S \to \mathbb{R} \}$ be the set of all the functions $f : S \to \mathbb{R}$. Let $S=\{ 1,2,...,n$}, now my train of thought was that the basis of V must be a linearly independent set that Span V so somehow show that there is a set of functions on S that can compose as a linear combination and map to any number in $\mathbb{R}$. Any idea how to get a general answer for basis for dimension of V?

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Choose $f_i\in V\;(i=1,\ldots, n)$ such that $f_i(j)=\delta_{ij}$, that is:

$$f_1\equiv\begin{cases}f_1(1)=1\\ f_1(2)=0\\ \ldots\\ f_1(n)=0\end{cases}\quad f_2\equiv\begin{cases}f_2(1)=0\\ f_2(2)=1\\ \ldots\\ f_2(n)=0\end{cases}\quad\ldots\quad f_n\equiv\begin{cases}f_n(1)=0\\ f_n(2)=0\\ \ldots\\ f_n(n)=1.\end{cases}$$ We can easily prove that $B=\{f_1,\ldots,f_n\}\subset V$ is a basis of $V.$

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Hint:

For $S=\{1,2,\dots,n\}$, V is simply $\mathbf R^n$. Do you know how to characterise families of vectors which are a basis of $\mathbf R^n$?

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    Are you asking about qualities that make a family of vectors a basis? In that case yes, but I'm understanding how to get a general/concrete answer without it sounding like I am making this basis up. – 2017-01-20
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    Isn't the determinant criterion a concrete answer? – 2017-01-20