Every vector space can be viewed as the space of functions GIVEN PROBLEM :

Let $V $ be a vector space over a field $F $,and suppose that $S $ is a basis for $V $.
Let $C(S,F)$ denote the vector space of all functions $f\in f(S,F) $ such that $f(s)=0$ for all but a finite number of vectors in $S $.
Let $\Phi:C (S,F) \to V $ be the function defined by $\phi (f)=\sum f(s)s $ such that $s\in S $ and $f (s) \ne 0$.
Prove that $\Phi $ is an isomorphism. Thus every nonzero vector space can be viewed as a space of function.

I have proved it but I am facing some conceptual problem,As there is a isomorphism $\Phi$ there the image of basis of $C(S,F) $ is also a basis  of $V $.
Here I guess $f $ such that $f(s)\ne 0$ is the basis element corresponds to $s $ in $C(S,F) $ under the inverse image of $\Phi $. Am I correct?
 A: The natural basis of your $C(S,F)$ are the functions whose value is $1$ at one $s\in S$ and $0$ at the rest of them:
$$ f_i(s) = \begin{cases} 1 & \text{if }s=s_i \\ 0 & \text{otherwise} \end{cases} $$
These are the functions such that $\Phi(f_i) = s_i$.
This may be what you mean by "$f$ such that $f(s)\ne 0$" -- but that is not a complete description of the function you're looking for.
A: Henning Makholm gave me a hint,For simplicity I assume $V $ is finite dimensional and $F=R $. I take ${e_1,e_2,e_3} $ as the basis $S $ of $V $.Take three different functions  $f_i $ as $f_i (s_j)=1$ if$ i=j$ and$ f_i (s_j)=0$ if $i\ne j.i,j=1,2,3$.
Now any functions gives value (may or may not distinct ) at 3 points as they maps from three vectors $e_1,e_2,e_3$ to $R $.Let $f $ gives $x,y,z $ correspondence to $e_1,e_2,e_3$ respectively. Then $f $ can be expressed as linear combination of $f_1,f_2,f_3$ as $f=xf_1+yf_2+zf_3$ ie. any vector of $C(S,F) $ can be expressed as the combination of $f_1,f_2,f_3$.ie. $C (S,F) $ has spanning set ${f_1,f_2,f_3} $ .To check linear independence let $r_1,r_2,r_3$ scalers such that $r_1f_1+r_2f_2+r_3f_3=O$, $O $ is the zero functional from $S $ to $R $.Multiplying both sides by $e_1$ we get $r_1=0$,Similarly using $e_2,e_3$ we get $r_2,r_3$ are $0$ scalers. Hence ${f_1,f_2,f_3}$ as defined above forms a basis of $C (S,F) $.
Using this idea we can construct a general basis of $C (S,F) $ $S ={f_1,f_2,f_3} $ defined as $f_i (e_j)=r ,r $ is any scaler   if $i=j $....and $f_i (e_j)=0$, if  $i\ne j$
