generalization for linear functionals I'm learning linear algebra on Kunze Hoffman book and stuck for a  long time by this problem.Please help me solve this. Thanks:

Let $S$ be a set, $F$ a field, and $V(S,F)$ the space of all functions from $S$ into $F$:
  $$ (f + g)(x) = f(x) +g(x)$$
  $$(cf)(x) = cf(x)$$
  Let $W$ be any $n$-dimensional subspace of $V(S,F)$ . Show that there exist points $x_1, x_2, ... x_n$ in $S$ and functions $f_1, f_2, ..., f_n$ in $W$ such that $f_i(x_j) = \delta_{ij}$. Here $$\delta_{ij} = 
\begin{cases}
1 ,  & \text {if $i = j$} \\
0 , & \text {if $i \neq j$} \\
\end{cases}
$$

 A: This can be shown by induction on $n$. For $n=1$ choose a nonzero function $f$ in $W$; since $f$ is not the zero function there is $x_1 \in S$ for which $f(x_1)$ is nonzero, now define $f_1=f/f(x_1)$ and you have what is claimed.
Suppose now the result holds at $n$, and let $W$ be an $n+1$ dimensional subspace. We can choose $n+1$ linearly independent functions in $W$, and using the first $n$ of them gives a set of $n$ linearly independent functions which span a space $V$ of dimension $n$.
By the inductive hypothesis there are $x_1,...,x_n$ in $S$ and functions $f_1,...,f_n$ in $V$ for which $f_i(x_j)=\delta_{i,j}$. Now since $W$ is of dimension $n+1$ the above $f_i$ do not span $W$, so that we may choose a function $g$ in $W$ which is not a linear combination of the $f_i$. Define, for $1 \le i \le n$, the numbers $k_i=g(x_i)$. Next define
$$h=g-k_1f_1-...-k_nf_n.$$
Note that $h$ is not a linear combination of the $f_i$ either, since $g$ isn't.
Then $h(x_i)=0$ for $1 \le i \le n$. Now if $h(x)=0$ for all other $x$ in $S$ then $h$ would be the zero function, against $h$ not being a combination of the $f_i$. So we may choose some $x_{n+1}$ for which $h(x_{n+1})$ is nonzero. Now define
$$f_{n+1}=\frac{1}{h(x_{n+1})}\cdot h.$$
Then we have now $f_i(x_j)=\delta_{i,j}$ for $1 \le i,j \le n+1$ to finish the induction.
EDIT: A slight adjustment is needed, but it still works. That is, for $1 \le i \le n$ we need to have $f_i(x_{n+1})=0$, which the above construction did not guarantee. However we may go back and subtract an appropriate muliple of the present $f_{n+1}$ from each of the first $n$ of the $f_i$, so as to ensure (for the redefined $f_i$) that $f_i(x_{n+1})=0$ as desired.
