TITLE: A question about an $n$-dimensional subspace of $\mathbb{F}^{S}$.
QUESTION [9 upvotes]: I am self-studying Hoffman and Kunze's book Linear Algebra. This is Exercise 3.6.3(Linear Transformation-The Double Dual) from page 111.

Let $S$ be a set, $\mathbb{F}$ a field, and $V(S,\mathbb{F})$ the
  space of all functions from $S$ into $\mathbb{F}:$
$$(f+g)(x)=f(x)+g(x)\hspace{0.5cm}(\alpha f)(x)=\alpha f(x).$$ Let $W$ be any $n$-dimensional subspace of $V(S,\mathbb F)$. Show
  that there exist points $x_{1},\ldots,x_{n}\in S$ and functions
  $f_{1},\ldots, f_{n}\in W$ such that $f_{i}(x_{j})=\delta_{ij}$.

Since $W$ is an $n$-dimensional subspace of $V(S,\mathbb{F})$ we can say find a basis $\mathcal{B}=\{f_{1},\ldots, f_{n}\}$. But I got stuck here. I don't know what to do from now on. I mean, what should I do in order to find those points $x_{1},\ldots,x_{n}\in S$ such that $f_{i}(x_{j})=\delta_{ij}$.

REPLY [3 votes]: For each $x$ in $S$ define $e_x\in W^*$ by $e_x(f):=f(x)$. 
Claim: the subspace $W'$ of $W^*$ generated by the $e_x$ is equal to $W^*$.
Proof of the claim: Let $W'^\perp$ be the orthogonal of $W'$ in $W$. By biduality, it suffices to show $W'^\perp=0$, which is clear.
Let $x_1,\dots,x_n$ be elements of $S$ such that the $e_{x_i}$ form a basis $B$ of $W^*$. The basis dual to $B$ is a basis of $W^{**}$, but, again by biduality, it can be viewed as a basis $f_1,\dots,f_n$ of $W$.