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 $f_i(x_j) = \delta_{ij}$.
This question has been asked before and the proof of the problem involved induction. However I think this can be solved without using induction.
Attempt :
Define $T_x : V(S;F)\to F$ as $T_x(f)=f(x)$ where $x\in S.$ It is trivial to see that $T_x$ is a linear transformation. Now consider the restriction of $T_x$ on $W.$
Let $\{f_1,\dots,f_n\}$ be a basis for $W.$ Using an elementary theorem we know there exists linear transformations $L_i : W\to F$ such that $L_i(f_j)=\delta_{ij}$ for each $i, 1\le i\le n.$
If I am able to show $L_i=T_{x_i}$ for some $x_i\in S$ then I am done.
I need help completing the proof this way OR please point a flaw showing why the proof cannot be going this way.
It is not always possible to show $L_i=T_{x_i}$ for some $x_i$.
Consider $S=\mathbb R = F$. Choose $W=\{f : f(1)=f(0), f(x)=0 \forall x\in \mathbb R\setminus\{0,1\} \}$ which is a $1$-dimensional subspace. A basis of $W$ is given by $f_1(x)=\delta_{x,1} + \delta_{x,0}$.
$L_i$ can be defined as $\frac12 (T_1+T_0)$. Thus it cannot be $T_{x_i}$ for some $i\in\mathbb R$.