Let $K$ be a field and let $V$ be a finite-dimensional $K$-vector space.
(a)Assume that $K$ is infinite. Show that $V$ is not the union of finitely many proper linear $K$-subspaces.
(b)Assume that $K$ is finite and $V$ is non-zero. Let $S$ be the set of affine hyperplanes of $V$. Let $g:V \longrightarrow \mathbb{R} $ be a function. The Radon transform $Rg:S \longrightarrow \mathbb{R} $ is defined by $(Rg)(\zeta)=\sum_{x \in \zeta}g(x)$ for $\zeta \in S$. Show that $Rg=0$ implies $g=0$.
(c)Let $v_1 , \cdots , v_n , w_1 , \cdots , w_n \in V$. Assume that for every $K$-linear map $f:V \longrightarrow K,(f(v_1) , \cdots , f(v_n))$ and $(f(w_1) , \cdots , f(w_n))$ coincide up to permutation of the indices. Deduce that $(v_1 , \cdots , v_n)$ and $(w_1 , \cdots , w_n)$ coincide up to permutation of the indices. Here we make no assumptions on $K$.
Assume (a) and (b) is proved. For (c), in the infinite case, by subtacting $v_1$ to all vectors, this is equivalent to prove: 'If $w_1 , \cdots , w_n \in V$ are such that for each $f\in V^*$ one of the $f(w_j)=0$, then one of the $w_j=0$.' Then we'll have $V^*=\cup_{i=1}^nAnn(w_i)$, and by (a) there exists a $j$ s.t. $Ann(w_j)=V^*$.
With this hint I tried to prove the finite case but this method is now no longer useful. I suppose this needs to use (b) but I don't know how to use it. How to solve the finite case of (c) ?
For the case when $K$ is finite, we can prove by induction on $n$. If $n=1$ this is trivial. If $v_i= w_j$ for some $i, j$, then we can apply induction hypothesis to conclude the proof. So we may assume that $v_i\neq w_j$ for any $i, j$ and we are going to find a contradiction. Define $g:V\to \mathbb{R}$ by $g(v_i)=1$, $g(w_j)=-1$, and $g(x)=0$ for other $x\in V$. Then for any affine hyperplane $\xi$, $$Rg(\xi)=\#\{v_i\mid v_i\in \xi\}-\#\{w_j\mid w_j\in \xi\}.$$ Recall that $\xi$ is defined by $\{x\in V\mid f(x)=a\}$ for some $f\in V^\vee$ and $a\in K$. Then the assumption implies that $Rg(\xi)=0$ for each $\xi.$ This implies that $g=0$, a contradiction.