Let $S$ be a set and $V$ a subspace of the $K$-vector space $K^S$. Suppose there exist a finite family $(s_i)_{1\leq i\leq p}$ of elements of $S$ and a finite family $(f_i)_{1\leq i\leq p}$ of elements of $V$ such that $f_i(s_j)=\delta_{ij}$ for all $i,j\in[1,p]$. I want to show that this implies $p\leq\text{dim}(V)$.
Let $(e_s)_{s\in S}$ be the canonical basis of $K^{(S)}$ and $(e_s^*)_{s\in S}$ the associated coordinate forms. There exists an isomorphism $\varphi:(K^{(S)})^*\rightarrow K^S$ sending $\varphi(e^*_s)$ to $(\langle e_{s'},e^*_s\rangle)_{s'\in S}$. It suffices to show that there exists a family $(x_i)_{1\leq i\leq p}$ of $K^{(S)}$ such that $$\langle x_i,\varphi^{-1}(f_j)\rangle=\delta_{ij}.$$ The book proposes to take $x_i:=e_{s_i}$. I don't see how $\langle e_{s_i},\varphi^{-1}(f_j)\rangle=\delta_{ij}$. (The book asserts that $f(s)=\langle e_s,\varphi^{-1}(f)\rangle$ for all $s\in S$ and $f\in K^S$.) Why does it hold?
Edit:
Silly mistake. Ofc $f=\varphi(\varphi^{-1}(f))=(\varphi^{-1}(f)(e_s))_{s\in S}$. Solved.
There is no need to go to the dual, we can show that the family $(f_i)_{1\le i\le p}$ is linearly independent directly. Assume we have a linear combination $$ 0 = \sum_{i=1}^p \lambda_i\, f_i. $$ Let $k\in\{1,\dots,p\}$ and evaluate both sides at $s_k$ to obtain $$ 0 = \left(\sum_{i=1}^p \lambda_i\, f_i\right)(s_k) = \sum_{i=1}^p \lambda_i\, f_i(s_k) = \sum_{i=1}^p \lambda_i\, \delta_{ik} = \lambda_k. $$ Hence $\lambda_k=0$ for $k=1,\dots,p$ and the $f_i$ are linearly independent.
Having found $p$ linearly independent vectors in $V$ we conclude that $p \le \dim(V)$.