Basis for Finite Submodule of Function Space

Theorem
Let $\left({R, +, \circ}\right)$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $A$ be a set.

For each $a \in A$, let $f_a: A \to R$ be defined as:
 * $\forall x \in A: f_a \left({x}\right) = \begin{cases}

1 & : x = a \\ 0 & : x \ne a \end{cases}$

Then $B = \left\{{f_a: a \in A}\right\}$ is a basis of the Finite Submodule of Function Space $R^{\left({A}\right)}$.

Proof
Let $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ be a sequence of distinct terms of $A$.

Let $\left \langle {\lambda_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ a sequence of scalars.

Then: $\displaystyle \sum_{k \mathop = 1}^n \lambda_k f_{a_k}$ is the mapping whose value at $a_k$ is $\lambda_k$ and whose value at any $x$ not in $\left\{{a_1, a_2, \ldots, a_n}\right\}$ is zero.

Hence $B$ is a generator of $R^{\left({A}\right)}$ which is linearly independent.

Thus, by definition, $B$ is a basis of $R^{\left({A}\right)}$.

If $A = \left[{1 \,.\,.\, n}\right]$, then $B$ is the standard basis of $R^n$.