If $a\in X$, let us define the function $\Delta_{a}\in C(X)$ by

These functions form a linearly independent basis for $C(X)$, i.e.,

Here, the space $\mathop{\mathrm{span}}\{\Delta_{a}\}_{{a\in X}}$ consists of all finite linear combinations of elements in $\{\Delta_{a}\}_{{a\in X}}$. It is clear that any element in $\mathop{\mathrm{span}}\{\Delta_{a}\}_{{a\in X}}$ is a member in $C(X)$. Let us check the other direction. Suppose $f$ is a member in $C(X)$. Then, let $\xi_{1},\ldots, \xi_{N}$ be the distinct points in $X$ where $f$ is non-zero. We then have

and we have established equality in equation 1.

To see that the set $\{\Delta_{a}\}_{{a\in X}}$ is linearly independent, we need to show that its any finite subset is linearly independent. Let $\{\Delta_{{\xi_{1}}},\ldots,\Delta_{{\xi_{N}}}\}$ be such a finite subset, and suppose $\sum_{{i=1}}^{N}\alpha_{i}\Delta_{{\xi_{i}}}=0$ for some $\alpha_{i}\in\mathbb{K}$. Since the points $\xi_{i}$ are pairwise distinct, it follows that $\alpha_{i}=0$ for all $i$. This shows that the set $\{\Delta_{a}\}_{{a\in X}}$ is linearly independent.

Let us define the mapping $\iota:X\to C(X)$, $x\mapsto\Delta_{x}$. This mapping gives a bijection between $X$ and the basis vectors $\{\Delta_{a}\}_{{a\in X}}$. We can thus identify these spaces. Then $X$ becomes a linearly independent basis for $C(X)$.

The mapping $\iota:X\to C(X)$ is universal in the following sense. If $\phi$ is an arbitrary mapping from $X$ to a vector space $V$, then there exists a unique mapping $\bar{\phi}$ such that the below diagram commutes:

Proof. We define $\bar{\phi}$ as the linear mapping that maps the basis elements of $C(X)$ as $\bar{\phi}(\Delta_{x})=\phi(x)$. Then, by definition, $\bar{\phi}$ is linear. For uniqueness, suppose that there are linear mappings $\bar{\phi},\bar{\sigma}:C(X)\to V$ such that $\phi=\bar{\phi}\circ\iota=\bar{\sigma}\circ\iota$. For all $x\in X$, we then have $\bar{\phi}(\Delta_{x})=\bar{\sigma}(\Delta_{x})$. Thus $\bar{\phi}=\bar{\sigma}$ since both mappings are linear and the coincide on the basis elements.$\Box$