Let $X$ be a set and $\mathbb{F}$ a field. For $Y \subset X$, let $\chi_Y: X \to \mathbb{F}$ be the characteristic function of $Y$, that is the function defined by
$$\chi_Y(x) = \begin{cases} 1 & x \in Y \\ 0 & x \not\in Y \end{cases}$$
Prove that $\{\chi_{\{x\}} \big| x \in X\}$ is a basis of $\mathcal{M}_{fin}(X,\mathbb{F})$
I want to give some context also:
For a function $f:X \to \mathbb{F}$, $spt(f) = \{x \in X \big| f(x) \neq 0\}$
$\mathcal{M}_{fin}(X, \mathbb{F}) = \{f:X \to \mathbb{F} \, \big| \hspace{0.2cm} |spt(f)| < \infty \}$
So, as we know $\mathcal{M}_{fin}(X,\mathbb{F})$ is the vector space consisting of all functions $f:X \to \mathbb{F}$ such that, for each $f$, there are finitely many $x \in X$ such that $f(x) \neq 0$.
My rough proof sketch looks like this:
I must use the characteristic functions to indicate when the support is finite or infinite. For example, if $\chi_{\{x\}} = 1$, then $spt(f)$ is finite and likewise if $\chi_{\{x\}} = 0$, $spt(f)$ is not finite. And I must do this somehow for every $x \in X$. Since supports are either finite or infinite, a mapping can be made using $1$ and $0$ respectively.
But a basis is a set, not a mapping. There are a lot of moving pieces here. Any hints or tips on how I should approach this problem?
If $f, g \in M_{fin}(X, \mathbb{F})$, $c \in \mathbb{F}$, we define $f + g \colon X \to \mathbb{F}$ by $(f + g)(x) = f(x) + g(x)$ and we define $cf \colon X \to \mathbb{F}$ by $(cf)(x) = cf(x)$. It is easy to show that $f + g, cf \in M_{fin}(X, \mathbb{F})$. Thus with these operations, $M_{fin}(X, \mathbb{F})$ is a vector space over $\mathbb{F}$.
If $f \colon X \to \mathbb{F}$ has finite support $\{x_1, \dots, x_N\}$, then $f = \sum_{i = 1}^{N}f(x_i)\chi_{\{x_i\}}$. Thus your set spans $M_{fin}(X, \mathbb{F})$. Showing linear independence is straightforward.