# Additive closure in the set of all functions $f \in \mathcal{F}(S,\mathbb{F})$ such that $f(s)=0$ for all but a finite number of elements of $S$

Let $S$ be a nonempty set and $\mathbb{F}$ a field. Let $\mathcal{C}(S,\mathbb{F})$ denote the set of all functions $f \in \mathcal{F}(S,\mathbb{F})$ such that $f(s)=0$ for all but a finite number of elements of $S$. Prove that $\mathcal{C}(S,\mathbb{F})$ is a subspace of $\mathcal{F}(S,\mathbb{F})$.

I know how to prove in general that a subset is a subspace (presence of additive identity, additive closure and closed under scalar multiplication). Some hints are also given like - It's closed under addition since the number of nonzero points of $f+g$ is less than the number of union of nonzero points of $f$ and $g$. It's closed under scalar multiplication since the number of nonzero points of $cf$ equals to the number of $f$. And zero function is in the set.

Here, I am not able to understand the additive closure part. Why is the number of nonzero points of $f+g$ is less than the number of union of nonzero points of $f$ and $g$? And how did we conclude additive closure from here?

Thanks.

• It shouldn't say "less." Rather, it should say "no greater." The answer by martini elucidates this very nicely. Commented May 8, 2015 at 10:49

Let $\def\F{\mathbf F}\def\Ft{\F^\times}\Ft := \F-\{0\}$. For $x \in f^{-1}[0] \cap g^{-1}[0]$ we have $$(f+g)(x) = f(x) + g(x) = 0 + 0 = 0$$ hence $f^{-1}[0] \cap g^{-1}[0] \subseteq (f+g)^{-1}[0]$, taking complements, this gives $$(f+g)^{-1}[\Ft] \subseteq f^{-1}[\Ft] \cup g^{-1}[\Ft]$$ by de Morgan's laws (it can also be seen directly: If $(f+g)(x) \ne 0$, we must have either $f(x) \ne 0$ or $g(x) \ne 0$). For the size of the sets, we have $$\def\abs#1{\left|#1\right|}\abs{(f+g)^{-1}[\Ft]}\le \abs{f^{-1}[\Ft] \cup g^{-1}[\Ft]} \le \abs{f^{-1}[\Ft]} + \abs{g^{-1}[\Ft]}$$ If $f, g \in C(S,\F)$, we have $\abs{f^{-1}[\Ft]}, \abs{g^{-1}[\Ft]} < \infty$, and hence $\abs{(f+g)^{-1}[\Ft]} < \infty$. So $f+g \in C(S,\F)$.