Exercise 1.3.14

Question

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

Jephian C.-H. Lin · Accepted Answer

\begin{proof}
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.
\end{proof}

Exercise 1.3.14

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Exercise 1.3.14

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