# Proof that $V=\text{Fun}(X, \mathbb{F})$ satisfies the existence of additive inverses.

I need to prove that $V=\text{Fun}(X, \mathbb{F})$ satisfies the existence of additive inverses. Where $V=\text{Fun}(X, \mathbb{F})$ is the set of all functions $X\rightarrow \mathbb{F}$ and $\mathbb{F}$ is some field. I need to explicitly state any field axioms used. Addition and multiplication is defined in the usual way.

Here is my attempt:

We want to show $\forall ~x\in V,~\exists -x~\in V$ such that $x+(-x)=0$.

Suppose $f \in \text{Fun}(X, \mathbb{F})$...

Here is where I don't know how to proceed obviously I want to show $-(f) \in \text{Fun}(X, \mathbb{F})$ but I can't see how to proceed.

Any help?

Edit: I appear to have made some progress we know that $f(x)$ is in the field for all $x\in X$ so by additive inverses in the field we have that there exists $-f(x)$ in the field satisfying the property we want, now how can I show this $-f(x)$ is in $V$?

Extra: Proof $\forall u\in V, 1\cdot u \in V$

Suppose $f \in V$ then $f(x) \in \mathbb{F} ~\forall x \in X$ So by existence of one in the field we have $1 \cdot f(x)=f(x)$ for all $x \in X$ now we define $g(x)=1$ for all $x \in X$ then $g(x) \in \mathbb{F}$ now by definition of the multiplication binary operation in the vector space we have $1 \cdot f(x)=f(x)$ for all $x \in X$ and so we are done.

• The last part is fine. – gamma Oct 9 '15 at 10:56

Take $f(x) \in Fun(X,\mathbb{F})$
For any $x \in X, f(x) \in \mathbb{F}$
As, $\mathbb{F}$ is a field $\implies -f(x) \in \mathbb{F}$
Define $g(x)=-f(x) ~\forall x~ \in X \implies g(x) \in Fun(X,\mathbb{F})$( One can check that $g$ is well defined as $f$ is well defined)$\implies g(x)$ is the additive inverse.
• I don't quite get the last step though (if you read my edit I managed to get down to the 4th line) how can we say that first implication on the 4th line? How do we even know $g$ is defined for all $x \in X$? And that it actually belongs to the function space? – Tim Oct 9 '15 at 10:40
• $g(x)$ is defined because $g(x) = -f(x) \forall x \in X$ since, $\mathbb{F}$ is a field $\implies -f(x) \in \mathbb{F}$ for that $x$ – gamma Oct 9 '15 at 10:44