Vector space function question

Question

Q:
Let $X$ be a set and $\Bbb F$ be a field. Consider the vector space $V=Fun(X,\Bbb F)$ of functions from $X$ to $\Bbb F$ with operations of addition and mult.
(i) Describe the zero element of V and show that it holds the condition of axiom vs3 (which is the existence of zero in vector space).
(ii) for each function f in V, describe the additive inverse -f and show that it satisfies the conditions of axiom vs4 ( which is the existence of an inverse vector)

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I am unsure how to do both parts but this is what I have. Please tell me what I'm doing wrong.

For i) I put if V has the operations of addition and mult then let there be f,-f in V and then (f+(-f))(x)=f(x)+(-f(x))=0 which is an element of V. So the axiom vs3 holds since there exists a 0 in V.

For ii) I have no idea...

Answer 1

Your answer for i) is a valid answer for (ii). For (i), it is enough to state that $f(x) = 0$ gives you the zero-vector.

Answer 2

To describe a function, you must define its value $f(x)$ for any $x$. Since $\mathbb F$ is a Field, it has a zero, which is the neutral element of the sum of $\mathbb F$. So, you can define the zero function $$z(x)=0 \ \ \forall x\in X$$

Clearly $z$ is the neutral element of the sum of $V$ as $$(f+z)(x)=f(x)+z(x)=f(x)+0=f(x)\qquad \forall x\in X$$

For $-f$ you need to find a function such that $$(-f+f)=z\qquad i.e. \qquad (-f+f)(x)=z(x)\ \ \forall x\in X$$

Since $\mathbb F$ is field, any element has an additive inverse. Thus, for any $f\in V$ you can define $$g_f(x)=-f(x)$$

Now you have $$(g_f+f)(x)=-f(x)+f(x)=0=z(x)\qquad \forall x\in X$$ Therefore $g_f=-f$ as elements of $V$.

Note that you can do all the above because of the properties of $\mathbb F$. If you try replace $\mathbb F$ with $\mathbb N$...