# Math Help - Prove that F* is a vector space over field F by verifying each axiom

1. ## Prove that F* is a vector space over field F by verifying each axiom

Let $F$ be a field and $X$ be a non-empty set. Let $F^*$ be the set of all functions $f:X\rightarrow F$ together with addition, defined by
$(f_1+f_2)(x)=f_1(x)+f_2(x)$, $\forall f_1,f_2\in F$ and all $x\in X$,
and scalar multiplication defined by
$(cf)(x)=cf(x)$, $\forall f\in F^*$ and $c\in F$.
Prove that $F^*$ is a vector space over $F$ by verifying each axiom in the definition.
You will need to refer to the axioms of the field $F$. It is important to make a distinction between a function $f\in F^*$ and one of its values $f(x)\in F$.

The main issue I'm having with this one is wording, and the fact that I just got back into this after the summer, so I'm virtually brain-dead until I can get settled in again. The fact that I only just started learning anything about algebra-related fields doesn't help. Any help on this would be appreciated.

2. So you have to prove all of these. What progress have you made on any of them?

3. ## Question

I just don't understand in what way when showing the axioms hold for this F* when know what is mean by make a difference between f and f(x) i know the difference but i don't see when I would have to use f(x) at all wouldn't i just use f? perhaps show how you would do some of the axioms?

4. ...what is mean by make a difference between $f$ and $f(x)$...
$f$ is a way, a handle, a label, for the function itself. Functions have three aspects to them: the domain, the range, and the rule of association. The domain of $f$ is the set of all allowable inputs to the function. The range is the set of all outputs that can be produced from the domain. The rule of association is the rule that tells us how we can get from elements of the domain to elements of the range. So, for example, let's let $f(x)=x^{2}.$ Since we haven't arbitrarily restricted the domain, and since our rule of association allows us to square any real number, the domain is all real numbers. That's also called the natural domain. The range is the semi-infinite interval $[0,\infty).$ The rule of association is the equation $f(x)=x^{2}.$ Does that make sense?

Two functions are equal if and only if their domains are the same and their rules of association are the same. (Since the range is uniquely determined by the domain and the rule of association, it is unnecessary to require that the ranges be the same). So, suppose I have two functions:

$\displaystyle{f(x)=x^{2}}$

$\displaystyle{g(x)=\begin{cases}x^{2}\quad &x\not=0\\ 1\quad &x=0\end{cases}.}$

These two functions have the same domain (all real numbers), but their rule of association differs at the origin. Hence they are not the same. However, if I have the following two functions:

$\displaystyle{f(x)=x^{2}}$

$\displaystyle{g(x)=\begin{cases}x^{2}\quad &x\not=0\\ 0\quad &x=0\end{cases},}$

then they are the same function. Let's take a look at these two functions:

$\displaystyle{f(x)=x^{2}}$

$\displaystyle{g(x)=x^{2},\quad x\not=0.}$

In the second function, I have artificially restricted the domain not to include the origin. Since the domains are not the same, the two functions are not the same.

In all these cases, once I've defined a function, I can refer to the whole shebang (the domain, the range, and the rule of association) by its label, $f$ or $g$. However, if I want to refer to the actual value that the function takes on for an element of its domain (so I'm wanting to refer to an arbitrary element of the function's range), then I'm going to write $f(x).$

I should point out that although what I've just written is technically correct, some mathematicians are a bit sloppy, and write one when they mean the other. You have to know your author and the conventions of the book you're reading.

So hopefully, this explanation will clear up the difference between $f$ and $f(x)$.

So, let's prove that there is an additive identity. Define the zero vector to be

$z(x)=0\;\forall\,x\in X,$ where $0\in F.$

We note that because $F$ is a field, it does have a unique additive identity $0$, to which we've just referred. (The previous sentence I just wrote is an example of "referring to the field axioms".)

Let $f\in F^{*}$ be an arbitrary function. Then

$f(x)+z(x)=f(x)+0=f(x)=0+f(x)=z(x)+f(x).$ So this candidate for the zero function does the job, and we've proven that axiom.

Does this all make sense?