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My books introduce the collection $\mathcal{F}(X,\mathbb{R})$ of all real-value functions defined on a non-empty set $X$ together with the operations satisfies the axiom of a real linear vector space. Firstly I don't really understand the significant of investigating the functions inside $\mathcal{F}$. Are those functions more special than the others? Also, at the other chapter about metric and normed space in my book, it said:

We shall frequently want to study classes of functions with certain other properties such as boundedness, continuity, etc. We shall use the following result from linear algebra: Let $\mathcal{A}(X,{R})$ be a non-empty subcollection (subspace) of $\mathcal{F}(X,\mathbb{R})$ satisfying the following two properties:

(1) If $f,g\in\mathcal{A}(X,\mathbb{R})$, then the sum $f+g\in\mathcal{A}(X,\mathbb{R})$

(2) If $f\in\mathcal{A}(X,\mathbb{R})$ and $k\in\mathbb{R}$, then the scalar multiple $kf\in\mathcal{A}(X,\mathbb{R})$

Then $\mathcal{A}(X,\mathbb{R})$ is, itself, a linear vector space.

I want to ask that why it is correct. For example it does not define the zero vector(function). I try to define it as $\mathbf{0}=0*f$ while it satisfies the second property, so the zero vector(function) is inside $\mathcal{A}(X,\mathbb{R})$. However in that way, it give rise to my next question: Since actually $\mathcal{F}(X,{R})$ is a non-empty subcollection (subspace) of $\mathcal{F}(X,\mathbb{R})$ itself. Then if it satisfies that two properties, $\mathcal{F}(X,\mathbb{R})$ is a linear vector space. In that way why it is needed to satisfy so many axiom (8 total) in order to check whether it is a vector space?

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up vote 4 down vote accepted

Recall that given a field $K$, a $K$-vector space $V$ is a set with an operation $+$ and an action $(\;\cdot\;):K\times V\to V$ such that $(V,+)$ is an abelian group, and the action satisfies $$\begin{align} a\cdot (v+w)&=a\cdot v+a\cdot w\\(a+_Kb)\cdot w&=a\cdot w+b\cdot w\\1_K\cdot v&=v\\ (a\cdot_K b)\cdot v&=a\cdot(b\cdot v)\end{align}$$

where I used the subcript $K$ to denote the operation or element is from the field $K$.

Let $V$ be a $K$-vector space and $S$ a subset. Since it is a subset in inherits all the properties (the axioms) which you have already checked. The three additional checkpoints

$$\begin{align}\tag 1 &S\neq\varnothing\\ \tag 2 x,y\in S&\implies x+y\in S\\ \tag 3\lambda\in K\;,\;x\in &S\implies \lambda x\in S\end{align}$$

ensure that $S$ is a vector space in its own right. Condition $(2)$ ensures $+$ is an operation $+:S\to S$ that is, that $S$ is "closed under $+$", while $(3)$ ensures $\left(\;\cdot\;\right)$ is an action $\left(\;\cdot\;\right) :K\times S\to S$. One can alternatively use the equivalent

$$\begin{align}\tag 1 &0\in S\\ \tag 2 x,y\in S&\implies x+y\in S\\ \tag 3\lambda\in K\;,\;x\in &S\implies \lambda x\in S\end{align}$$

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You can't apply the result you have quoted until you have proved that $F(X,\mathbb{R})$ is a vector space first. In order to do that, you need to check all the axioms.

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