# prove F(S,R) is a vector space over R by checking the axioms.

I have a question for checking vector space axiom 3 (There exists an element in V denoted by 0 such that x+0=x for each x in V.)

My proof for this axiom is that define function H:S$$\to$$0 and $$x \in V$$. By field axiom A4 (F contains an element 0 such that 0+x=x for every $$x \in$$ F), we get x+0=x $$\rightarrow$$ 0+0=0. Hence proved. Is it ok?

• Fourth axiom is the same as the third axiom? Jan 9, 2020 at 4:25
• third axiom is for vector space, which is the same as the fourth axiom in field axiom. Jan 9, 2020 at 4:26
• Oh. I'm sorry. I didn't see the word axiom. Jan 9, 2020 at 4:27

You just forgot to define the function $$H$$ (which I will denote it by $$\mathbf 0$$). And, in this case $$F=\mathbb R$$, so, the element $$0$$ such that $$0+x = x+0 = x$$ is simply the real number zero. Hence, you would to define $$\mathbf0 \in \mathcal{F}(S,\mathbb R)$$ just as $$\mathbf0(s) = 0$$, that is, $$H$$ is the function that sends every $$s\in S$$ to $$0$$.
Therefore, for every $$f\in \textsf{V} = \mathcal{F}(S,\mathbb R)$$, we have $$(f+\mathbf0)(s) = f(s) + \mathbf0(s) = f(s) + 0 = f(s)$$ so, $$f+\mathbf0 =f$$ as we want to show.