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?

  • $\begingroup$ Fourth axiom is the same as the third axiom? $\endgroup$
    – azif00
    Jan 9, 2020 at 4:25
  • $\begingroup$ third axiom is for vector space, which is the same as the fourth axiom in field axiom. $\endgroup$
    – Beacon
    Jan 9, 2020 at 4:26
  • $\begingroup$ Oh. I'm sorry. I didn't see the word axiom. $\endgroup$
    – azif00
    Jan 9, 2020 at 4:27

1 Answer 1


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.


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