Proving that the set of all real numbers sequences is a vector space. Let $S(R)$ be the set of all sequeces of real numbers $(a_1,a_2,a_3,...)$
show that $S(R)$ is a R-Vectorspace
So do 1 need to prove each of the 8 conditions for a vector space, and how exactly do i do this. 
Or is there an easier way? 
 A: Yes, you need to show that all the axioms of vector space hold. But, first of all, you need to define $+\,\colon S(\Bbb R)\times S(\Bbb R)\to S(\Bbb R)$ and $\cdot\,\colon \Bbb R\times S(\Bbb R)\to S(\Bbb R)$:
$$(a_n)_{n\in\Bbb N} + (b_n)_{n\in\Bbb N} = (a_n+b_n)_{n\in\Bbb N}$$
$$\alpha\cdot (a_n)_{n\in\Bbb N} = (\alpha a_n)_{n\in\Bbb N}$$


*

*Associativity:


$((a_n)_{n\in\Bbb N}+(b_n)_{n\in\Bbb N})+(c_n)_{n\in\Bbb N} = (a_n+b_n)_{n\in\Bbb N}+(c_n)_{n\in\Bbb N} = ((a_n+b_n)+c_n)_{n\in\Bbb N} = (a_n+(b_n+c_n))_{n\in\Bbb N} = (a_n)_{n\in\Bbb N}+(b_n+c_n)_{n\in\Bbb N} = (a_n)_{n\in\Bbb N}+((b_n)_{n\in\Bbb N}+(c_n)_{n\in\Bbb N})$


*Identity:


$(a_n)_{n\in\Bbb N}+(0)_{n\in\Bbb N} = (a_n+0)_{n\in\Bbb N} = (a_n)_{n\in\Bbb N} = (0+a_n)_{n\in\Bbb N} = (0)_{n\in\Bbb N}+(a_n)_{n\in\Bbb N}$


*Additive inverse:


$(a_n)_{n\in\Bbb N}+(-a_n)_{n\in\Bbb N} = (a_n+(-a_n))_{n\in\Bbb N} = (0)_{n\in\Bbb N} = (-a_n+a_n)_{n\in\Bbb N} = (-a_n)_{n\in\Bbb N}+(a_n)_{n\in\Bbb N}$


*Commutativity:


$(a_n)_{n\in\Bbb N}+(b_n)_{n\in\Bbb N} = (a_n+b_n)_{n\in\Bbb N} = (b_n+a_n)_{n\in\Bbb N} = (b_n)_{n\in\Bbb N}+(a_n)_{n\in\Bbb N}$


*Compatibility:


$ \alpha\cdot (\beta\cdot(a_n)_{n\in\Bbb N}) = \alpha\cdot(\beta a_n)_{n\in\Bbb N} = (\alpha(\beta a_n))_{n\in\Bbb N} = ((\alpha\beta)a_n)_{n\in\Bbb N} = (\alpha\beta)\cdot(a_n)_{n\in\Bbb N}$


*Distributivity (of scalar multiplication with respect to vector addition):


$\alpha\cdot((a_n)_{n\in\Bbb N}+(b_n)_{n\in\Bbb N}) = \alpha\cdot(a_n+b_n)_{n\in\Bbb N} = (\alpha(a_n+b_n))_{n\in\Bbb N} = (\alpha a_n+\alpha b_n)_{n\in\Bbb N}= \\= (\alpha a_n)_{n\in\Bbb N} + (\alpha b_n)_{n\in\Bbb N} = \alpha\cdot (a_n)_{n\in\Bbb N}+\alpha\cdot (b_n)_{n\in\Bbb N}$


*Distributivity (of scalar multiplication with respect to field addition):


$(\alpha + \beta)(a_n)_{n\in\Bbb N} = ((\alpha + \beta)a_n)_{n\in\Bbb N} = (\alpha a_n+\beta a_n)_{n\in\Bbb N} = (\alpha a_n)_{n\in\Bbb N} + (\beta a_n)_{n\in\Bbb N}= \\= \alpha\cdot (a_n)_{n\in\Bbb N}+\beta\cdot (a_n)_{n\in\Bbb N}$


*Identity:


$1\cdot (a_n)_{n\in\Bbb N} = (1a_n)_{n\in\Bbb N} = (a_n)_{n\in\Bbb N}$
Note that we used all the appropriate properties of real numbers essentially.
This is special case of more general construction: let $V = \{f\colon X\to\Bbb R\}$ and define \begin{align}(f+g)(x) &= f(x)+g(x),\\(\alpha\cdot f)(x) &= \alpha f(x).\end{align} Then, $(V,+,\cdot)$ is a vector space and we prove it the same way as above. Notice that $S(\Bbb R) = V$ when $X = \Bbb N$. While this is not an easier way to do your exercise, it is certainly easier on the eyes.
