Showing A Set is Vector Space 
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*Check that the set $\mathbb{R}^\omega$ of all infinite-tuples of real numbers


$$x=(x_1,x_2,...)$$
is a vector space under component-wise addition and scalar multiplication.
My Attempt. Let $x,y\in\mathbb{R}^\omega$. We need to show:
1) $x+y=(x_1,x_2,...)+(y_1,y_2,...)=(x_1+y_1,x_2+y_2,...)\in\mathbb{R}^\omega$
2)$cx=c(x_1,x_2,...)=(cx_1,cx_2,...)\in\mathbb{R}^\omega$.
How do we know $1)$ and $2)$ is true? Can you explain, can you show $1),2)?$
 A: Let $x,y\in\mathbb{R}^\omega$. Since we know
1) $x+y=(x_1,x_2,...)+(y_1,y_2,...)=(x_1+y_1,x_2+y_2,...)\in\mathbb{R}^\omega$
2)$cx=c(x_1,x_2,...)=(cx_1,cx_2,...)\in\mathbb{R}^\omega$,
then $\mathbb{R}^\omega$ is a vector space.
A: At first we have to explain why $(x_1 + y_1, x_2 + y_2, \dots)$ and $(cx_1, cx_2, \dots)$ are elements of $\mathbb R^\omega$. Note that this is true because the sum of two real numbers is again a real number and the product of two real numbers is again a real number.
So what we now have to do is to verify the eight vector space axioms (see e. g. the wiki page). I will give you here the proof of two of them and let you the proof of the remaining ones as an exercise. 
Commutativity of addition: Let $x = (x_1, x_2, \dots), y = (y_1, y_2, \dots)$ as above. Then
$$x + y = (x_1 + y_1, x_2 + y_2, \dots)\ {\color{red} = }\ (y_1 + x_1, y_2 + x_2, \dots) = y + x.$$
Here, the crucial point is to explain why the red marked equality holds. This is the case because of the commutativity of the addition in $\mathbb R$: We have $a + b = b + a $ for all $a, b \in \mathbb R$. (Remark: You will need the properties of $\mathbb R$ for showing that $\mathbb R^\omega$ is a vector space, everything follows from this.)
Existence of a zero vector: We have to show that there is a $0_{\mathbb R^\omega}$ such that $x + 0_{\mathbb R^\omega} = x = 0_{\mathbb R^\omega} + x$ for all $x \in \mathbb R^\omega$. Note that I write $0_{\mathbb R^\omega} $ more because of pedagogical reasons, this distinguishes the zero vector from $0 \in \mathbb R$. So take
$$0_{\mathbb R^\omega} = (0, 0, \dots) \in \mathbb R^\omega.$$
Then 
\begin{align}
x + 0_{\mathbb R^\omega} &= (x_1 + 0, x_2 + 0, \dots ) = (x_1, x_2, \dots) = x , \\
0_{\mathbb R^\omega} + x &= (0 + x_1, 0 + x_2, \dots ) = (x_1, x_2, \dots) = x.
\end{align}
Here we use again a fact of the real numbers: The zero element is there $0$, so we have $x_1 + 0 = 0 + x_1 = x_1$ for every real $x_1$. We have established the existence of a zero vector, you now have to prove the remaining six axioms.
