Basis of space of arithmetic sequences $$U= \lbrace \text{sequences}~~~ x_n~|~x_n ~~\text{is an arithmetic progression}\rbrace$$
It is known that U is a subspace of the vector space of all sequences.
Find a basis and the dimension of U.

$$x_n=x_0+dn$$
Every arithmetic sequence is determined by two things: $x_0$ and $d$. So i think the dimension should be 2.
When writing the basis, should I write it a vector form, in this case as sequences. Basis is $x_0$ and $dn$, how to write in a sequence form ?
My idea:
$x_0$ is an constant sequence and $dn$ is also sequence. Can I end the problem like this or do i have to write it more formally ?
 A: You are correct that the basis of $U$ depends on $x_0$ and $d$.
Let us define what it means to sum two (infinite) sequences, and what it means to multiply a sequence with a constant $c \in \mathbb{R}$. 
If $\mathbf{x}$ and $\mathbf{y}$ are sequences, we have that
$$
\begin{eqnarray}
\mathbf{x} + \mathbf{y} &=& (x_0 + y_0, x_1 + y_1, x_2+ y_2, ...), \\
c\mathbf{x} &=& (cx_0, cx_1, cx_2, ...).
\end{eqnarray}
$$
Recall that a basis $\mathcal{B}$ for the vector space $U$, is a set of elements such that


*

*Any element in $U$ can be written as a linear combination of elements in $\mathcal{B}$.

*The elements in $\mathcal{B}$ are linearly independent.


Consider now the sequences
$$
\mathbf{b_1} = (1,1,1,...), \\
\mathbf{b_2} = (0,1,2,...).
$$
As you said your self, any arithmetic progression is unique from the two values $x_0$ and $d$. Given these two values, we can find its sequence by looking at the linear combination
$$
\begin{eqnarray}
x_0\mathbf{b_1} + d\mathbf{b_2} &=& (x_0, x_0, x_0, ...) + (0,d,2d,...) \\
&=& (x_0 +0, x_0 + d, x_0+2d, ...).
\end{eqnarray}
$$
Since the sequence of any arithmetic progression can be written as a linear combination of elements in $\mathcal{B} = \{ \mathbf{b_1}, \mathbf{b_2} \}$, and it is clear that the two elements of $\mathcal{B}$ can not be written as a multiple of the other, we can conclude that $\mathcal{B}$ is a basis for $U$. Since $\mathcal{B}$ has $2$ elements, it follows that $\text{dim }U = 2$!
Note: It might be easier to think of arithmetic progressions as vectors in $\mathbb{R}^2$, where a vector $(a,b) \in \mathbb{R}^2$ represents the arithmetic progression $x_0 = a, d = b$. Any vector in $\mathbb{R}^2$ can be written as a linear combination of $\mathbf{e_1} = (1,0)$ and $\mathbf{e_2} = (0,1)$. To find a basis in terms of sequences, we simply "translate" the vectors $\mathbf{e_1}$ and $\mathbf{e_2}$ into their equivalent sequence form. 
