Which of the following statements are true/false?Justify your answer. (i) R2 has infinitely many non-zero,...

(i) We can construct such a set of subspaces:
1) ${\displaystyle \mathrm{\forall }r\in \mathbb{R}}$, let: ${\displaystyle V_r=\{(x,rx)\in {\mathbb{R}}^2\mid x\in \mathbb{R}\}}$.
[Geometrically, ${\displaystyle V_r}$ is the line through origin of ${\displaystyle {\mathbb{R}}^2}$, os slope r.]
2) We will check that these subspaces justify assertion (i).
3) Clearly: ${\displaystyle V_r\subseteq {\mathbb{R}}^2.}$
4) Check that: ${\displaystyle V_r}$ is a proper subspace of ${\displaystyle {\mathbb{R}}^2.}$.
Let: ${\displaystyle u,v\in V_r,\alpha ,\beta \in \mathbb{R}}$. Verify that: ${\displaystyle \alpha u+\beta v\in V_r}$
${\displaystyle u,v\in V_r,\Rightarrow u=(x_1,r{x}_1),v=(x_2,r{x}_2);}$ for some ${\displaystyle x_1,x_2\in \mathbb{R}}$
${\displaystyle \therefore \alpha u+\beta v=\alpha (x_1,r{x}_1)+\beta (x_2,r{x}_2)}$
${\displaystyle =\alpha (x_1,r{x}_1)+\beta (x_2,r{x}_2)}$
${\displaystyle =(\alpha x_1,\alpha r{x}_1)+(\beta x_2,\beta r{x}_2)}$
${\displaystyle =(\alpha x_1+\beta x_2,\alpha r{x}_2+\beta r{x}_2)}$
${\displaystyle =(\alpha x_1+\beta x_2,r(\alpha x_1+\beta x_2)}$
${\displaystyle =(x_3,r{x}_3)\in V_r}$; with ${\displaystyle x_3=\alpha x_1+\beta x_2}$
So: ${\displaystyle u,v\in V_r,\alpha ,\beta \in \mathbb{R}\Rightarrow \alpha u+\beta v\in V_r}$.
Thus: ${\displaystyle V_r}$ is a subspace of ${\displaystyle {\mathbb{R}}^2}$
To see that ${\displaystyle V_r}$ is non-zero, note that:
${\displaystyle (1,r)\in V_r}$ and ${\displaystyle (1,r)\ne (0,0)}$.
To see that ${\displaystyle V_r}$ is proper, note that ${\displaystyle (1,r+1)\notin V_r}$:
${\displaystyle (1,r+1)\in V_r\Rightarrow }$ (by construction of ${\displaystyle V_r}$) ${\displaystyle r\cdot 1=r+1}$
${\displaystyle \Rightarrow r=r+1}$, plainly impossible.
Thus: ${\displaystyle V_r}$ is a non-zero, proper subspace of