(a) Prove that the only subspaces of $ \Bbb{R}$ are $ \Bbb{R}$ and the zero subspace.

(b) Prove that a subspace of $ \Bbb{R}^2$ is $ \Bbb{R}^2$ , or the zero subspace, or consist of all scalar multiples of some fixed vector in $ \Bbb{R}^2$ .

My attempt: (a) Assume towards contradiction, $ \exists A$ such that $ A\neq \{0\},\Bbb{R}$ and $ A\leq \Bbb{R}$ . Since $ A\neq \{0\}$ , $ \exists a\in A$ such that $ a\neq 0$ . By axiom of multiplication in fields, $ \exists \frac{1}{a}\in \Bbb{R}$ such that $ a\cdot \frac{1}{a}=1$ . Since $ A\neq \Bbb{R}$ , $ \exists r\in \Bbb{R}$ such that $ r\notin A$ . Define $ p=\frac{1}{a}\cdot r\in \Bbb{R}$ . Since $ A\leq \Bbb{R}$ , $ p\cdot a=(\frac{1}{a}\cdot r)\cdot a=r\in A$ . Thus we reach contradiction. Is this proof correct?

(b) Claim: $ L_x=\{c\cdot x|c\in \Bbb{R}\}\leq \Bbb{R}^2$ , $ \forall x\in \Bbb{R}^2$ . Proof: let $ x\in \Bbb{R}^2$ . (1) $ 0_v=0\cdot x\in L_x$ . (2) $ c_1\cdot x+c_2\cdot x=(c_1+c_2)\cdot x\in L_x$ , since $ c_1+c_2\in \Bbb{R}$ . (3) $ c’\cdot(c\cdot x)=(c’\cdot c)\cdot x\in L_x$ , since $ c’\cdot c\in \Bbb{R}$ . Hence $ L_x\leq \Bbb{R}^2$ , $ \forall x\in \Bbb{R}^2$ . How to prove these are the only subspaces of $ \Bbb{R}^2$ ? I tryed using contradiction but made no progress. I haven’t yet studied matrix theory.