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(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.

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    $\begingroup$ I am not sure what tools you have at your disposal. Every subspace has a basis. The question is really about the dimension of the subspace. $\endgroup$
    – copper.hat
    Commented Jun 19, 2022 at 18:48
  • $\begingroup$ This answer gives a good explanation $\endgroup$
    – Dragonoverlord3000
    Commented Jun 19, 2022 at 18:50
  • $\begingroup$ @copper.hat why we wanna use basis and dimension concept? $\endgroup$
    – user264745
    Commented Jun 19, 2022 at 18:51
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    $\begingroup$ Suppose $S \subset \mathbb{R}^2$ is a subspace. If $L=\{0\}$ you are finished. Otherwise pick $x_0 \neq 0$ from $L$. If $L=\{ t x_0\}_{t \in \mathbb{R}}$ then you are finished. Otherwise pick $x_2 \notin \{ t x_0\}_{t \in \mathbb{R}}$. Then $x_0,x_1$ is a basis of $\mathbb{R}^2$ and so $L=\mathbb{R}^2$. Basically just saying the dimension can be $0,1,2$. $\endgroup$
    – copper.hat
    Commented Jun 19, 2022 at 18:55
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    $\begingroup$ @Dragonoverlord3000 Thank you for sharing that link. I will read that answer. $\endgroup$
    – user264745
    Commented Jun 19, 2022 at 19:15

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