The only subspaces of R over R are zero space and R itself.

Let W be a subspace of R. If W is finite then W has to be Zero space since otherwise it won't be closed under scalar multiplication. If W is infinite, we want W=R. Claim: W' is empty Pf: if W' is non-empty then there exists some x in W'. Therefore, we can choose a scalar C for a given y in W such that C.y=x. Which means x is in W. Therefore W' is empty hence W=R Is this proof correct?

• if $R$ is $\mathbb{R}$ and $W'$ means the complementary of $W$, yes your proof is correct. However I would suggest to say the following if $W=\{0\}$ then we are good, if it is not, it must contain some $y\in W$ with $y\neq 0$. Now any $x\in \mathbb{R}$ is $\frac{x}{y}y\in W$ since $W$ is closed by scalar multiplication so $W$ is the set of real numbers... The alternative is not between finite/infinite, but either it contains only the $0$ element either it contains one that is not zero. Nov 5, 2015 at 8:42
• Oh yes. I basically just gave a sketch of what I wanted to do. I will fill the details Nov 5, 2015 at 8:53

If $$W \subset \mathbb{R}$$ is finite ($$W \neq \emptyset$$). Then exist elements $$x,y \in W$$ (and $$x,y \in \mathbb{R}$$) such that $$cx + y \in W$$ But, if $$x \neq y \Longrightarrow x \leq cx + y$$ and $$y\leq cx+y$$

The idea is:

$$W=\{1,2,3\} \subset \mathbb{R}^+$$, then $$2c+3$$ is bigger than any element of $$W$$. So $$2c+3$$ is a new element in $$W$$

Following this process we have that $$W = \mathbb{R}$$ because $$W$$ have some element in $$\mathbb{R}$$ but $$\mathbb{R}$$ is closed. So the new element ($$cx+y$$) has to be in $$W$$ (because is subspace).

If $$W$$ have just one element, we would like that $$cx+y \in W$$, but the only element in $$\mathbb{R}$$ who makes this is 0. So $$x=y=0 \Longrightarrow W=\{0\}$$

Sorry for my english, I hope can help with this.