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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?

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    $\begingroup$ 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. $\endgroup$ Nov 5, 2015 at 8:42
  • $\begingroup$ Oh yes. I basically just gave a sketch of what I wanted to do. I will fill the details $\endgroup$
    – Non-Being
    Nov 5, 2015 at 8:53

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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.

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