Let $V$ be a vector space over a field $\mathbb{K}$ (either $\mathbb{C}$ or $\mathbb{R}$) that has an infinite linearly independent subset. Prove that if $B$ and $B'$ are two bases for $V,$ then $B$ and $B'$ have the same cardinality. Denote the unique cardinality of a basis for $V$ as $\dim_{\mathbb{K}}V,$ the dimension of $V$ over the field $\mathbb{K}.$ Determine without appealing to the continuum hypothesis, $\dim_{\mathbb{Q}}\mathbb{R}.$
For any set $S,$ let $\mathcal{F}(S)$ denote the set of finite subsets of $S.$ Let $S$ be an infinite linearly independent subset of $V.$ Then for any basis $B$ for $V,$ since $B$ is maximally linearly independent, $|S|\leq |B|. $ Since $S$ is infinite, $\aleph_0\leq |S|.$ Let $B$ and $B'$ be two bases for $V.$ I know that $|\mathcal{F}(B)| = |B|$ and $|\mathcal{F}(B')| = |B'|,$ so it suffices to show that $|\mathcal{F}(B)| = |\mathcal{F}(B')|,$ but I'm not sure how to show this.
In the vector space $\mathbb{R}$ over $\mathbb{Q},$ I think, but I'm not sure how to show, that $\{\sqrt{2},\sqrt{2}^\sqrt{2}, \sqrt{2}^{\sqrt{2}^\sqrt{2}},\cdots \}$ is an infinite linearly independent subset. I'm not sure how to determine the cardinality of this vector space. Is it $\aleph_1 = 2^{\aleph_0},$ and if so, is there some proof for this?
Best Answer
$\aleph_1$ is irrelevant: the statement that $\aleph_1=2^{\aleph_0}$ is the continuum hypothesis, which is independent of $\mathsf{ZFC}$ and which you are not supposed to use.
What you need to show is that $\dim_{\Bbb Q}\Bbb R=2^{\aleph_0}$. Suppose that $B$ is a base for $\Bbb R$ over $\Bbb Q$. $B$ is infinite, so it has $|B|$ finite subsets, and $\Bbb Q$ has $\aleph_0$ finite subsets, so there are at most $|B|\cdot\aleph_0$ linear combinations of elements of $B$ with rational coefficients. $B$ spans $\Bbb R$, so $$2^{\aleph_0}=|\Bbb R|\le\aleph_0\cdot|B|=|B|\le|\Bbb R|=2^{\aleph_0}\,,$$ and therefore $|B|=2^{\aleph_0}$.