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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?

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  • $\begingroup$ See here. It is a consequence of the Baire Category Theorem that any Hamel basis for an infinite-dimensional vector space is uncountable. The cardinality of $\mathbb{R}$ over $\mathbb{Q}$ is then at least the cardinality of $\mathbb{R}$, so I would guess that you might be able to contradict the maximal linear independence of this Hamel basis if the cardinality is bigger than $|\mathbb{R}|$. $\endgroup$
    – JWP_HTX
    Dec 30, 2020 at 21:32
  • $\begingroup$ I suppose one could show that $\dim_{\mathbb{Q}}\mathbb{R}$ that way, but that doesn't answer my first question of why the two bases $B$ and $B'$ have the same cardinality. Could you provide some help for that @JWP_HTX? $\endgroup$
    – Gord452
    Dec 30, 2020 at 21:41
  • $\begingroup$ Right, perhaps this is more relevant to your question. $\endgroup$
    – JWP_HTX
    Dec 30, 2020 at 21:42
  • $\begingroup$ Thanks. That almost solves my problem; I still can't really formally show that $\dim_{\mathbb{Q}}\mathbb{R} = \aleph_1 := 2^{\aleph_0}$ though. $\endgroup$
    – Gord452
    Dec 31, 2020 at 0:06

1 Answer 1

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$\aleph_1$ 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 $|B|=2^{\aleph_0}$.

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  • $\begingroup$ Thanks. Just to be sure, though, isn't the number of linear combinations of elements of $B$ at most $|B|$ (since one can choose the singleton element) and at least $|B|\aleph_0$ (since given a linear combination $r_1b_1+\cdots + r_n b_n$, we can assume WLOG that the $r_i$'s are nonzero as otherwise we can exclude them and map that combination to the element $(\{r_1,\cdots, r_n\}, \{b_1,\cdots, b_n\}) \in \mathcal{F}(B)\times \mathcal{F}(\mathbb{Q})$)? $\endgroup$
    – Gord452
    Dec 31, 2020 at 1:03
  • $\begingroup$ I'm saying at most because it might not include elements such as $\{b_1, b_2, b_3\}$ and $\{r_1, r_2, r_3, r_4\}$ (basically when the number of (nonzero) rational numbers is greater than the number of basis elements specified by an element of $\mathcal{F}(B)\times \mathcal{F}(\mathbb{Q})$). $\endgroup$
    – Gord452
    Dec 31, 2020 at 1:04
  • $\begingroup$ @Gord452: You have your inequalities backwards: the number of distinct lin. combs. is at least $|B|$, since it includes the singletons from $B$, and it is at most $\aleph_0\cdot|B|$ for the reason that you give. But these are equal, so we’re done; I just didn’t bother to write out that part, as I figured that it was more familiar. $\endgroup$
    – Brian M. Scott
    Dec 31, 2020 at 1:09
  • $\begingroup$ @Gord452: I’ve added at least to make it clearer. $\endgroup$
    – Brian M. Scott
    Dec 31, 2020 at 1:16

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