# dimension of infinite dimensional vector space

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?

• 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 $|\mathbb{R}|$. Dec 30, 2020 at 21:32
• 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? Dec 30, 2020 at 21:41
• Right, perhaps this is more relevant to your question. Dec 30, 2020 at 21:42
• 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. Dec 31, 2020 at 0:06

$$\aleph_1$$ 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 $$|B|=2^{\aleph_0}$$.
• 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 $(\{r_1,\cdots, r_n\}, \{b_1,\cdots, b_n\}) \in \mathcal{F}(B)\times \mathcal{F}(\mathbb{Q})$)? Dec 31, 2020 at 1:03
• 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 $\mathcal{F}(B)\times \mathcal{F}(\mathbb{Q})$). Dec 31, 2020 at 1:04
• @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. Dec 31, 2020 at 1:09