# Take $\mathbb{R}$ as a vector space over $\mathbb{Q}$, then a basis for $\mathbb{R}$, then $|B|=2^{\aleph_0}$

I am trying to prove that Given $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and a basis for $\mathbb{R}$, then $|B|=2^{\aleph_0}$.

Proof: Suppse that $|B|<2^{\aleph_0}$, then $\mathbb R = span\{ b_i ; b_i \in B \}$ So $\mathbb R = \{ \Sigma_{i=1}^{\infty}\alpha_i b_i ; \alpha_i \in \mathbb Q, b_i \in B, i \in \mathbb N \} = \bigcup_{n=1}^{\infty}\{\Sigma_{i=1}^n \alpha_ib_i\}$. Which is a contradiction since a countable union of countable union of countable sets is countable and $\mathbb R$ is not.

What do you think?

Thank you, Shir

• If you assume the continuum hypothesis, okay, but if you don't assume that, you need a different argument. It boils down to $\lvert B\rvert \cdot \aleph_0 = \lvert B\rvert$ for $\lvert B\rvert \geqslant \aleph_0$. – Daniel Fischer Nov 27 '13 at 19:00
• @Daniel: It’s not quite right even under $\mathsf{CH}$: Shir is taking infinite linear combinations. – Brian M. Scott Nov 27 '13 at 19:08
• @BrianM.Scott Oy, I only saw the finite sums on the right. – Daniel Fischer Nov 27 '13 at 19:10

## 2 Answers

The general idea is on the right track, but it’s not quite right, because $|B|<2^{\aleph_0}$ does not imply that $B$ is countable, and because you should be looking only at finite linear combinations.

Suppose that $B$ is a basis for $\Bbb R$ over $\Bbb Q$; then for each $x\in\Bbb R$ there are a finite set $\{b_1^{(x)},\ldots,b_{n_x}^{(x)}\}\subseteq B$ and a finite set $\{q_1^{(x)},\ldots,q_{n_x}^{(x)}\}\subseteq\Bbb Q$ such that

$$x=q_1^{(x)}b_1^{(x)}+\ldots+q_{n_x}^{(x)}b_{n_x}^{(x)}\;.$$

Let $\mathscr{F}$ be the family of finite subsets of $B$; clearly $B$ is infinite, so $|\mathscr{F}|=|B|$. For each $F=\{b_1,\ldots,b_n\}\in\mathscr{F}$ there are $|\Bbb Q^n|=\aleph_0^n=\aleph_0$ linear combinations $q_1b_1+\ldots+q_nb_n$ with each $q_k\in\Bbb Q$, so

$$2^{\aleph_0}=|\Bbb R|=|\operatorname{span}(B)|\le|\mathscr{F}|\cdot\aleph_0=|B|\cdot\aleph_0=|B|\;.$$

Since $|B|$ is obviously at most $2^{\aleph_0}$, this shows that $|B|=2^{\aleph_0}$.

• @Shir: You’re welcome. – Brian M. Scott Nov 28 '13 at 16:14

HINT:

If $V$ is a vector space over $K$, and $B$ is a basis for $V$, then $|V|=|K\times B|^{<\omega}$. If either $K$ or $B$ is infinite, then this is just $|K\times B|$.

• Yes, I get what you mean. Thank you!! – topsi Nov 28 '13 at 14:10
• Just to clarify something, shouldn't it be $|V| \leq |K \times B|^{<\omega}$? For example $\{q_0r_0 : q_i \in \mathbb{Q}, r_i \in B \} \subseteq \{q_0r_0 + q_1 r_1 : q_i \in \mathbb{Q}, r_i \in B \}$ so there will be repeated elements between different values in $\omega$. Hope my question isn't nonsensical! – gowrath Sep 4 '17 at 2:44