TITLE: Continuum-many independent vectors over Q in R as a Q-vector space
QUESTION [9 upvotes]: Let's put ourselves in the framework of ZF. Is it true that if we think of the set of real numbers as a rational vector space, there are continuum-many linearly independent vectors? I feel that we could then use this to make an injection from $\mathbb{R}$ to $\mathbb{R}/\mathbb{Q}$, the quotient vector space by the 1-dimensional subspace $\mathbb{Q}$. Does this strategy work? This is to help me think about the result that (in ZF) $|\mathbb{R}| \leq |\mathbb{R}/\mathbb{Q}|$, proved by Mycielski in 1964. I haven't looked at his paper, just trying to imagine how one would prove this result.

REPLY [15 votes]: Let $f: 2^{\mathbb{N}} \rightarrow \mathbb{R}/\mathbb{Q}$ be the function given by
$$ f((a_i)_{i \in \mathbb{N}}) = \text{the equivalence class of }\sum_{k=0}^{\infty} \frac{b_k}{2^{(k+1)!}}$$
where $(b_i)_{i \in \mathbb{N}}=(a_0,a_0,a_1,a_0,a_1,a_2,\dots)$. Notice that a real number of the form $$\sum_{k=0}^{\infty} \frac{c_k}{2^{(k+1)!}}$$ where $(c_i)_{i \in \mathbb{N}} \in 2^{\mathbb{N}}$ is rational if and only if the sequence $(c_k)$ is eventually zero. Consequently, $f((a_i)_{i \in \mathbb{N}})=f((a'_i)_{i \in \mathbb{N}})$ if and only if the corresponding sequences $(b_i)_{i \in \mathbb{N}}$ and $(b'_i)_{i \in \mathbb{N}}$ are eventually equal if and only if $(a_i)_{i \in \mathbb{N}}=(a'_i)_{i \in \mathbb{N}}$. Finally, compose this function with your favorite explicit injection from $\mathbb{R}$ to $2^{\mathbb{N}}$. This gives you an injection from $\mathbb{R}$ to $\mathbb{R}/\mathbb{Q}$.
If you want to find a $\mathbb{Q}$-linearly independent subset of $\mathbb{R}$ of size continuum, consider the image of the map $g: \mathcal{A} \rightarrow \mathbb{R}$ given by
$$ g(S)=\displaystyle \sum_{k=0}^{\infty} \frac{\chi_{S}(k)}{2^{(k+1)!}}$$
where $\mathcal{A} \subseteq \mathcal{P}(\mathbb{N})$ is an almost disjoint family of size continuum. For example, enumerate the vertices of the full binary tree of height $\omega$ by $\mathbb{N}$ and let $\mathcal{A}$ be the set of (labels of) branches. To see why this set is linearly independent, see this nice answer of Tim Gowers on another MO question.
Note that both of these constructions can be done in ZF.