$\mathbb{R}^3$ \ $\mathbb{Q}^3$ is union of disjoint lines. The lines are not in an axis diretion.

I have the following question:

$\mathbb R^3\setminus \mathbb Q^3$ is a union of disjoint lines. This exercise is in the book: Set theory for the working mathematician of Krzysztof Ciesielski.

someone can give me a help on how to proceed?

Solutions Collecting From Web of "$\mathbb{R}^3$ \ $\mathbb{Q}^3$ is union of disjoint lines. The lines are not in an axis diretion."

Let $\mathbb{R}^3\setminus\mathbb{Q}^3=(p_\alpha)_{\alpha<\frak{c}}$. We will find lines $r_\alpha\subset\mathbb{R}^3\setminus\mathbb{Q}^3$ for all $\alpha<\frak{c}$ such that $r_\alpha\cap r_\beta=\emptyset$ for $\alpha\ne\beta$ and $p_\alpha\in\bigcup_{\beta\leq\alpha}r_\beta$ $-$ we will allow that some (but not all, of course) of the $r_\alpha$ be empty. For $r_0$ choose any line with no rational points that contains $p_0$. Now, suppose you already have defined $r_\beta$ for all $\beta<\alpha$ for some $\alpha<\frak{c}$. If $p_\alpha\in\bigcup_{\beta<\alpha}r_\beta$, then let $r_\alpha=\emptyset$. If not, notice that there exists at least one plane $\mathcal{P}$ such that $p_\alpha\in \mathcal{P}$ and $r_\beta\not\subset \mathcal{P}$ for all $\beta<\alpha$ (this can be done because there are $\frak{c}$ many planes passing through $p_\alpha$ and, in the “worst case” there are $|\alpha|<\frak{c}$ many pairwise “non-coplanar” lines). Finally, notice two things:

1) the plane $\mathcal{P}$ may intercept each $r_\beta$ at most one time;

2) there are at most countable many rational points in $\mathcal{P}$.

Since the lines in $\mathcal{P}$ that pass through $p_\alpha$ are determined by the points in $\mathcal{P}\setminus\{p_\alpha\}$, by (1) and (2) certainly there is a line $r_\alpha\subset \mathcal{P}\cap\mathbb{R}^3\setminus\mathbb{Q}^3 $ such that $p_\alpha\in r_\alpha$ and $r_\alpha\cap(\bigcup_{\beta<\alpha}r_\beta)=\emptyset$.

Thus, there exists a decomposition of $\mathbb{R}^3\setminus\mathbb{Q}^3$ in pairwise disjoint lines.