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Problem: What is the cardinality of all lines $l$ on $\mathbb R^{2}$ which do not contain a point $(x,y)\in l$ where $x, y \in \mathbb Q$ (call it $A$).

My solution: I was thinking of using CB theorem for this problem. It's easy to show that the cardinality of all lines in $\mathbb R^{2}$ is $2^{\aleph_0}$, so it's obvious that $|A|\le 2^{\aleph_0}$, but I'm having trouble of showing that the opposite direction ($|A|\ge 2^{\aleph_0}$). I thought about this injective function ($f:\mathbb R \rightarrow A$)

$\forall r \in \mathbb R$ $f(r)=\left\{\begin{matrix} (r,0), r \in \mathbb R-\mathbb Q & \\ (g(r),0), r \in \mathbb Q& \end{matrix}\right. $

where $g(r) = min{(x\in\mathbb R-\mathbb Q, x \lt r)}$

Is that injective correct? Thanks!

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    OK. Then the function f(r) needs to output a line for any real number r. The function you wrote gives a point, like (r,0). (If that represents a line, it's not clear how.) If I plug in a number like 1, or pi, I need to get a line.2011-02-10

2 Answers 2

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Consider how many horizontal lines there are with the $y$ coordinate irrational, such lines certainly can't intersect $\mathbb{Q}\times\mathbb{Q}$.

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    Yep, though Ma.H did mention that the upper bound was obvious, so I didn't elaborate. (Though never hurts to give the complete argument).2011-02-12
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Here is my approach (it may not be the best one, but it works):

To each line passing through each point $(x,y)\in\mathbb{R}^2$ we can assign a value $0\leq \theta<\pi$ which is the angle of the line with respect to the horizontal. Then we are thinking of each line as a triple $(x,y),\theta$, and we have at most $|\mathbb{R}^3|$ lines. As $\mathbb{R}^3$ has cardinality of the continuum, this gives the upper bound.

Next choose any point not equal to the given $(x,y)$, call it $(w,z)$ Then there is one and only one value of theta such that the line through $(w,z)$ passes through $(x,y)$. This means all of the other triples $(z,w),\theta$ will be in our set $A$. Since the interval $[0,\pi)$ with one point missing has the cardinality of the continuum the problem is finished.