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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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    By A, do you mean the set of all lines not containing (x,y)? (So elements of A are lines.) Or is A the set of points other than (x,y) on a line l? (So elements of A are points.)2011-02-10
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    Also, there seem to be two big problems with the definition of g(r). There is no minimum x that's _less than_ r. (You probably wanted to use >.) A more fundamental problem: Is there a least real number that is greater than 0?2011-02-10
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    @Jonas: A is the set of all lines which do not contain (x,y).2011-02-10
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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

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