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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

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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    Sorry, but I don't follow your hint. How can I get the cardinality?2011-02-10
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    How many irrationals are there? (If there are $\aleph_0$ rationals and $2^{\aleph_0}$ total real numbers, then ...) There is one horizontal line for each irrational. This will give you a lower bound (since there are lots of non-horizontal lines that miss rational points) on the number of lines you are looking for.2011-02-10
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    And since it takes two numbers $b$ and $m$ to define a line $y = mx + b$, there are at most as many lines as pairs of real numbers. Together with your reasoning, this demonstrates that the cardinality of the set Ma.H asked for is the cardinality of the continuum.2011-02-11
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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.