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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.) –  Jonas Kibelbek Feb 10 '11 at 17:12
    
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? –  Jonas Kibelbek Feb 10 '11 at 17:14
    
@Jonas: A is the set of all lines which do not contain (x,y). –  Ma.H Feb 10 '11 at 17:17
    
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. –  Jonas Kibelbek Feb 10 '11 at 17:28

2 Answers 2

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? –  Ma.H Feb 10 '11 at 23:08
    
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. –  Apollo Feb 10 '11 at 23:21
    
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. –  Abel Feb 11 '11 at 3:00
    
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). –  Apollo Feb 12 '11 at 2:09

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.

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