A curve that does not meet rational points

[wrobel]

This is just to recall a nice fact:

Any two points $A,B\in\mathbb{R}^n\backslash\mathbb{Q}^n,\quad n>1$ can be connected with a $C^\infty$-smooth curve that does not intersect $\mathbb{Q}^n$.

The proof is surprisingly simple: see the attachment

 

[Office_Shredder]

Sadly the pdf doesn't open on my phone.

 

[soloenergy]

That's a really cool fact! It's always interesting to see how seemingly unrelated concepts like smooth curves and irrational numbers can be connected. Can you explain the proof a bit more? It looks like the attachment is just a picture.

 

[WWGD]

That's a really cool fact! It's always interesting to see how seemingly unrelated concepts like smooth curves and irrational numbers can be connected. Can you explain the proof a bit more? It looks like the attachment is just a picture.

Edit: Just search for "$\mathbb R^n - \mathbb Q^n$ is path-connected, for $n>1$There is a search engine that allows Latex, though can't remember which.
Here's for n=2.

https://math.stackexchange.com/questions/748103/how-is-mathbb-r2-setminus-mathbb-q2-path-connected

Though it doesn't address the part about the path being $C^{\infty}$.

 

[Svein]

I think the relevant answer is the "bump function".

 

[WWGD]

I think the relevant answer is the "bump function".

I think not too hard to give it a bump, Edit: given $\mathbb R^n -\mathbb Q^n$ is dense in $\mathbb R^n$.

 

[mathwonk]

I don't know how to prove this, especially the smooth part, but it is not too shocking. E.g. any line whatsoever passing through a point with coordinates (a,b) where a is rational and b is irrational, e.g. (1,π), meets at most one rational point. (If it meets at least one, it has irrational slope; but if it meets at least two, it has rational slope.)

well, let (a,b) be any non rational point. Then since rational points are countable, only countably many lines through (a,b) meet one. If (c,d) is any other non rational point, then almost any two lines, one through (a,b) and one through (c,d), will miss all rational points and will intersect.

not quite smooth, but almost.

 

[WWGD]

I think not too hard to give it a bump, Edit: given $\mathbb R^n -\mathbb Q^n$ is dense in $\mathbb R^n$.

But I will look for something more rigorous.

 

[WWGD]

I don't know how to prove this, especially the smooth part, but it is not too shocking. E.g. any line whatsoever passing through a point with coordinates (a,b) where a is rational and b is irrational, e.g. (1,π), meets at most one rational point. (If it meets at least one, it has irrational slope; but if it meets at least two, it has rational slope.)

well, let (a,b) be any non rational point. Then since rational points are countable, only countably many lines through (a,b) meet one. If (c,d) is any other non rational point, then almost any two lines, one through (a,b) and one through (c,d), will miss all rational points and will intersect.

not quite smooth, but almost.

Yes, you can multiply by a bump function to smooth things out, but you must then show the result function won't intersect $\mathbb Q^2$.

 

[mathwonk]

ok, how's this? given any two non rational points P,Q, and any third point R not collinear with PQ, there is a unique plane containing them all, and a unique circle in that plane containing them all. The countable number of rational choices for R thus determine the only countably many circles containing both P,Q and a rational point.
Now choose any plane at all containing P,Q and look at the perpendicular bisector of the segment PQ in that plane. The uncountably many points on that bisector are each the center of a different circle containing P and Q, but only countably many of them can contain a rational point.

Thus there exist a huge infinity of circles each containing P and Q and not containing any rational points. Each circle is smooth.

I did not realize how common it is for simple curves to contain no rational points. E.g. the circle centered at (0,0) with radius cuberoot(2) does not contain any.

In reference to my previous example, the union of two line segments meeting at an angle at one point, that non- smooth looking curve is actually the image of an infinitely smooth map from the unit interval to the plane. I.e. a smooth map from [0,1] to the plane can have an image "curve" that does not look smooth. Of course the functions x(t), y(t) parametrizing this image "curve" will have derivative zero at the point of [0,1] mapping to the vertex of the angle. So one has to define what one means by a "smooth curve". But if it suffices for the "curve" to have a smooth parametrization, then my earlier example also works.

E.g. using functions like e^(-1/x^2), one can define a smooth x(t) to equal zero on [0,1/2], and increasing, from 0 to 1, on [1/2,1]. If one defines a smooth y(t) to be decreasing, from 1 to 0, on [0,1/2] and zero on [1/2,1], the resulting parametrization is smooth with image "curve" a right angle, made of the segments joining the points (0,1), (0,0), and (1,0). n Composing with a linear function takes this to a parametrization of any angular segment.
(I have not looked at the linked pdf file.)
(And I have just noticed this was first posted 3 years ago!)