Circle set of points

grobber

Could you just tell me what's wrong? I'm sooo tired of trying to find counterexamples... Smile

Pleeease?

**Posted:** Thu Jan 06, 2005 3:26 pm

Myth

I think, I was wrong. Your construction contains advanced details, so it is correct. I will remove my remark.

**Posted:** Thu Jan 06, 2005 3:34 pm

_________________
Myth is out of here

pbornsztein

I didn't know what you was refusing in Grobber's construction.

Pierre.

**Posted:** Thu Jan 06, 2005 3:41 pm

grobber

Myth asked me it this thread about the construction for the circle.

By the bijection $f:[0,1)\to [0,\pi)$ given by $f(x)=e^{\pi ix}$ color the points on the upper semi-circle according to the coloring of $[0,1)$ given in the message about the coloring of $\mathbb R$ . Then color the lower semi-circle $e^{\pi ix}$ corresponding to $x\in [1,2)$ according to the same scheme, only that each color on the lower semi-circle is different from each color on the upper semi-circle.

I think this takes care of the circle. First of all, since the colors on the lower semi-circle are different from those on the upper semi-circle, any monochromatic isosceles triangle must either be situated in the upper or lower semi-circle. However, on the upper semi-circle, for example, thare are no monochromatic triangles $e^{\pi ia},e^{\pi ib},e^{\pi ic}$ , because that would make $a,b,c\in [0,1)$ three numbers in arithmetic progression on the real line, which are colored the same, and our coloring of $\mathbb R$ does not allow that.

Is it Ok?

**Posted:** Mon Jan 24, 2005 2:22 am

pbornsztein

Looks cool Smile

Pierre.

**Posted:** Mon Jan 24, 2005 4:23 am

grobber

Just an observation: I think such a construction (the one for the reals) can be turned into a proof that there exist sets of the real line which are not Lebesgue measurable.

Since we have partitioned the real line in countably many sets (according to color), assuming that they are all measurable shows that one of them has positive measure. However, it shouldn't be very hard to show that any Lebesgue measurable real set with positive measure contains three points in arithmetic progression, so we get a contradiction.

**Posted:** Fri Feb 04, 2005 9:42 am

DusT

Sorry, grobber, but I don't get something!
I don't want to look stupid, but still:
About the decomposition of reals into a base over $\mathbb Q$ , how do you know you will have a countable base? Probably it is well-known, but I would like to know if there is a countable base of $\mathbb R$ over $\mathbb Q$ .
I think the way you colored the numbers, no two will have the same color, but also, there is no bijection between reals and something countable, such as the number of coloring!
Sorry, maybe I misunderstood your solution!
Blush

**Posted:** Sun Feb 06, 2005 3:59 am

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A step forward can sometimes be the result of a kick in the @ss. Mr. Green

grobber

But there is no countable base of $\mathbb R$ over $\mathbb Q$ . I did not use the countability of the base, but the countability of the set of sets of finitely many rationals. The coloring depends on the rational coefficients that you assign to the reals.

**Posted:** Sun Feb 06, 2005 4:03 am

DusT

Ok, but why then any real can be written as a sum of FINITELY many elements of a base??? Can we choose such a base??
If we would have such a base, couldn't we construct, as you did, a bijection between the rationals and the reals???

**Posted:** Sun Feb 06, 2005 4:09 am

_________________
A step forward can sometimes be the result of a kick in the @ss. Mr. Green

grobber

Because that's what a base means: every real can be written as a finite linear combination with rational coefficients of elements in the base, Ok? Smile

How elese could we deal with bases then? We're talking about the reals now, but what about other vector spaces? What would an infinite sum mean in general?

**Posted:** Sun Feb 06, 2005 4:18 am

DusT

Ok!!!
I think I have finally understood the solution Blush

I can say it is great!! Smile

Still, is there any known (I mean somehow describeable) base of $\mathbb Q$ over $\mathbb R$ ??? I would really like to know!

**Posted:** Sun Feb 06, 2005 8:09 am

_________________
A step forward can sometimes be the result of a kick in the @ss. Mr. Green

grobber

I don't know if anyone ever made the attempt to describe one Smile

. By using Zorn's Lemma it can be shown that any vector space has a basis. That's reason enough for us to know that there is one of $\mathbb R$ over $\mathbb Q$ .

**Posted:** Sun Feb 06, 2005 8:19 am