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Show that the sets $S(x)=\{(m,n):m\in\mathbb{Z},n\in\mathbb{N}^+,|x-\frac{m}{n}|\leq\frac{1}{n^3}\}$ where $x\in\mathbb{R}$ are finite for almost all $x\in\mathbb{R}$ in the sense of Lebesgue measure.

mathman
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  • Your set describes at most countable set, so It's Lebesgue measure is 0. (If we can call it finite, it seems to me is enough.) – kolobokish Nov 10 '16 at 19:13
  • ok it has finite measure but is it the same thing that the question called finite for almost all $x$? – mathman Nov 10 '16 at 19:16
  • The problem is to show that $S(x)$ is finite for ae. $x$. – copper.hat Nov 10 '16 at 19:17
  • This might have some relevant info: https://www.encyclopediaofmath.org/index.php/Diophantine_approximation,_metric_theory_of (from http://mathoverflow.net/questions/43381/what-numbers-can-be-approximated-pretty-well-by-rationals). – copper.hat Nov 14 '16 at 17:08

2 Answers2

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It seem to me I have found a possible proof. Let's take first the case $x>0$ and is not rational. Than we have the following alternatives , either $$\frac{m}{n} \leq x \leq \frac{m}{n}+\frac{1}{n^{3}}$$ or $$\frac{m}{n}-\frac{1}{n^{3}} \leq x \leq \frac{m}{n}$$. For the first case,by Archimedean axiom there are finitely many $\frac{m}{n}$, sutisfying the left part of the inequility with $(m,n) =1$. However there can be infinitely many pairs $(m,n)$ sutisfying the left part taking for example $(km, kn)$ which gives the same value. However let's show that it can't be done for infinitely many $k$-s. Here we can use the right part of the inequility. Let's suppose it can be done for infinitely many $k$-s, than it will mean that $\sup(x) = \frac{m}{n}$(taking x satisfying the inequality.). But on the other hand $x \geq \frac{m}{n}$. So we would have $x=\frac{m}{n}$, which can't be so as we have taken only irrational ones. So there are finitely many pair $(m,n)$, (though there can be some of the form $(km,kn)$ with $(k \neq 1)$).
For the second inequality, the situation is exactly the same. So while $x$ is irrational, we have only finitely many pairs $(m,n)$. However for all rationals there are infinitely many such pairs. Taking $x=\frac{p}{q}$, take $m=kp$, $n=kq$. Than for all $k$-s, the satisfies the conditions. For rationals there are infinitely many pairs, but exactly only for rationals. Their Lebesgue measure is 0, so the condition is satisfied almost everywhere in $R$.

Maybe I'm missing something. If this has little sense, I'm sorry.

kolobokish
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  • But you assumed that x is positive and went on,ı am trying to ruin it for the negative case.. – mathman Nov 10 '16 at 20:43
  • I don't think so. Look, in what I did, no matter $x>0$, or $x<0$, the inequilities will be of the same form. (so finiteness for positive ones imply finiteness for negative ones.). – kolobokish Nov 10 '16 at 20:45
  • I see,the proof makes sense but ı try to be sure,so thank you for your effort. – mathman Nov 10 '16 at 20:56
  • Thank you. It seem to me, this is not quite rigorous proof. (So I myself would like to find what others will answer. )) ) – kolobokish Nov 10 '16 at 20:59
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The relevant result is Khinchin's theorem. (I haven't looked at the proof, but suspect it can be proved by an application of the Borel Cantelli lemma.)

Here $\psi(q) = {1 \over q^2}$, and $\sum_q \psi(q) $ is convergent, hence $A(x)=\{ {p \over q} \,|\, |x - { p \over q} | \le {\psi(q) \over q} \}$ is finite for ae. $x$.

The only detail is to translate this into showing that $S(x)$ is finite for ae. $x$.

Let $F$ be the set of $x$ for which $A(x)$ is finite, then $mF^c = 0$.

Then for $x \in F \setminus \mathbb{Q}$, the set $S(x)$ is finite.

To see this, note that if $(m,n) \in S(x)$ then ${m \over n} \in A(x)$. Suppose $S(x)$ is infinite, then there is some $(p,q) \in S(x)$ such that $(n_k p,n_k q) \in S(x)$ for some infinite sequence $n_k$. In particular, this means that $|x-{p\over q}| \le {\psi(n_kq) \over n_kq}$ for all $n_k$, and so $x={p\over q}$, which is a contradiction. Hence $S(x)$ is finite.

copper.hat
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