Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose we have an infinite sequence of intervals $[a_n,b_n]$, where for all n, $a_n<b_n$. Then is there some $x\in \mathbb{R}$ such that for each n there is some $k\in\mathbb{Z}$ such that $kx\in [a_n,b_n]$?

It seems unlikely, but I'm not sure where to begin with the proof. Thank you.

share|improve this question

1 Answer 1

up vote 9 down vote accepted

Sketch of a proof: Take $a_n = 2^{-n}$ and $b_n = 2^{-n} + 2^{-n-2}$. Suppose such $x$ exists, it must be non-zero since none of the intervals intersect the origin. WLOG assume it is positive. Then take $N$ sufficiently large such that $2^{-N + 1} < x$. Then there cannot be any integer $k$ such that $a_N < k x$.

Edit Since the OP asked about sequences of intervals bounded away from 0. Clearly there exists sequences such that the desired property is true: just take $[a_n,b_n]$ to be intervals containing $n$. But there can also be counter examples. Let $2> C_n > 1$ be a decreasing sequence of numbers, and $\delta_n < C_n - C_{n+1}$ a decreasing sequence of numbers tending to 0. $C_n$ will be prescribed at the end of the proof.

Consider the intervals $a_n = C_n - \delta_n$, $b_n = C_n$. By construction the intervals do not intersect. Let $m$ be an integer. If there were to exist some $x$ such that there exists $k,k'\in \mathbb{Z}$ such that $kx \in [a_m,b_m]$ and $k'x \in [a_{m+1},b_{m+1}]$, this will require

$$ x \in [a_m / k, b_m/k] \cap [a_{m+1}/k', b_{m+1}/k'] \subset [a_{m+1}/k, b_m/k] \cap [a_{m+1}/k', b_m/k'] $$

So we are led to consider intersections of annuli with rescalings of itself. Now, if $\frac{b_m}{a_{m+1}} - 1 < 1 / N$, then for any $k,k' < N$, the intersection $[a_{m+1}/k, b_m/k] \cap [a_{m+1}/k', b_m/k'] = \emptyset$.

Now, let $C_n = \frac{n+1}{n}$. Then

$$b_m - a_{m+1} < b_m - b_{m+2} = \frac1m \frac2{m+2} < \frac1m b_{m+1} < \frac1m a_{m+1}$$

So this means that for $x$ to have multiples belonging to both $[a_m,b_m]$ and $[a_{m+1},b_{m+1}]$, said multiplication factor needs to be at least $m$ (actually at least $m^2$, but $m$ suffices for the proof). But since $1 < a_m,b_m < 2$, this implies that $x < 2/m$.

Now take $m\to \infty$, we see that $x$ must equal to 0, which is absurd.

share|improve this answer
1  
If you know a priori that the lengths of the intervals are bounded by below, then such x will exist: just choose x to be, say, one third of the minimal length of the interval. –  Willie Wong Nov 14 '10 at 22:09
    
OK, thanks. Is there any way to modify the proof to deal with sequences of intervals bounded away from 0? –  Ben Derrett Nov 15 '10 at 0:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.