# For a sequence of intervals can we find some x such that each interval contains a multiple of x?

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.

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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.

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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