This problem is supposed to be very trivial, but I'm losing my mind over it. Given that $j_0 >j_1$ and $k_0 > k_1$, one and only one of these three cases must hold:

$1)\quad \left[\dfrac{k_0}{2^{j_{0}}},\dfrac{k_0+1}{2^{j_{0}}} \right) \subseteq \left[\dfrac{k_1}{2^{j_{1}}},\dfrac{k_1+1}{2^{j_{1}}} \right) $

$2)\quad \left[\dfrac{k_0}{2^{j_{0}}},\dfrac{k_0+1}{2^{j_{0}}} \right) \supseteq \left[\dfrac{k_1}{2^{j_{1}}},\dfrac{k_1+1}{2^{j_{1}}} \right) $

$3)\quad \left[\dfrac{k_0}{2^{j_{0}}},\dfrac{k_0+1}{2^{j_{0}}} \right) \cap \left[\dfrac{k_1}{2^{j_{1}}},\dfrac{k_1+1}{2^{j_{1}}} \right) = \emptyset$

My vain idea is that if the supremum of one the intervals is less than the infimum of the other interval, then the job is done. Shockingly I can't show this because I can't compare the boundaries. Any help will be appreciated.

  • 1
    $\begingroup$ Note that the interval on the right is longer than the one on the left, so the second case can't hold. $\endgroup$ – Cameron Buie Jun 22 '17 at 11:09
  • $\begingroup$ Right. I might as well delete the second case. $\endgroup$ – Erfan Jun 22 '17 at 11:35

Consider the following inequalities: $$\frac{k_1}{2^{j_1}}\leq\frac{k_0}{2^{j_0}}<\frac{k_1+1}{2^{j_1}}\tag{$\heartsuit$}$$ $$\frac{k_1}{2^{j_1}}<\frac{k_0+1}{2^{j_0}}\leq\frac{k_1+1}{2^{j_1}}\tag{$\diamondsuit$}$$

Your goal should be to show that $(\heartsuit)$ holds if and only if $(\diamondsuit)$ holds. From this, you can show that the first case holds if and only if the third case fails to hold.

The claim (as it stands) is fairly awkwardly phrased and over-specific. More generally, we can say the following:

Suppose $k_0,k_1$ are integers and that $j_0,j_1$ are non-negative integers such that $$\left[\frac{k_0}{2^{j_0}},\frac{k_0+1}{2^{j_0}}\right)\neq \left[\frac{k_1}{2^{j_1}},\frac{k_1+1}{2^{j_1}}\right).$$ Then exactly one of the following holds: $$\left[\frac{k_0}{2^{j_0}},\frac{k_0+1}{2^{j_0}}\right)\subsetneq \left[\frac{k_1}{2^{j_1}},\frac{k_1+1}{2^{j_1}}\right)\tag{1}$$ $$\left[\frac{k_0}{2^{j_0}},\frac{k_0+1}{2^{j_0}}\right)\supsetneq \left[\frac{k_1}{2^{j_1}},\frac{k_1+1}{2^{j_1}}\right)\tag{2}$$ $$\left[\frac{k_0}{2^{j_0}},\frac{k_0+1}{2^{j_0}}\right)\cap\left[\frac{k_1}{2^{j_1}},\frac{k_1+1}{2^{j_1}}\right)=\emptyset.\tag{3}$$ In particular, $(1)$ cannot happen unless $j_0>j_1,$ and $(2)$ cannot happen unless $j_0<j_1,$ so $(3)$ must happen if $j_0=j_1.$


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