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.

• Note that the interval on the right is longer than the one on the left, so the second case can't hold. – Cameron Buie Jun 22 '17 at 11:09
• Right. I might as well delete the second case. – 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.
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 $j_0=j_1.$