- (a) If $n$ is fixed, each $x \in \Bbb{R}^k$ lies in one and only one member of $\Omega_n$.
- (b) If $Q \in \Omega_n$, $Q' \in \Omega_r$, and $r < n$, then either $Q \subseteq Q'$ or $Q \cap Q' = \emptyset$.
- (c) If $Q \in \Omega_r$, then $\mbox{vol}(Q) = 2^{-rk}$; and if $n > r$, the set $P_n$ has exactly $2^{(n-r)k}$ points in $Q$.
- (d) Every nonempty open set in $\Bbb{R}^k$ is a countable union of disjoint boxes belonging to $\Omega_1 \cup \Omega_2 \cup ...$
The following are definitions of some of the symbols appearing in the above text:
$$P_n = \{(2^{-n}z_1,...,2^{-n}z_k) \mid z_i \in \Bbb{Z} \}$$
$$\Omega_n = \{Q(x,2^{-n}) \mid x \in P_n \}$$
$$Q(a,\delta) = \{x \in \Bbb{R}^k \mid a_i \le x_i < a_i + \delta \}$$
The parts I am having trouble with are (b) and the second half of (c). For (b), I attempted a proof-by-contradiction and tried to show, e.g., there exists a point that belongs to two sets in $\Omega_n$, but I couldn't how to do this, nor if it is even possible (I tried other ideas, but none worthy of note). For the second half of part (c), I tried drawing pictures on graph paper for $k=2$ and various values of $n$ and $r$, but I couldn't see a clear path to a solution. I gather that since $2^{(n-r)k} = (2^{n-r})^k$, each dimension must "contain" $2^{n-r}$ points, and since there are $k$ dimensions the total number of points is $(2^{n-r})^k$; but I couldn't make any sense of this intuition in my drawings (indeed, it be a mistaken intuition).
I could use some help.