Rudin Principles Theorem 2.40: Every k-cell is compact. In the proof $I$ is a $k$-cell whose coordinates are bounded by $a_{j}\le x_{j}\le b_{j}$ where $1\le j\le k$. From the proof: Put $c_{j}=(a_{j}+b_{j})/2$.
The intervals $[a_{j},c_{j}]$ and $[c_{j},b_{j}]$ then determine
$2^{k}$ $k$-cells $Q_{i}$ whose union is $I$.
What does each of the $Q_{i}$ look like?
 A: Each $Q_i$ looks exactly like $I$, except that each of its dimensions is half as big. For $k=1$, $k=2$, and $k=3$ you can draw pictures. When $k=1$, $I$ is a closed interval, $Q_1$ is the lefthand half of $I$, and $Q_2$ is the righthand half. For instance, if $I=[0,1]$, the $Q$’s are $[0,1/2]$ and $[1/2,1]$. If $k=2$, $I$ is a square, and the four $Q$’s divide it into four quarters, each square in shape, like this: $\boxplus$. When $k=3$, $I$ is a cube, and the eight $Q$’s are cubes half as long on each side. Take four cubes and arrange them in a square, then place another four cubes directly on top of the first layer to form a bigger cube.
A: As an example, look at the 3-cell $I=[0,1]\times[10,20]\times[0,10]$. Then we get $c_1=1/2, c_2=15$ and $c_3=5$. So we can create $2^3=8$ new 3-cells,
$$\begin{align*}
Q_1 &= [0,1/2]\times[10,15]\times[0,5] \\
Q_2 &= [0,1/2]\times[10,15]\times[5,10] \\
\vdots \\
Q_8 &= [1/2,1]\times[15,20]\times[5,10],
\end{align*}$$
whose union is the original 3-cell, $I$.
