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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?

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    It h$a$s $b$een over two years since I've used Rudin, but certainly for this proof if you draw the case for only an interval (which perhaps mirrors the 'standard' proof of the Bolzano Weistrass theorem for R in a first course of analysis), then you can easily see how to extend this. I say this so that when you try to prove this result, the idea and proof will be natural if you stay in the simpler interval scenario.2012-01-15

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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, $I$.

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    Ah... forgot about that.2012-01-15
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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.

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    brilliant explanation, thanks2016-05-14