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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 has been 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. – Adam Jan 15 '12 at 0:52
up vote 4 down vote accepted

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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Thanks so much. – MathMajor Jan 15 '12 at 0:43
You're welcome, @MathMajor. But you can say thanks by up-voting or, after you feel the question has gotten enough attention, accepting the "best" answer. – dls Jan 15 '12 at 0:44
Ah... forgot about that. – dls Jan 15 '12 at 0:54

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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Thanks so much. – MathMajor Jan 15 '12 at 0:43
@Brian M. Scott How would you prove such a statement? I'm trying to prove this for every $k$ and tried induction but didn't get anywhere. – BU982T May 27 '13 at 15:35
@Bryan: I’m not sure exactly which statement you’re trying to prove. – Brian M. Scott May 27 '13 at 17:24
We have $2^k$ k- cells $Q_i$ whose union is I.Atleast one of these $Q_i$ call $I_1$. How can we say that $I_1$ can not be covered by finite subcollection of $\{ G_\alpha \}$. Suppose $I_1 \subseteq \cup_{i=1}^n G_{\alpha_i}$, then $I \subseteq \cup_{i=1}^n G_{\alpha_i} \cup I_1^c$. Is $I_1^c$ is open I – Struggler May 20 '14 at 11:18
@ dls: How does prove that if $I_1$ is covered by a finite subcollection of $\{ G_{\alpha} \}$, then I has covered by finite subcollectiom of $\{ G_{\alpha} \}$. where $I_1$ is any one of the $Q_i$. – user120386 May 26 '14 at 10:34

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