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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    $\begingroup$ 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. $\endgroup$
    – Adam
    Jan 15, 2012 at 0:52

2 Answers 2


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, $I$.

  • $\begingroup$ 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. $\endgroup$
    – dls
    Jan 15, 2012 at 0:44
  • $\begingroup$ Ah... forgot about that. $\endgroup$
    – dls
    Jan 15, 2012 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.

  • $\begingroup$ @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. $\endgroup$
    – user70962
    May 27, 2013 at 15:35
  • $\begingroup$ @Bryan: I’m not sure exactly which statement you’re trying to prove. $\endgroup$ May 27, 2013 at 17:24
  • $\begingroup$ 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 $\endgroup$
    – Struggler
    May 20, 2014 at 11:18
  • $\begingroup$ @ 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 $Q_i$. $\endgroup$
    – user120386
    May 26, 2014 at 10:34
  • $\begingroup$ brilliant explanation, thanks $\endgroup$
    – Dorgham
    May 14, 2016 at 12:41

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