In Theorem 2.40, Rudin talks about a $k$-cell $I$ formed by the intervals $[a_1, b_1], \ldots, [a_k, b_k]$. We split each interval at its midpoint $c_j = \frac{a_j + b_j}{2}$ and end up with $2k$ intervals $$ [a_1, c_1], [c_1, b_1], \ldots, [a_k, c_k], [c_k, b_k]. $$ He then says that the intervals $\{[a_j, c_j]\}$ and $\{[c_j, b_j]\}$ determine $2^k$ $k$-cells $Q_j$ whose union is the $k$-cell $I$.

Now, from an intuitive point of view, this is not hard to see. A line will be split into 2 regions, a square into 4 regions, a cube into eight regions, etc. For each dimension, we must choose either the interval $[a_j, c_j]$ or the interval $[c_j, b_j]$, and since there are $k$ binary choices, there are $2^k$ total of these cells.

But what does "determine" mean formally? Is it inconsequential for the rest of the proof? I tried showing it with induction, but without knowing what I'm really trying to show, a proof is fruitless.


1 Answer 1


I think you answered your own question. The intervals $[a_j, c_j]$ and $[c_j, b_j]$ for $1 \le j \le k$ determine $2^k$ $k$-cells, that is, there are $2^k$ $k$-cells

$$J_1 \times J_2 \times \dots\times J_k$$

where $J_i$ is either $[a_i, c_i]$ or $[c_i, b_i]$, such that their union is $I$.

  • $\begingroup$ Thank you for pointing that out, you're right. Any idea how one would prove that rigorously? $\endgroup$ Apr 12, 2014 at 19:47
  • $\begingroup$ @AmadeusDrZaius Prove what rigorously? $\endgroup$
    – Ink
    Apr 12, 2014 at 20:17
  • $\begingroup$ That "there are $2^k$ $k$-cells $J_1 \times J_2 \times \cdots \times J_k$ where $J_i$ is either $[a_i, c_i]$ or $[c_i, b_i]$ such that their union is $I$." $\endgroup$ Apr 12, 2014 at 20:19
  • $\begingroup$ @AmadeusDrZaius The statement I wrote was kind of redundant. There are $2^k$ $k$-cells of that form, and the union of all such $k$ cells is $I$. There's nothing to prove. Please don't think too much into this. $\endgroup$
    – Ink
    Apr 12, 2014 at 20:37

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