# Relationship Between K-Cells and Intervals in Baby Rudin

1. Jan 5, 2012

### gwsinger

First consider the following definitions from Baby Rudin:

Interval: A set of real numbers of the form $[a,b]$ where for all $x \in [a,b]$ we have $a \le x \le b$.

K-Cell: A set of k-dimensional vectors of the form $x = (x_1, ...,x_k)$ where for each $x_j$ we have $a_j \le x_j \le b_j$ for each $j$ from $1 \le j \le k$.

Clearly, a one-dimensional k-cell is an interval. But I'm confused about the relationship between a multi-dimensional k-cell and an interval. For example, in Theorem 2.40, Rudin speaks of "subdividing" some k-cell $I$ into smaller intervals $Q_i$, such that the union of $Q_i$ is precisely $I$.

So suppose we are dealing with a multi-dimensional k-cell (i.e., let $I$ be a k-cell with $k > 1$). And suppose further we fix $c_j = (a_j + b_j)/2$ to then construct the two intervals $[a_j,c_j]$ and $[c_j,b_j]$. According to Rudin, we have then just created $2^k$ k-cells named $Q_i$. But it seems to me that since the k-cells of $Q_i$ are precisely intervals, that the union of these intervals could not possibly equal $I$ since the union of intervals must be another interval which $I$ is not.

What am I missing? How does the union of $Q_i$ equal $I$ in this case.

2. Jan 5, 2012

### lavinia

the midpoints span 2^k new cells. E.g. for a 2 cell, the midpoints define 4 intervals, 2 along the x axis, 2 along the y axis and each pair spans a 2 cell half the size of the original.

3. Jan 6, 2012

### gwsinger

I see that we have such intervals which are "taken" from the x-axis, y-axis, etc. But literally such intervals are now taken to the reals (as per definition of an interval being a set of real numbers), so that when united, do not form anything else but another interval.

Is Rudin just being loose here? Is he just saying "think of the intervals along each x-axis, y-axis and so forth and then just force them to form $2^k$ k-cells with your clear intuition of what I mean"?

4. Jan 6, 2012

### lavinia

Subdivision of an interval means breaking it inot subintervals that overlap only at end points. The union of the intervals in a subdivison is the whole original interval. These subintervals, as I described above, span new k-cless that form a subdivsion of the first k-cell.

Maybe it would be helpful for you to think of this in the reverse direction. Start with a k-cell and subdivide it into sub k-cells that are parallel to the coordinate axes. Each of these sub k cells is spanned by a k-tuple of points on the axes.

5. Jan 6, 2012

### gwsinger

Thanks for your response lavinia. I'm still having trouble with the language here, so let me construct an example that I hope gets to the heart of my misunderstanding.

Suppose we're dealing with a 2-cell $I$ which contains all points $x = (x_1, x_2)$ such that $a_1 \le x_1 \le b_1$ and $a_2 \le x_2 \le b_2$ where

$a_1 = 0$
$b_1 = 4$

$a_2 = 0$
$b_2 = 2$

Then the resulting 2-cell $I$ is a 4-by-2 rectangle placed up against the x and y axis.

Now if we want to "subdivide" $I$ using $c_j = (a_j + b_j)/2$ into the intervals $[a_j, c_j]$ and $[c_j, b_j]$ for $j =1, 2$ then we have the following $2^k = 2^2 = 4$ intervals:

$[a_1, c_1]$ and $[c_1, b_1]$: $[0,2]$ and $[2,4]$
$[a_2, c_2]$ and $[c_2, b_2]$: $[0,1]$ and $[1,2]$

Then if we take the union of these intervals we have another interval, namely $[0,4]$, which is not $I$!

Now intuitively, I think what we want to happen when we "subdivide" $I$ (a 4-by-2 rectangle) is to form 4 smaller rectangles which taken together make up $I$, but the language used to state this (i.e., splitting a k-cell into intervals) seems imprecise since subdividing a 2-cell into intervals (1-cells) and taking the union of intervals can produce nothing more than another interval, not a k-cell of any $k$ greater than 1!

What am I missing here?

6. Jan 6, 2012

### lavinia

I do not have baby Rudin but in my opinion your description of how the subdivision works is correct. It seems that the language is confusing in Rudin. In any case, the right answer is clear.

7. Jan 6, 2012

### gwsinger

Thanks for your help. I attached the actual language from Rudin (Theorem 2.40) in case anybody wants to take a look at it.

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8. Jan 6, 2012

### lavinia

I in this is the k cell so the union o the smaller k cells, the Q's is equal to the large k cell, I.

So the language seems clear and says the same thing that you were saying.