$k$-cells: Why $a_i < b_i$ instead of $a_i \le b_i$
Andrew Henderson 
In Rudin, The Principles of Mathematical Analysis, there is the following definition:
Definition: If $a_i < b_i$ for $i=1,2,...,k$, the set of all points, $ \boldsymbol{x} = ( x_1, x_2, ..., x_k )$ in $ \mathbb{R} ^ k $ whose coordinates satisfy the inequalities $ a_i \le x_i \le b_i$ $( 1 \le i \le k )$ is called a $k$-cell.
The examples that are provided give a $1$-cell as an interval and a $2$-cell as a rectangle.
What I am trying to figure out is why the definition is given this particular way. In particular, why the definition is not phrased the same way, with the exception that $a_i \le b_i$.
Upon consideration, the only difference between the two situations is what may arise in a particular $k$-cell.
For example, consider a $3$-cell. This would provide the set of all $\boldsymbol{x} = (x_1, x_2, x_3)$ where each $a_i \le x_i \le b_i$. Such a characterization gives a cuboid.
However, by allowing for the possibility that $a_i = b_i$, we allow for the possibility that at least one $x_i$ is fixed in the set. Thus in the $3$-cell case we would be allowed to have cuboids, planes, lines, and points. Thus the restriction $a_i \le b_i$ seems to be more flexible.
$\endgroup$ 1Thus my question is, why is the more restrictive $a_i < b_i$ used rather than $a_i \le b_i$.
1 Answer
$\begingroup$Well, think about the specific case of a $2$-cell under your less-restrictive definition. For instance, we could take $\vec{a}=(0,0)$ and $\vec{b}=(0,1)$. The result is actually the line segment $\{(0,y)\mid 0\leq y\leq 1\}$. This is a 1-dimensional object... and yet we would be calling it a 2-cell.
The big idea with Rudin's definition of a $k$-cell is that we have to have room to move in each of the $k$ directions - so that the object is actually $k$-dimsional, rather than at most $k$-dimensional.
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