Why Must Cuts in the Rationals Have No Largest Element?

[Demon117]

I'm having some trouble understanding why cuts are defined with property 3 below:

"A cut in Q is a pair of subsets A, and B of Q such that

(a) $A\cup B =Q$, $A\ne ∅$, $B \ne ∅$, $A\cap B = ∅$

(b) If $a \in A$ and $b \in B$, then $a < b$

(c) A contains no largest element." (Pugh, 2001)

I'm not quite sure why property three must be in place. Why does A have "no largest element" or I guess, what does it mean by "no largest element" in this context?

 

[R136a1]

It means that there must not be an $a\in A$ such that $a\geq x$ for all $x\in A$. Equivalent to this is demanding that for each $a\in A$, there is a $b\in A$ such that $a<b$.

Why do we demand this? Well, the intuition is the following. With any cut $(A,B)$ we will define a real number. Intuitively, the number associated with the cut $(A,B)$ will be $\textrm{sup}(A)$. For example, if you take

$$A = \{x\in \mathbb{Q}~\vert~x^2<2\}~\text{and}~B=\mathbb{Q}\setminus A$$

then this will correspond to the real number $\sqrt{2}$.

But this leaves us with some technical problem if (3) were not there. Indeed, if we only demand (1) and (2) then both

$$A = \{x\in \mathbb{Q}~\vert~x<0\}~\text{and}~B=\mathbb{Q}\setminus A$$
and
$$A^\prime = \{x\in \mathbb{q}~\vert~x\leq 0\}~\text{and}~B^\prime=\mathbb{Q}\setminus A$$

are valid cuts. But both of these correspond to $\textrm{sup}(A) = \textrm{sup}(A^\prime) =0$. So both cuts would define the same number. This is an unwanted situation. The goal of (3) is to eliminate $A^\prime$ from being a valid cut.