Dedekind cut/additional property Consider the following lemma and its proof. My question follows.
Let $(P,<)$ be a dense unbounded linearly ordered set. Then there is a complete unbounded linearly ordered set $(C,\prec)$ such that:
(i) $P\subseteq C$, and $<$ and $\prec$ agree on $P$;
(ii) $P$ is dense in $C$.
Proof: A Dedekind cut in $P$ is a pair $(A,B)$ of disjoint nonempty subsets of $P$ such that
(i) $A\cup B=P$
(ii) $a<b$ for any $a\in A$ and $b\in B$
(iii) $A$ does not have a greatest element.
Let $C$ be the set of all Dedekind cuts in $P$ and let $(A_1,B_1)\preceq(A_2,B_2)$ if $A_1\subseteq A_2)$ the set $C$ is complete.
For $p\in P$, let
$$A_p=\{x\in P:x<p\},\quad B_p=\{x\in P:x\geq p\}.$$
Then $P'=\{(A_p,B_p:p\in P)\}$ is isomorphic to $P$ and is dense in $C$.
My question is why there is the condition (iii) ($A$ doesn't have the greatest element) in the above proof?
I think that this is somehow connected with the definition of $A_p$, but I don't know clearly enough how.
 A: Consider the rationals as a dense subset of the reals.
There are two sorts of lower sets (those A's).
Those with an upper bound and those without.
For a rational q, there is all the rationals less than q 
and all the rational less than or equal to q.
For an irrational r, the lower set of the all rationals less than r does not have a maximum and that set is the same as all the rationals less than or equal to r.
To have an order isomorphism between the reals and the lower sets requires that there cannot be two different lower sets for the same rational.
Thus the lower sets with with a maximum are selected to be tossed out.
That's why the A's are not allow to have a maximum, to avoid double representation.
A: The reason why condition (iii) is desired is because one wants $C$ to also be dense. Suppose you take out condition (iii). Then $(A_p,B_p)$ has an immediate successor, namely, $(A_p\cup\{p\},B_p\setminus\{p\})$. That is to say there is no element of $C$ strictly in between. I.e. $C$ is not dense. Therefore $P'$ cannot be dense in $C$. The conclusion (ii) is not satisfied.
