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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.

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  • $\begingroup$ @Aasaf Why the question from Jech's book on pure set theory cannot have the tag Set theory?? $\endgroup$
    – user122424
    Commented Apr 8, 2019 at 9:01
  • $\begingroup$ Kanamori put in his book a chess riddle. Does that qualify it as set theory as well? $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 8, 2019 at 9:03
  • $\begingroup$ Beware that in order theory, a complete poset is something else. In particular, every complete poset is bounded. But maybe it's clear from the context what you mean with complete... $\endgroup$
    – amrsa
    Commented Apr 8, 2019 at 16:39
  • $\begingroup$ @amrsa Jech says The $C$ is complete: if $\{(A_i,B_i):I\in I\}$ is a non-empty bounded subset of $C$ , then it has supremum. $\endgroup$
    – user122424
    Commented Apr 8, 2019 at 17:45
  • $\begingroup$ @user122424 OK, as you can see in the linked page, there are, indeed, several types of completeness. My comment was not intended to be a critic, but just a warning about the possible ambiguity. In the context of posets, I think that the most common notion is that a complete poset is one in which there is a supremum of every subset. It then follows that it is a lattice, but there are several ones that don't imply that. See for example, Complete partial order. $\endgroup$
    – amrsa
    Commented Apr 8, 2019 at 20:07

2 Answers 2

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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.

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  • $\begingroup$ What do you mean by an order isomorphism $f$? $x\leq y$ iff $f(x)\leq f(y)$ OR $x< y$ iff $f(x)< f(y)$ $\endgroup$
    – user122424
    Commented Apr 9, 2019 at 19:12
  • $\begingroup$ An order ismorphism is a an order preserving bijection. @user122424 $\endgroup$
    – William Elliot
    Commented Apr 9, 2019 at 21:51
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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.

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