Definition 1. Let $(X,\leq)$ be a poset. A cut is a pair $(A,B)$ of subsets of $X$ satisfying
- $A=B^l$
- $B=A^u$
where $A^u$ (resp. $B^l$) denotes the set of all upper (resp. lower) bounds of $A$ (resp. $B$).
Note that $A\cap B$ contains at most one element, namely $\sup A$. If $\leq$ is total then we also have $A\cup B=X$.
Definition 2. Let $(X,\leq)$ be a poset. A cut is a pair $(A,B)$ of subsets of $X$ satisfying
- $A\cup B=X$, $A\cap B=\emptyset$
- $A$ is downward closed : $x\in A$ and $y\leq x$ imply $y\in A$.
- $A$ does not contain a greatest element
If $\leq$ is total then conditions $1,2,3$ above can be replaced by
- $A\cup B=X$
- $a\in A$ and $b\in B$ imply $a<b$.
- $A$ does not contain a greatest element
I want to show the following proposition:
Proposition. Let $(X,\leq)$ be a totally ordered set, and let $\mathcal C_1,\mathcal C_2$ denote the sets of all cuts according to definitions $1$ and $2$ respectively. If $\leq $ is dense in $X$, then $\mathcal C_1$ and $\mathcal C_2$, ordered by set inclusion on the lower set of the cut, are order isomorphic.
Proof. Define $f:\mathcal C_1\to\mathcal C_2$ by $f(A,B)=(A\setminus B,B)$. I claim that $f$ is an order isomorphism.
First we check that $f$ is well-defined. Let $(A,B)\in \mathcal C_1$. Since $A\setminus B $ and $B$ are disjoints, using definition $1$ we get $a<b$ whenever $a\in A \setminus B$ and $b\in B$. There are two cases:
Case 1. $A\cap B=\emptyset$. In this case $A\setminus B=A$, and $A$ cannot have a greatest element $a$ for otherwise we would have $a=\sup A \in A\cap B$. Hence $f(A,B) \in \mathcal C_2$.
Case 2. $A\cap B\neq\emptyset$. Then $\sup A$ exists and $\sup A\in A\cap B$. If $A\setminus B$ had a greatest element $a$ then we would have $a<\sup A $. By density then there would exists $x\in X$ satisfying $a<x<\sup A$. But then $x\notin A\setminus B$ and $x\notin B$, a contradiction. Hence $A\setminus B$ does not contain a greatest element, and $f(A,B) \in \mathcal C_2$.
Next we check that $f$ is one-to-one. If $f(A_1,B_1)=f(A_2,B_2)$ then $B_1=B_2$ and $A_1=B^l_1=B^l_2=A_2$ from definition 1.
Next we check that $f$ is onto. Let $(A,B)\in \mathcal C_2$. There are two cases:
Case 1. $B$ does not have a smallest element. Then $A=B^l$ and $B=A^u$. Hence $(A,B)\in \mathcal C_1$ and $f(A,B)=(A,B)$ since $A,B$ are disjoints.
Case 2. $B$ has a smallest element $b$. Then $A\cup\{b\}=B^l$ (using density) and $B=(A\cup\{b\})^u$. Hence $(A\cup\{b\},B)\in\mathcal C_1$ and $f(A\cup\{b\},B)=(A,B)$.
Finally we check that $f$ preserves order. If $(A_1,B_1),(A_2,B_2) \in \mathcal C_1$ with $A_1\subset A_2$ then $B_2=A^u_2\subset A_1^u=B_1$ from definition 1. Hence $A_1\setminus B_1 \subset A_2\setminus B_2$. Conversely if $A_1\setminus B_1 \subset A_2\setminus B_2$ then $B_2\subset B_1$, and so $A_1 =B^l_1\subset B^l_2=A_2$ from definition 1.
Am I missing something?
Thanks a lot for your help.