A question about cut sets in posets and comparable elements 4.3 Definition Let A be a partially ordered class. Two elements x and y in A are said to be comparable if either x $\leq$  y or y$\leq$ x..
4.7 Definition If A is a partially ordered class, then a cut of A is pair (L, U) of nonempty subclasses of A with the following properties:
i) L∩U=ØandL∪U=A.
ii) Ifx∈Land y$\leq$x,theny∈L.
iii) Ifx∈Uand y$\geq$x,theny∈U.
My question is:
By 4.3 can I use x $\leq$ y instead of y $\leq$ x in (i i)
When I look at (i i), I suspect it is a chain in A.
 A: No, you cannot change $y\le x$ to $x\le y$ in (ii). The condition is intended to ensure that the set $L$ is downward-closed, meaning that if some element of $A$ is in $L$, so is every element of $A$ below (i.e., less than or equal to) that element. In other words, if $x\in L$, then every $y\le x$ is also in $L$. $L$ is in general not a chain in $A$. An example may help.

Example: Let $A=\Bbb R\times\Bbb R$, and for $\langle x_0,y_0\rangle,\langle x_1,y_1\rangle\in A$ let $\langle x_0,y_0\rangle\le\langle x_1,y_1\rangle$ if and only if $x_0\le x_1$ and $y_0\le y_1$. Let $L=\{\langle x,y\rangle\in A:x,y\le 0\}$, the closed third quadrant of the plane, and let $U=A\setminus L$.
If $\langle a,b\rangle\in L$, and $\langle x,y\rangle\le\langle a,b\rangle$, then $x\le a\le 0$ and $y\le b\le 0$, so $\langle x,y\rangle\in L$, and $L$ satisfies (ii). It’s clear that $L$ and $U$ satisfy (i). Finally, suppose that $\langle a,b\rangle\in U$ and $\langle a,b\rangle\le\langle x,y\rangle$. Since $\langle a,b\rangle\notin L$, either $a>0$ or $b>0$. If $a>0$, then $x\ge a>0$, so $\langle x,y\rangle\notin L$ and hence $\langle x,y\rangle\in U$. Similarly, if $b>0$, then $y\ge b>0$, and again $\langle x,y\rangle\in U$. Thus, $U$ satisfies (iii), and $\langle L,U\rangle$ is a cut in $A$.
And $L$ is definitely not a chain, since $\langle -1,0\rangle,\langle 0,-1\rangle\in L$, $\langle -1,0\rangle\not\le\langle 0,-1\rangle$, and $\langle 0,-1\rangle\not\le\langle -1,0\rangle$.

