"intertwined" subset in linearly ordered set I am trying to solve an exercise in utility representation theory which leads me to the following question. 
Take any (nonempty) linearly ordered set $(X,\succeq)$. Can we construct a set $Y\subset X$ such that, for each $x,y \in X\backslash Y$ with $x\succ y$, there exists a $z\in Y$ such that $x\succ z \succ y$,
edit: AND for each $x,y \in  Y$ such that $x\succ y$, there exists a $z\in X\backslash Y$ such that $x\succ z \succ y$ ?
If such a set cannot be constructed, is there another way to show that it must exist?
(by the way, is there a way to call such subsets? "intertwined" is obviously something I made up...)
 A: You look for a partition $X=Y\cup Z$ into disjoint dense subsets (that is $y_1\succ y_2$ with $y_1,y_2\in Y$ implies $y_1\succ z\succ y_2$ for some $z\in Z$ and the same with the roles of $Y,Z$ reversed). 
One has to distinguish "$\mathbb Z$-like" and "$\mathbb Q$-like" parts.
Say $x\sim y$ if there are only finitely many elements of $z$ between $x$ and $y$. This defines an equivalence relation. 
Let $C=\{\,x\in X\mid \exists y\ne x\colon x\sim y\,\}$ and let $S$ be the set of pairs $(A,B)$ with $A,B\subseteq X$, $A\cap B=A\cap C=B\cap C=\emptyset$ and for $x,y\in A$ with $x\succ y$ there is $z\in B\cup C$ with $x\succ z\succ y$ and for $x,y\in B$ with $x\succ y$ there is $z\in A\cup C$ with $x\succ z\succ y$.
Then $S$ is partially ordered via $(A,B)\le (A',B')\iff A\subseteq A'\land B\subseteq B'$. If $\{(A_i,B_i)_{i\in I}$ is a totally ordered subset, consider $(A,B)=(\bigcup_{i\in I}A_i,\bigcup_{i\in I}B_i)$. Then one verifies that $(A,B)\in S$. Therefore Zorn's lemma applies to $S$. Let $(A_m,B_m)$ be a maximal element of $S$.
Assume there exists $x\in X\setminus(A_m\cup B_m\cup C)$. 
By maximality of $(A_m,B_m)$, the pair $(A_m\cup\{x\},B_m)$ is not in $S$, that is 
there exists $a\in A_m$ such that there is no element of $B_m\cup C$ between $a$ and $x$. Likewise,ther exists $b\in B_m$ such that there is no element of $A_m\cup C$ between $b$ and $x$. Then wlog. $a\succ x\succ b$.
As $x\notin C$, there exist elements $a',b'\in X$ with $a\succ b'\succ x\succ a'\succ b$.
We immediately have $a'\notin A_m\cup C$, but also $a'\notin B_m$ as otherwise there'd be an element of $A_m\cup C$ between $a'$ and $b$ and hence between $x$ and $b$. So $a'\in X\setminus (A_m\cup B_m\cup C)$. Likewise, $b'\in X\setminus(A_m\cup B_m\cup C)$. Then $(A_m\cup\{a'\},B_m\cup \{b'\})$ is in $S$ contradicting maximality of $(A_m,B_m)$.
Therefore $X$ is the disjoint union of $A_m,B_m$, and $C$.
Each equivalence class in $C$ is order-isomorphic to a contiguos interval of $\mathbb Z$, that is to either $\mathbb Z$ or $\mathbb N$ or $-\mathbb N$ or $\{1,\ldots,n\}$ with $n\ge2$.Pick such isomorphisms and let $C_0$ be the set of $c\in C$ mapped to even integers by this.
Let $Y=A_m\cup C_0$.
Then $Y$ has the desired property (why?).
A: Yes, $Y$ can be constructed by removing any one element from $X$, and it is trivially true.  Also we can let $Y=X \setminus \{a,b\}$ as long as between $a$ and $b$ is some other element of $X$.  This is related to the idea of a dense order and one can say $Y$ is dense in $X$.
