Successor Sets of Linearly Ordered Set Induced by Convex Component Partition
Theorem
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Let $A$ and $B$ be separated sets of $T$.
Let $A^*$ and $B^*$ be defined as:
- $A^* := \ds \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}$
- $B^* := \ds \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}$
where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$.
Let $A^*$, $B^*$ and $\relcomp S {A^* \cup B^*}$ be expressed as the union of convex components of $S$:
- $\ds A^* = \bigcup A_\alpha, \quad B^* = \bigcup B_\beta, \quad \relcomp S {A^* \cup B^*} = \bigcup C_\gamma$
where $\relcomp S X$ denotes the complement of $X$ with respect to $S$.
Let $M$ be the linearly ordered set:
- $M = \set {A_\alpha, B_\beta, C_\gamma}$
as defined in Partition of Linearly Ordered Space by Convex Components is Linearly Ordered Set.
Then each of the sets $A_\alpha \in M$ has an immediate successor in $M$ if $A_\alpha$ intersects the closure of $S_\alpha$, the set of strict upper bounds for $A_\alpha$.
Similarly for $B_\beta$.
That immediate successor ${C_\alpha}^+$ to $A_\alpha$ is an element in $\set {C_\gamma}$.
Proof
Let $A_\alpha \cap {S_\alpha}^- \ne \O$.
Then $A_\alpha \cap {S_\alpha}^-$ contains exactly $1$ point, say $p$.
This belongs to the complement in $S$ of the closed set $\paren {B^*}^-$.
Hence there exists a neighborhood $\openint x y$ of $p$ which is disjoint from $\paren {B^*}^-$.
Then:
- $\openint x y \cap S_\alpha \ne \O$
and so:
- $\openint p y \ne \O$
But $\openint p y$ is disjoint from both $A^*$ and $B^*$.
Thus there must exist some $C_\gamma$ which contains $\openint p y$.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $39$. Order Topology: $5$