Successor Sets of Linearly Ordered Set Induced by Convex Component Partition

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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 $\openint p y$.

$\blacksquare$


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