Successor Sets of Linearly Ordered Set Induced by Convex Component Partition

Theorem
Let $T = \left({S, \preceq, \tau}\right)$ be a linearly ordered space.

Let $A$ and $B$ be separated sets of $T$.

Let $A^*$ and $B^*$ be defined as:
 * $A^* := \displaystyle \bigcup \left\{ {\left[{a \,.\,.\, b}\right]: a, b \in A, \left[{a \,.\,.\, b}\right] \cap B^- = \varnothing}\right\}$
 * $B^* := \displaystyle \bigcup \left\{ {\left[{a \,.\,.\, b}\right]: a, b \in B, \left[{a \,.\,.\, b}\right] \cap A^- = \varnothing}\right\}$

where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$.

Let $A^*$, $B^*$ and $\complement_S \left({A^* \cup B^*}\right)$ be expressed as the union of convex components of $S$:
 * $\displaystyle A^* = \bigcup A_\alpha, \quad B^* = \bigcup B_\beta, \quad \complement_S \left({A^* \cup B^*}\right) = \bigcup C_\gamma$

where $\complement_S \left({X}\right)$ denotes the complement of $X$ with respect to $S$.

Let $M$ be the linearly ordered set:


 * $M = \left\{ {A_\alpha, B_\beta, C_\gamma}\right\}$

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 $\left\{ {C_\gamma}\right\}$.

Proof
Let $A_\alpha \cap {S_\alpha}^- \ne \varnothing$.

Then $A_\alpha \cap {S_\alpha}^-$ contains exactly $1$ point, say $p$.

This belongs to the complement in $S$ of the closed set $\left({B^*}\right)^-$.

Hence there exists a neighborhood $\left({x \,.\,.\, y}\right)$ of $p$ which is disjoint from $\left({B^*}\right)^-$.

Then:
 * $\left({x \,.\,.\, y}\right) \cap S_\alpha \ne \varnothing$

and so:
 * $\left({p \,.\,.\, y}\right) \ne \varnothing$

But $\left({p \,.\,.\, y}\right)$ is disjoint from both $A^*$ and $B^*$.

Thus there must exist some $C_\gamma$ which contains $\left({p \,.\,.\, y}\right)$.