User:Dfeuer/Definition:Convex Component (Order Theory)
swap def 1 and def 2.et $\left({S, \preceq}\right)$ be a totally ordered set.
Let $T \subseteq S$.
Define a relation $\sim$ on $T$ by letting $a \sim c$ iff
- For some subset $C$ of $T$ which is convex in $S$: $a,c \in C$.
Then the $\sim$ equivalence classes of $T$ are the convex components of $T$.
Lemma
$\sim$ is an equivalence relation.
Proof
Reflexive:
Let $a \in T$.
By Singleton is Convex Set (Order Theory), $\left\{{a}\right\}$ is convex.
Thus $a \sim a$.
Symmetric:
If $a \sim b$, then for some $C \subseteq T$ which is convex in $S$: $a, b\in C$.
Then $b, a \in C$, so $b \sim a$.
Transitive:
Let $a \sim b$ and $b \sim c$.
Then for some $C, D \subseteq T$ which are convex in $S$: $a,b \in C$ and $b,c \in D$.
By Union of Overlapping Convex Sets in Toset is Convex, $C \cup D$ is convex.
But $a,c \in C \cup D$, so $a \sim c$.
$\Box$
Lemma 2
Convex components of $T$ in $S$ are convex in $S$.
Proof
Let $C$ be a convex component of $T$ in $S$.
Let $a, c \in C$, $b \in S$, and $a \preceq b \preceq c$.
Since $a, c \in C$, $a \sim c$.
Thus there is a set $D \subseteq T$ such that $a,c \in D$ and $D$ is convex in $S$.
Thus $b \in D$, so $b \in T$.
$\Box$
Definition 2
Let $\left({S, \preceq}\right)$ be a totally ordered set.
Let $T \subseteq S$.
Define a relation $\sim$ on $T$ by letting $a \sim c$ iff
- For all $b \in S$: if $a \preceq b \preceq c$ or $c \preceq b \preceq a$ then $b \in T$.
Then the $\sim$ equivalence classes of $T$ are the convex components of $T$.
Lemma 1
If $x \sim y$ and $x \preceq y$, then $[x..y] \cap T$ is convex in $S$.
Proof
Suppose $a,c \in [x..y] \cap T$, $b \in S$, and $a \preceq b \preceq c$.
Then $x \preceq a \preceq b \preceq c \preceq y$.
Since $x \sim y$, $b \in [x..y] \cap T$.
Since this holds for all $a,b,c$, $[x..y] \cap T$ is convex in $S$.
$\Box$
Lemma 2
$\sim$ is the same as the relation defined in Definition 1, and is therefore an equivalence relation.
Proof
Let $\sim_1$ be the equivalence relation from Definition 1.
Suppose that $a \sim_1 c$.
Then for some $C \subseteq T$ which is convex in $S$, $a, c \in C$.
Let $b \in S$.
Then if $a \preceq b \preceq c$ or $c \preceq b \preceq a$, we must have $b \in C$, so $b \in T$.
Since this holds for all $b$, $a \sim c$.
Suppose instead that $a \sim c$ and assume WLOG that $a \preceq c$
Then by Lemma 1, $[a..c] \cap T$ is convex in $S$.
Thus $a \sim_1 c$.
$\Box$
Definition 3
Let $\left({S, \preceq}\right)$ be a totally ordered set.
Let $T \subseteq S$.
Let $a \in T$.
Then the convex component of $a$ is the largest convex subset of $T$ that contains $a$.
Lemma
The other definitions satisfy this one.