Union of Nest of Orderings is Ordering

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Theorem

Let $S$ be a set.

Let $C$ be a nonempty nest of orderings on $S$.


Then $\bigcup C$ is an ordering on $S$.


Proof

Let $\preceq$ be an arbitrary element of $C$.

Let $\sim \, = \, \bigcup C$.

Checking in turn each of the criteria for an ordering:


Let $a, b \in S$.

Let $a \sim b$ and $b \sim a$.

Since $a \sim b$ and $b \sim a$, there exist $\preceq_1, \preceq_2 \, \in \, C$ such that $a \preceq_1 b$ and $b \preceq_2 a$.

Since $C$ is a chain:

$\preceq_1 \, \subset \, \preceq_2$

or:

$\preceq_2 \, \subset \, \preceq_1$

Without loss of generality, suppose $\preceq_1 \, \subset \, \preceq_2$.

Then it must hold that $a \preceq_2 b$.

Since:

$a \preceq_2 b$
$b \preceq_2 a$
$\preceq_2 \, \in \, T$ is an ordering

it follows that:

$a = b$

and so $\sim$ is antisymmetric.

$\Box$


Reflexivity

Let $a \in S$.

Because $\preceq$ is an ordering:

$a \preceq a$

As $\preceq \, \subseteq \, \sim$:

$a \sim a$

and so $\sim$ is reflexive.

$\Box$


Transitivity

Let $a, b, c \in S$.

Let $a \sim b$ and $b \sim c$.


Thus there are elements $\preceq_1$ and $\preceq_2$ of $C$ such that:

$a \preceq_1 b$ and $b \preceq_2 c$

Since $C$ is a chain, it must hold (as above) that:

$a \preceq_2 b$ or $b \preceq_1 c$

Without loss of generality, suppose $a \preceq_2 b$

Since:

$a \preceq_2 b$
$b \preceq_2 c$
$\preceq_2 \, \in \, T$ is an ordering

it follows that:

$a \preceq_2 c$

Since $\preceq_2 \, \subseteq \, \sim$:

$a \sim c$


So $\sim$ has been shown to be transitive.

$\Box$


$\sim$ has been shown to be reflexive, transitive and antisymmetric.

Hence by definition it is an ordering.

$\blacksquare$