# Union of Chain of Orderings is Ordering

## Theorem

Let $S$ be a set.

Let $C$ be a non-empty chain 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.

$\blacksquare$