Mapping is Constant iff Increasing and Decreasing

Theorem

Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be ordered sets.

Let $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ be a mapping.

Suppose $\phi$ is a constant mapping.

Then:

$\forall x, y \in S: \phi \left({x}\right) = \phi \left({y}\right)$

So:

$\forall x, y \in S: \phi \left({x}\right) \mathop{\preceq_2} \phi \left({y}\right)$

$\forall x, y \in S: \phi \left({y}\right) \mathop{\preceq_2} \phi \left({x}\right)$

and so $\phi$ is both increasing and decreasing.

$\Box$

Suppose $\phi$ is both increasing and decreasing.

Let $x, y \in S$.

Then:

$\phi \left({x}\right) \mathop{\preceq_2} \phi \left({y}\right)$

$\phi \left({y}\right) \mathop{\preceq_2} \phi \left({x}\right)$

As $\preceq$ is an ordering, by definition $\preceq$ is antisymmetric.

This means thtat:

$\phi \left({y}\right) = \phi \left({x}\right)$

As this holds for any $x, y \in S$ it follows for all $x, y \in S$ by Universal Generalisation.

Thus:

$\forall x, y \in S: \phi \left({x}\right) = \phi \left({y}\right)$

and so $\phi$ is a constant mapping.

$\blacksquare$