Mapping is Constant iff Increasing and Decreasing

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

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

Then $\phi$ is a constant mapping iff $\phi$ is both increasing and decreasing.

Sufficient Condition
Suppose $\phi$ is a constant mapping.

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

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

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

Necessary Condition
Suppose $\phi$ is both increasing and decreasing.

Let $x, y \in S$.

Then:


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

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

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

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.

Hence the result.