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

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

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

Let $$x, y \in S$$.

Then:


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

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.