Constant Mapping both 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.

Proof

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.

$\Box$

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.

$\Box$

Hence the result.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 12.1$