Strictly Increasing Mapping is Increasing

Theorem

A mapping that is strictly increasing is an increasing mapping.

Proof

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 strictly increasing.

From Strictly Precedes is a Strict Ordering:

$x \,\preceq_1\, y \implies x = y \lor x \,\prec_1\, y$

So:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \phi \left({x}\right)\)	\(=\)	\(\displaystyle \phi \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Mapping
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \phi \left({x}\right)\)	\(\preceq_2\)	\(\displaystyle \phi \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $\preceq_2$, being an ordering, is reflexive

This leaves us with:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(\prec_1\)	\(\displaystyle y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \phi \left({x}\right)\)	\(\prec_2\)	\(\displaystyle \phi \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Strictly Increasing
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \phi \left({x}\right)\)	\(\preceq_2\)	\(\displaystyle \phi \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Strictly Precedes is a Strict Ordering

$\blacksquare$

Also see

• Strictly Decreasing Mapping is Decreasing

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 14$