Strictly Increasing Mapping is Increasing

From ProofWiki
Jump to navigation Jump to search

Theorem

A mapping that is strictly increasing is an increasing mapping.


Proof

Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.

Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be strictly increasing.


From Strictly Precedes is Strict Ordering:

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

So:

\(\ds x\) \(=\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds \map \phi x\) \(=\) \(\ds \map \phi y\) Definition of Mapping
\(\ds \leadsto \ \ \) \(\ds \map \phi x\) \(\preceq_2\) \(\ds \map \phi y\) as $\preceq_2$, being an ordering, is reflexive


This leaves us with:

\(\ds x\) \(\prec_1\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds \map \phi x\) \(\prec_2\) \(\ds \map \phi y\) Definition of Strictly Increasing Mapping
\(\ds \leadsto \ \ \) \(\ds \map \phi x\) \(\preceq_2\) \(\ds \map \phi y\) Strictly Precedes is Strict Ordering

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Strictly_Increasing_Mapping_is_Increasing&oldid=523626"