B The Relationship Between Function and Ordering Relation

[littlemathquark]

TL;DR

What is the necessary and sufficient condition for a function to induce an ordering relation (partial or total)?

I think, the necessary and sufficient condition for a function to induce an ordering relation (specifically a partial or total order) on its domain is that it must be compatible with the ordering defined on the codomain (i.e., it must be order-preserving).

How can we express this necessary and sufficient condition more clearly? Thank you.

 

[fresh_42]

The term of an order inducing function doesn't make much sense. We have order preserving functions, and order reversing functions if domain and codomain allow an ordering. But what would be an ordering solely created by a function? If a domain is ordered, say $\mathbb{N},$ then any function $f\, : \,\mathbb{N}\to \mathbb{N}$ defines another order, simply by defining $f(n)< f(n+1).$ You can get any order on $\mathbb{N}$ this way since all permutations can be $f.$

 

[pasmith]

Is the question: if &lt;_Y is an order on Y, what conditions must f: X \to Y satisfy in order that we can define &lt;_X via f(x) &lt;_Y f(y) \Leftrightarrow x &lt;_X y?

 

[WWGD]

TL;DR Summary: What is the necessary and sufficient condition for a function to induce an ordering relation (partial or total)?

I think, the necessary and sufficient condition for a function to induce an ordering relation (specifically a partial or total order) on its domain is that it must be compatible with the ordering defined on the codomain (i.e., it must be order-preserving).

How can we express this necessary and sufficient condition more clearly? Thank you.

For one, your function would have to be defined in ordered domains and codomains, I would guess, with a nice -enough relationship between the two orderings.

 

[littlemathquark]

I think that is true:
Let (S,⪯) be an ordered set. Then given any set X and a function f:X→S , we can define an order $⪯_X$ on X by $x_1⪯_Xx_2$ if and only if $f(x_1)⪯f(x_2)$

 

[fresh_42]

I think that is true:
Let (S,⪯) be an ordered set. Then given any set X and a function f:X→S , we can define an order $⪯_X$ on X by $x_1⪯_Xx_2$ if and only if $f(x_1)⪯f(x_2)$

Only on $f^{-1}(S)$ not on entirely $X.$

 

[pasmith]

Only on $f^{-1}(S)$ not on entirely $X.$

X is the domain of f : X \to S. By definition, each element x \in X has a unique image f(x) \in S.

 

[fresh_42]

X is the domain of f : X \to S. By definition, each element x \in X has a unique image f(x) \in S.

This discussion is about precision and wording. The assumption that $X$ is the entire domain has to be stated, in my opinion. People speak of real functions and $x\mapsto 1/x$ at the same time without distinguishing the two. If we continue to be as sloppy as the thread's start already is, then this entire discussion becomes even more ridiculous. Moreover, the example in post #5 does not "induce" an order by the function $f$, it pulls back an order!

This thread is along the pattern: choose a subject everybody has something to say about, be as vague as possible in your question, occasionally contribute a few crumbs, and enjoy the resulting nonsense.

 

[WWGD]

It seems the issue here of the pullback of an order relation. Then we say $x_1$~$x_2$ iff(def.) $f(x_1)$~$f(x_2)$. Given a set S with given structure and a bijection $f: X\rightarrow S$, the structure can be pulled back into $X$.