Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Any function to a (total) order induces a (total) preorder on its domain. What can be said about the total preorders induced by a continuous function from a "nice" topological space (for instance, a Euclidean space) to the reals?

share|improve this question
1  
What kind of things are you looking for to be said? –  Henning Makholm Aug 6 '12 at 12:57

1 Answer 1

up vote 2 down vote accepted

It is possible to characterize all preorders induced by a real-valued function. The following result is essentially due to Debreu and builds on previous work by Cantor.

Theorem: Let $(T,\preceq)$ be a totally preordered set. The the following are equivalent:

  1. There exists a function $u:T\to\mathbb{R}$ such that $u(x)\leq u(y)$ if and only if $x\preceq y$.

  2. There exists a countable set $C\subseteq T$ such that for all $x\prec y$, there is $c\in C$ with $x\preceq c\preceq y$.

Proof:

(1$\implies$2)$~~~$ Let $u$ be such a function. Call an interval $(r,s)$ a gap if there exists $x,y\in T$ with $u(x)=r$, $u(y)=s$ and $u^{-1}\big((r,s)\big)=\emptyset$. Any two gaps are disjoint and contain a rational number. So there are only countably many gaps. Pic for each gap $(r,s)$ elements $x,y\in T$ with $u(x)=r$, $u(y)=s$ and collect these elements in the countable set $C_1$. For each open interval with rational endpoints $(p,q)$ such that $u^{-1}\big((p,q)\big)\neq\emptyset$, pick an element $x\in u^{-1}\big((p,q)\big)$ and collect them in the countable set $C_2$. Then $C=C_11\cup C_2$ has the desired properties.

(2$\implies$1)$~~~$ Let $C=\{c_1,c_2,\ldots\}$ be such a countable set. For $x\in T$, let $$L_x=\big\{n\in\mathbb{N}:c_n\prec x\big\}$$ and $$U_x=\big\{n\in\mathbb{N}:c_n\succ x\big\}.$$ By letting $$u(x)=\sum_{n\in L_x}\frac{1}{2^n}-\sum_{n\in U_x}\frac{1}{2^n}$$ for all $x\in T$, we get the desired function. $~~\blacksquare$

If $u$ is continuous, we get that the sets $\{y\in T:y\succ x\}$ and $\{y\in T:y\prec x\}$ are open for all $x\in T$. If $T$ is a second countable topological space with a complete preorder on it, this condition in sufficient for the preorder being induced by continuous real valued function. The proof, due to Debreu, is quite complicated, see here (JSTOR). This settles of course the case in which $T$ is a Euclidean space. A book that contains a lot of material around these themes is "Representations of preferences orderings" by Bridges and Mehta.

share|improve this answer
    
Thanks! This completely answers my main question and lets me know what field it is part of. –  Apostolos Suvici Aug 6 '12 at 15:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.