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On $\mathbb R^2$, define $x=(x_1,x_2)>(y_1,y_2)=y$ if $x_i\geq y_i$ for all $i$ and $x_j>y_j$ for some $j$.

Let $\sim $ be an equivalence relation that $x\sim y$ implies $x\not> y$.

Define preorder $\succ$ this way:

  1. $x>y\implies x\succ y$, and
  1. $x\sim y>w\sim z$ implies $x\succ z$ and similar for $\prec$.

That is, $\succsim$ is extended from $\geq$ and $\sim$ relation.

Question: Do we need additional assumption to conclude that $\succsim$ is represented by a strictly increasing utility function?

Perhaps we need some continuity for $\sim$?

Solution attempt:

One assumption on $\sim$ seems both sufficient and necessary:

A1. The set $\{x:x\not\sim y\}$ is the union of two disjoint open sets.

Proof idea: Suppose $x_1>x_2$ and $x=(x_1,x_2).$

$(x_1,x_1)\succ x\succ (x_2,x_2)$.

A1 implies that, for each $x$, there exists a real number $a$ such that $x\sim (a,a)$. Then define $u(x)=a$.

Improve this question
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  • $\begingroup$ Is the first line meant to say that $\succ$ extends the usual greater than relation on $\mathbb{R}$? I don't get the question. $\endgroup$
    – Michael Greinecker
    Jun 11, 2022 at 11:33
  • $\begingroup$ @MichaelGreinecker Yes you are right. I'll add more explanation. $\endgroup$
    – dodo
    Jun 11, 2022 at 12:03
  • $\begingroup$ If I understand you correctly, there is no proper extension like that. $\endgroup$
    – Michael Greinecker
    Jun 11, 2022 at 18:12
  • $\begingroup$ @MichaelGreinecker Is it possible that an equivalence relation is represented by a function: $x\sim y\iff u(x)=u(y)$? $\endgroup$
    – dodo
    Jun 12, 2022 at 6:41
  • $\begingroup$ Yes. All you need is that there are no more equivalence classes than real numbers. $\endgroup$
    – Michael Greinecker
    Jun 12, 2022 at 7:07

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