Representation theorem for $\succsim\supset>\cup\sim$

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 $$u(x)=a$$.

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