3
$\begingroup$

Let $\succsim$ a complete, reflexive and transitive binary relation defined on $X$, a non-degenerated (i.e not identical to a singleton) convex compact subset of $\mathbb{R}^n_{++}$ (the set of n-dimensional vectors with positive components). Suppose $\succsim$ is continuous for the order topology and strictly monotonic (i.e, it preserves the usual partial order of $\mathbb{R}^n_{++}$).

Are the associated indifference sets connected ?

For a preference relation defined on the whole convex cone $\mathbb{R}^n_{++}$ the properties I introduced are indeed sufficient, as shown in this post Are Indifference Curve graphs continuous given the preferential definition of continuity?

Can you think of a counter-example or, on the contrary, provide the proof of the sufficiency of these conditions ?

Thank you very much !

$\endgroup$

1 Answer 1

5
$\begingroup$

No. Actually, standard preferences restricted to a budget line will not have this property.

For concreteness, take the preferences on $\mathbb{R}^2_{+}$ defined by the utility function given by $u(x_1,x_2)=\sqrt{x_1}+\sqrt{x_2}$. Clearly, these preferences have all the desired properties. Now let $$X=\{(x_1,x_2)\in \mathbb{R}^2_{+}\mid x_1+x_2=1\}.$$ Then the indifference curve corresponding to $(1,0)$ when restricted to $X$ is $\{(1,0),(0,1)\}$. Indeed, the preferences are strictly convex, and any other point on the line must be, as a proper convex combination, strictly better and, therefore, not indifferent.

$\endgroup$
6
  • $\begingroup$ Thanks a lot for this counter example ! It answers perfectly to my question. $\endgroup$
    – Peter
    Jul 8, 2023 at 9:21
  • $\begingroup$ May I ask if you are aware of additional conditions that would imply this connectedness (apart from convexity of indifference sets obviously) ? $\endgroup$
    – Peter
    Jul 8, 2023 at 15:31
  • $\begingroup$ I don't have a proof, but I think that $X$ contains with every points all larger points might work. Note that monotonicity has absolutely no bite on the $X$ in my example, no point there is comparable according to size. $\endgroup$ Jul 8, 2023 at 16:15
  • $\begingroup$ By “X contains with every points all larger points”, you mean “if x belongs to X, then all points dominating x also belong to X” am I right ? In that case, the proof consisting in constructing the homeomorphism between each indifference class and the simplex that I mentioned in the original proof would work, but, obviously, X would not be compact. I would be looking for conditions on the relation defined on a convex compact. Anyway, thanks a lot for your answers ! $\endgroup$
    – Peter
    Jul 8, 2023 at 17:27
  • 1
    $\begingroup$ I have the example of a relation that can be represented by a function which is the restriction to X of a linear function. But this satisfies the stronger property of having convex, and not simply connected, indifference sets. $\endgroup$
    – Peter
    Jul 8, 2023 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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