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 !


1 Answer 1


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.

  • $\begingroup$ 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

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