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Let $u$ be a continuous utility function on $\mathbb R^2_+\setminus\{0\}$. Consider the following three conditions:

  1. Local non satiation says that for any $x \in X$ and $\epsilon > 0$, there exists $y \in X$ such that $d(x,y) < \epsilon$ and $U(x) < U(y)$.
  2. Local non satiation* says that for any $x \in X$ and $\epsilon > 0$, there exists $y \in X$ such that $d(x,y) < \epsilon$ and $U(x) \neq U(y)$.
  3. The indifference sets of $U$ are curves.

As a standard result, (1) implies (3) and (3) implies(1).

Obviously, (1) implies (2) so (3) also implies (2).

Can (2) also imply (1) and (3)?

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1 Answer 1

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(2) does not imply (1). Consider a utility function with "circular indifference curves", e.g. $u(x,y)=-(x-1)^2-(y-1)^2$. At the bliss point $(1,1)$, the function satisfies (2) but violates (1).

(2) does imply (3), and the proof should be similar to the one showing that (1) implies (3).

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    $\begingroup$ Herr's first example also shows that (3) does not imply (1), for example, if $x=(1,1)$. $\endgroup$
    – Regio
    Commented Jun 14, 2019 at 17:40
  • $\begingroup$ @Regio You are right. Local non-satiation seems to be only a sufficient but not necessary condition for indifference "curve". $\endgroup$
    – High GPA
    Commented Apr 20, 2021 at 0:08

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