A weaker definition of local non-satiation can also imply indifference “curve”

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)？

1 Answer

(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).

• Herr's first example also shows that (3) does not imply (1), for example, if $x=(1,1)$. – Regio Jun 14 '19 at 17:40