Consider an agent with consumption set $X \subseteq \mathbb R^L$ (where $0 < L < \infty$) and rational preferences $\succeq$ defined on $X$, and define monotonicity and local nonsatiation as follows:

  1. $\succeq$ is monotone iff for all $x, y \in X$, $x \gg y$ (that is, $x_\ell > y_\ell$ for all $\ell = 1, \dots, L$) implies $x \succ y$.
  2. $\succeq$ is locally nonsatiated iff for all $x \in X$ and all $\varepsilon > 0$, there exists $y \in X$ such that $\left\lVert x - y \right\rVert \le \varepsilon$ and $y \succ x$.

(These are definitions 3.B.2 and 3.B.3 in Mas-Colell/Whinston/Green, hereafter MWG).

It's sometimes claimed that monotonicity of a (rational) preference relation implies local nonsatiation. This is true for special cases such as $X = \mathbb R^L_+$ (see e.g. this answer, but false in general. A simple counterexample is $X = [0, 1] \cup [2, \infty[$ with $\ge$ as the preference relation, for which $x = 1$ has no strictly preferred $x'$ in any $\varepsilon$-environment when $\varepsilon < 1$. (MWG themselves, BTW, do not limit themselves to $X = \mathbb R^L_+$, but do not give conditions under which the implication (which they leave as an exercise) holds.)

My question is: what are sufficient condition on $X \subseteq \mathbb R^L$ for monotonicity of a rational preference relation to imply local nonsatiation? It's been suggested to me that it is sufficient that $X$ is "open from above", which strikes me as sensible in light of the previous counterexample, but I'd like to hear if there's anything on this in the literature. For that matter, what are necessary conditions?


1 Answer 1


As we focus on the property of monotonicity, I will consider $X \subset \mathbb{R}^L$. Consider the following condition:

Assumption A: For every $x \in X$, and every $\varepsilon > 0$ there is an $y \in X$ such that $y \gg x$ and $\|x - y\| < \varepsilon$.

Proposition: Assumption A is satisfied, if and only if, (for all utility functions $U: X \to \mathbb{R}$, if $U$ is monotone, then it is locally non-satiated).

Proof: Assume assumption $A$ is satisfied and take any $U: X \to \mathbb{R}$. Let $U$ be monotone and let $x \in X$. Let $\varepsilon > 0$ then from assumption $A$, there is an $y \in X$ with $\|x - y\| < \varepsilon$ and $y \gg x$. Monotonicity implies $U(y) > U(x)$, so $U$ is also locally non-satiated.

For the reverse direction assume, by contradiction, that all monotone utility functions are also locally non-satiated and that assumption $A$ is not satisfied.

This means that

  • $\text{There is an $x \in X$ and a $\varepsilon > 0$ such that for all $y \in X$ with $\|x - y\| < \varepsilon$ we have $y \not \gg x$} \tag{1}$

Consider the following utility function: $$ U(y) = \min_{i \le L} \{y_i - x_i\}. $$ This function is clearly monotone and we have that $U(x) = 0$. From the premises, it must be locally non-satiated. In particular, for all $\varepsilon > 0$ there must be a $y \in X$ such that $\|x - y\| < \varepsilon$ and $U(y) > U(x)$. The latter implies that $y_i > x_i$ for all $i \le L$ so $y \gg x$. But this contradicts $(1)$.


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