<p>Monotonicity of preferences is a <strong>stronger</strong> condition than local nonsatiation. Monotonicity implies local nonsatiation, but <em>not</em> the other way around.</p>

<p>To see this:</p>

<p><strong>Claim:</strong> Let <span class="math-container">$\succsim$</span> be a monotonic preference relation over <span class="math-container">$\mathbb{R}^n_{+}$</span>. Therefore, <span class="math-container">$\succsim$</span> is locally nonsatiated.</p>

<p><strong>Proof:</strong> Fix some <span class="math-container">$\varepsilon &gt; 0$</span>. Let there be an arbitrary <span class="math-container">$x \in \mathbb{R}^n_{+}$</span> and let <span class="math-container">$\mathbf{1} \in \mathbb{R}^n_{+}$</span> be the unit vector. For any <span class="math-container">$\lambda &gt; 0$</span>, we also have <span class="math-container">$x + \lambda \mathbf{1} \in \mathbb{R}^n_{+}$</span>. Since clearly <span class="math-container">$x + \lambda \mathbf{1} \gg x,$</span> <span class="math-container">$x + \lambda \mathbf{1} \succ x$</span> by monotonicity. Consider the following metric over <span class="math-container">$\mathbb{R}^n_{+}$</span>:</p>

<p><span class="math-container">$$
d(x+\lambda \mathbf{1}, x) = ||x+\lambda \mathbf{1} -x|| = \lambda||\mathbf{1}|| = \lambda \sqrt{n}.
$$</span></p>

<p>Thus for <span class="math-container">$\lambda &lt; \frac{\varepsilon}{\sqrt{n}}$</span>, <span class="math-container">$d(x+\lambda \mathbf{1}, x) &lt; \varepsilon$</span> yet <span class="math-container">$x + \lambda \mathbf{1} \succ x$</span>. Since <span class="math-container">$x$</span> was arbitrary, the existence of such a point implies that <span class="math-container">$\succsim$</span> is locally nonsatiated. <span class="math-container">$\blacksquare$</span></p>

<p>To show that locally nonsatiated preferences do <em>not</em> imply monotonic preferences, you can come up with a utility function <span class="math-container">$u(\cdot)$</span> over various goods that strictly increases with respect to some of the goods but reaches a <em>satiation point</em> for at least one of the others. For example, in <span class="math-container">$\mathbb{R}^2_{+}$</span>: </p>

<p><span class="math-container">$$
u(x_1,x_2) = x_1 - |1-x_2|.
$$</span></p>

<p>A consumer with such preferences is satiated with respect to <span class="math-container">$x_2$</span> at <span class="math-container">$x_2=1$</span>.</p>

CC BY-SA 4.0
2018-10-15T15:04:27.380Z
2018-10-15T20:41:26.197Z
2018-10-15T20:41:26.197Z
Kenneth Rios
<p>Is there a practical difference between local non-satiation and montonicity? Can one exist in a utility function without the other?</p>

CC BY-SA 4.0
2018-10-15T16:31:12.927Z
Do you mean [weak or strong monotonicity](https://en.wikipedia.org/wiki/Monotone_preferences)?
Giskard
<microeconomics><consumer-theory>
Whats the difference between local non-satiation and monotonicity?
EconJohn
# Whats the difference between local non-satiation and monotonicity?

(Asked by: username_0 on 2018-10-15)

Is there a practical difference between local non-satiation and montonicity? Can one exist in a utility function without the other?

**Tags:** microeconomics, consumer-theory
> **Comments:**
> **username_1**: Do you mean [weak or strong monotonicity](https://en.wikipedia.org/wiki/Monotone_preferences)?  

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## 1 answers

### Answer by username_2 on 2018-10-15. Score: 6 | Accepted answer

Monotonicity of preferences is a **stronger** condition than local nonsatiation. Monotonicity implies local nonsatiation, but *not* the other way around.

To see this:

**Claim:** Let $\succsim$ be a monotonic preference relation over $\mathbb{R}^n\_{+}$. Therefore, $\succsim$ is locally nonsatiated.

**Proof:** Fix some $\varepsilon > 0$. Let there be an arbitrary $x \in \mathbb{R}^n\_{+}$ and let $\mathbf{1} \in \mathbb{R}^n\_{+}$ be the unit vector. For any $\lambda > 0$, we also have $x + \lambda \mathbf{1} \in \mathbb{R}^n\_{+}$. Since clearly $x + \lambda \mathbf{1} \gg x,$ $x + \lambda \mathbf{1} \succ x$ by monotonicity. Consider the following metric over $\mathbb{R}^n\_{+}$:

$$
d(x+\lambda \mathbf{1}, x) = ||x+\lambda \mathbf{1} -x|| = \lambda||\mathbf{1}|| = \lambda \sqrt{n}.
$$

Thus for $\lambda < \frac{\varepsilon}{\sqrt{n}}$, $d(x+\lambda \mathbf{1}, x) < \varepsilon$ yet $x + \lambda \mathbf{1} \succ x$. Since $x$ was arbitrary, the existence of such a point implies that $\succsim$ is locally nonsatiated. $\blacksquare$

To show that locally nonsatiated preferences do *not* imply monotonic preferences, you can come up with a utility function $u(\cdot)$ over various goods that strictly increases with respect to some of the goods but reaches a *satiation point* for at least one of the others. For example, in $\mathbb{R}^2\_{+}$:

$$
u(x\_1,x\_2) = x\_1 - |1-x\_2|.
$$

A consumer with such preferences is satiated with respect to $x\_2$ at $x\_2=1$.


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