<p>Yes. Suppose not, then there exists some <span class="math-container">$x \in X, \varepsilon &gt; 0$</span> such that <span class="math-container">$u(x) &gt; u(x') \; \forall \;x' : \|x - x'\| &lt; \varepsilon$</span>, which is the definition of a local maximum and contradicts strict monotonicity.</p>

CC BY-SA 4.0
2022-12-13T19:52:56.523Z
2022-12-13T19:52:56.523Z
kmf
<p>How to prove this? I understand monotonicity implies local non-satiation but does strongly monotone also imply it? How to prove it like this - <a href="https://felixmunozgarcia.files.wordpress.com/2017/08/recitation_1.pdf" rel="nofollow noreferrer">https://felixmunozgarcia.files.wordpress.com/2017/08/recitation_1.pdf</a> (page 4 answer(b))?</p>

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2022-10-07T14:49:33.073Z
Yes because Strong monotonicity implies monotonicity.
Amit
<microeconomics><preferences>
Does strongly monotone preference imply local non-satiation?
reindeer
# Does strongly monotone preference imply local non-satiation?

(Asked by: username_0 on 2022-10-07)

How to prove this? I understand monotonicity implies local non-satiation but does strongly monotone also imply it? How to prove it like this - <https://felixmunozgarcia.files.wordpress.com/2017/08/recitation_1.pdf> (page 4 answer(b))?

**Tags:** microeconomics, preferences
> **Comments:**
> **username_1**: Yes because Strong monotonicity implies monotonicity.  

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## 1 answers

### Answer by username_2 on 2022-12-13. Score: 2

Yes. Suppose not, then there exists some $x \in X, \varepsilon > 0$ such that $u(x) > u(x') \; \forall \;x' : \|x - x'\| < \varepsilon$, which is the definition of a local maximum and contradicts strict monotonicity.


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