<p>Let <span class="math-container">$\succeq$</span> be a relation on <span class="math-container">$\mathbb{R}^l_+$</span> such that <span class="math-container">$x\gg y$</span> implies <span class="math-container">$x\succeq y$</span> for all <span class="math-container">$x,y\in\mathbb{R}^l_+$</span>, and such that all upper contour sets are closed. Then <span class="math-container">$x\geq y$</span> implies <span class="math-container">$x\succeq y$</span>.</p>
<p>Proof: Assume that <span class="math-container">$x\geq y$</span>. Let <span class="math-container">$\mathbf{1}=(1,1,\ldots,1)\in\mathbb{R}^l_+$</span>. Then for all <span class="math-container">$n\geq 1$</span>, we have <span class="math-container">$n^{-1}\mathbf{1}+x\gg y$</span>. By assumption, this implies <span class="math-container">$n^{-1}\mathbf{1}+x\succ y$</span> and, therefore, also <span class="math-container">$n^{-1}\mathbf{1}+x\succeq y$</span>. Therefore, <span class="math-container">$n^{-1}\mathbf{1}+x$</span> lies in the upper contour set <span class="math-container">$\{z\in \mathbb{R}^l_+\mid z\succeq y\}$</span> for all <span class="math-container">$n$</span>. But <span class="math-container">$\lim_{n\to\infty}n^{-1}\mathbf{1}+x=x$</span>. Since upper contour sets are closed, the limit <span class="math-container">$x$</span> must also lie in <span class="math-container">$\{z\in \mathbb{R}^l_+\mid z\succeq y\}.$</span> Consequently, <span class="math-container">$x\succeq y$</span>.</p>

CC BY-SA 4.0
2024-05-19T19:12:07.053Z
2024-05-19T19:12:07.053Z
Michael Greinecker
<p>In MWG, the definition of weak preference is for all <span class="math-container">$x,y \in X$</span>, <span class="math-container">$y&gt;&gt;x$</span> implies <span class="math-container">$y\succ x$</span> . But I have read some other articles that define weak preference as <span class="math-container">$y\geq x\implies y\succeq x$</span>.
Notice here, <span class="math-container">$y\geq x$</span> means <span class="math-container">$y_i\geq x_i$</span> for all <span class="math-container">$i$</span>, and <span class="math-container">$y&gt;&gt;x$</span> means <span class="math-container">$y_i&gt;x_i$</span> for all <span class="math-container">$i$</span>.</p>
<p>These two definitions are not directly equivalent. I think the second one should imply the first MWG one, but I don't know how to prove it. It is clear that if <span class="math-container">$y&gt;&gt;x$</span> then <span class="math-container">$y\geq x$</span> so <span class="math-container">$y\succeq x$</span>. But then I need to prove <span class="math-container">$\lnot(x\succeq y)$</span>, I don't know how to approach.
Additionally, from  <a href="https://economics.stackexchange.com/questions/58052/understanding-the-definition-of-monotone">Understanding the definition of monotone</a>, one of the answers gives a proof that if the preference is continuous and rational, then <span class="math-container">$y&gt;&gt;x\implies y\succ x$</span> implies <span class="math-container">$y\geq x\implies y\succeq x$</span>. But I don't quite understand the proof. How should I approach this question if I want to use continuous definition of closed lower and upper contour sets?</p>

<microeconomics><consumer-theory><preferences><decision-theory>
Equivalence of two definitions of monotone preference
Nonenicht
# Equivalence of two definitions of monotone preference

(Asked by: username_0 on 2024-05-19)

In MWG, the definition of weak preference is for all $x,y \in X$, $y>>x$ implies $y\succ x$ . But I have read some other articles that define weak preference as $y\geq x\implies y\succeq x$.
Notice here, $y\geq x$ means $y\_i\geq x\_i$ for all $i$, and $y>>x$ means $y\_i>x\_i$ for all $i$.

These two definitions are not directly equivalent. I think the second one should imply the first MWG one, but I don't know how to prove it. It is clear that if $y>>x$ then $y\geq x$ so $y\succeq x$. But then I need to prove $\lnot(x\succeq y)$, I don't know how to approach.
Additionally, from [Understanding the definition of monotone](https://economics.stackexchange.com/questions/58052/understanding-the-definition-of-monotone), one of the answers gives a proof that if the preference is continuous and rational, then $y>>x\implies y\succ x$ implies $y\geq x\implies y\succeq x$. But I don't quite understand the proof. How should I approach this question if I want to use continuous definition of closed lower and upper contour sets?

**Tags:** microeconomics, consumer-theory, preferences, decision-theory


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## 1 answers

### Answer by username_1 on 2024-05-19. Score: 3 | Accepted answer

Let $\succeq$ be a relation on $\mathbb{R}^l\_+$ such that $x\gg y$ implies $x\succeq y$ for all $x,y\in\mathbb{R}^l\_+$, and such that all upper contour sets are closed. Then $x\geq y$ implies $x\succeq y$.

Proof: Assume that $x\geq y$. Let $\mathbf{1}=(1,1,\ldots,1)\in\mathbb{R}^l\_+$. Then for all $n\geq 1$, we have $n^{-1}\mathbf{1}+x\gg y$. By assumption, this implies $n^{-1}\mathbf{1}+x\succ y$ and, therefore, also $n^{-1}\mathbf{1}+x\succeq y$. Therefore, $n^{-1}\mathbf{1}+x$ lies in the upper contour set $\{z\in \mathbb{R}^l\_+\mid z\succeq y\}$ for all $n$. But $\lim\_{n\to\infty}n^{-1}\mathbf{1}+x=x$. Since upper contour sets are closed, the limit $x$ must also lie in $\{z\in \mathbb{R}^l\_+\mid z\succeq y\}.$ Consequently, $x\succeq y$.


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