<p>Def 3.B.2: The preference relation <span class="math-container">$\succsim$</span> on <span class="math-container">$X$</span> is monotone if <span class="math-container">$x\in X$</span> and <span class="math-container">$y\gg x$</span> implies <span class="math-container">$y\succ x$</span>.</p>
<p>[1] If <span class="math-container">$\alpha(x)\geq \alpha(y)$</span>, then there are two possibilities:</p>
<p>(i) <span class="math-container">$\alpha(x)&gt;\alpha(y)$</span>. In this case, <span class="math-container">$\alpha(x)e=(\alpha(x),\alpha(x),\ldots,\alpha(x))\gg (\alpha(y),\alpha(y),\ldots,\alpha(y))=\alpha(y)e$</span>. So, by monotonicity, <span class="math-container">$\alpha(x)e\succ\alpha(y)e$</span>. Therefore, <span class="math-container">$\alpha(x)e\succsim\alpha(y)e$</span>.</p>
<p>(ii) <span class="math-container">$\alpha(x)=\alpha(y)$</span>. In this case, <span class="math-container">$\alpha(x)e=(\alpha(x),\alpha(x),\ldots,\alpha(x))= (\alpha(y),\alpha(y),\ldots,\alpha(y))=\alpha(y)e$</span>. So by completeness (reflexivity), <span class="math-container">$\alpha(x)e\succsim\alpha(y)e$</span></p>
<p>So, we don't need continuity to show that if <span class="math-container">$\alpha(x)\geq \alpha(y)$</span> holds then <span class="math-container">$\alpha(x)e\succsim\alpha(y)e$</span>.</p>
<p>[2] If <span class="math-container">$\alpha(x)e\succsim \alpha(y)e$</span>, then either <span class="math-container">$\alpha(x) &gt; \alpha(y)$</span> or <span class="math-container">$\alpha(x) = \alpha(y)$</span>. This is because if that is not the case and we have <span class="math-container">$\alpha(x) &lt; \alpha(y)$</span> for some <span class="math-container">$x,y$</span>, then by monotonicity, <span class="math-container">$\alpha(y)e\succ\alpha(x)e$</span>, contradicting <span class="math-container">$\alpha(x)e\succsim \alpha(y)e$</span>.</p>
<p>Contrapositive of monotonicity: The preference relation <span class="math-container">$\succsim$</span> on <span class="math-container">$X$</span> is monotone if <span class="math-container">$x\succsim y$</span> then either <span class="math-container">$x\notin X$</span> or <span class="math-container">$\neg y\gg x$</span>.</p>

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2024-05-13T17:19:44.783Z
Thanks a lot! About question 1, I forgot that $\alpha(\cdot)\in\mathbb{R}_+$ instead of $\mathbb{R}^L_+$.
Champa
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2024-05-13T21:47:09.233Z
(Please correct me if I'm wrong.) After a second thought, I think your (ii) for question [1] depend on the ***lemma*** from the original question, and thus continuity. Completeness doesn't say anything about $\gg$ or $\leq$ relations.
Champa
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2024-05-14T03:48:35.380Z
It follows from completeness of $\succsim$ that $\succsim$ is reflexive, and therefore, $\forall x\in X$, $x\succsim x$ holds. Since $\alpha(x)=\alpha(y)$ implies that $\alpha(x)e=\alpha(y)e$ (i.e. both are the same bundle), so by reflexivity, $\alpha(x)e\succsim \alpha(x)e$ which is just another way of writing $\alpha(x)e\succsim \alpha(y)e$
Amit
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2024-05-13T14:20:05.267Z
2024-05-13T14:27:02.473Z
2024-05-13T14:27:02.473Z
Amit
<p>Continue from <a href="https://economics.stackexchange.com/questions/58380/question-about-the-step-involving-the-connectedness-of-mathbbr-in-the-pro">this question</a>, the book <em>Microeconomic Theory</em> by Mas-Colell et al. said</p>
<blockquote>
<p>We now take <span class="math-container">$\alpha(x)$</span> as our utility function; that is, we assign a utility value <span class="math-container">$u(x)=\alpha(x)$</span> to every <span class="math-container">$x$</span>. This utility level is also depicted in Figure 3.C.1. We need to check two properties of this function: that it represents the preference <span class="math-container">$\succsim$</span> (i.e., that <span class="math-container">$\alpha(x)\geq\alpha(y)$</span> if and only if <span class="math-container">$x\succsim y$</span>) and that it is a continuous function.</p>
<p>That <span class="math-container">$\alpha(x)$</span> represents preferences follows from its construction. Formally, suppose first that <span class="math-container">$\alpha(x)\geq\alpha(y)$</span>. <strong>By monotonicity, this implies that <span class="math-container">$\mathbf{\alpha(x)e\succsim\alpha(y)e}$</span>.</strong> Since <span class="math-container">$x\sim\alpha(x)e$</span> and <span class="math-container">$y\sim\alpha(y)e$</span>, we have <span class="math-container">$x\succsim y$</span>. Suppose, on the other hand, that <span class="math-container">$x\succsim y$</span>. Then <span class="math-container">$\alpha(x)e\sim x\succsim y\sim\alpha(y)e$</span>; and so <strong>by monotonicity, we must have <span class="math-container">$\mathbf{\alpha(x)\geq\alpha(y)}$</span></strong>. Hence, <span class="math-container">$\alpha(x)\geq\alpha(y)\iff x\succsim y$</span>.</p>
</blockquote>
<p>I have two questions about this second paragraph (as highlighted in bold).</p>
<ol>
<li>I don't think monotonicity is sufficient to draw the conclusion. We also need continuity, right? See <a href="https://economics.stackexchange.com/questions/12403/continuous-rational-and-monotone-preference-relation-implies-x-succsim0?rq=1">this post</a> and <a href="https://math.stackexchange.com/questions/4914969/prove-for-a-monotone-continuous-and-rational-preference-relation-succsim-on">this post</a> for a related issue as well as the following lemma:</li>
</ol>
<blockquote>
<p><strong>Lemma</strong><span class="math-container">$\quad$</span> <em>For a monotone, continuous, and rational preference relation <span class="math-container">$\succsim$</span> on <span class="math-container">$X=\mathbb{R}^L_+$</span>, <span class="math-container">$y\geq x$</span> implies <span class="math-container">$y\succsim x$</span>.</em></p>
</blockquote>
<ol start="2">
<li>&quot;...by monotonicity, we must have <span class="math-container">$\mathbf{\alpha(x)\geq\alpha(y)}$</span>.&quot; Does this follow from the contrapositive of the definition of monotonicity?</li>
</ol>
<blockquote>
<p><strong>Definition</strong><span class="math-container">$\quad$</span> The preference relation <span class="math-container">$\succsim$</span> on <span class="math-container">$X$</span> is <em>monotone</em> if <span class="math-container">$x\in X$</span> and <span class="math-container">$y\gg x$</span> implies <span class="math-container">$y\succ x$</span>.</p>
</blockquote>
<p>of which the contrapositive is</p>
<blockquote>
<p>The preference relation <span class="math-container">$\succsim$</span> on <span class="math-container">$X$</span> is <em>monotone</em> if <span class="math-container">$x\succsim y$</span> implies <span class="math-container">$x\geq y$</span>.</p>
</blockquote>
<hr />
<p><strong>Reference</strong> <em>Microeconomic Theory</em> by Andreu Mas-Colell, Michael D. Whinston, and Jerry R. Green.</p>

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2024-05-13T06:53:24.333Z
Could you include an explicit reference to the book (authors and title) by editing your post?
Richard Hardy
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2024-05-13T12:58:07.793Z
@RichardHardy Yes, I added a reference.
Champa
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2024-05-13T13:29:05.140Z
Your contrapositive is not correct. It can be the case that neither $y\gg x$ nor $x\geq y$ holds. For example, if $y=(0,1)$ and $x=(0,0)$.
Michael Greinecker
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2024-05-13T14:21:25.947Z
@MichaelGreinecker So do we need to prove it by contradiction: Assume to the contrary that $\alpha(y)>\alpha(x)$. Then $\alpha(y)e\gg\alpha(x)e$ and so $y\sim\alpha(y)e\succ\alpha(x)e\sim x$, a contradiction.
Champa
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2024-05-13T17:13:13.870Z
Yes, that should work.
Michael Greinecker
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2024-05-13T17:16:14.197Z
@MichaelGreinecker Thank you very much!
Champa
<microeconomics><mathematical-economics><utility><preferences><theory>
Question About the Step Proving $\alpha(x)$ Represents Preferences in the Proof of Proposition 3.C.1 from MWG
Champa
# Question About the Step Proving $\alpha(x)$ Represents Preferences in the Proof of Proposition 3.C.1 from MWG

(Asked by: username_0 on 2024-05-12)

Continue from [this question](https://economics.stackexchange.com/questions/58380/question-about-the-step-involving-the-connectedness-of-mathbbr-in-the-pro), the book *Microeconomic Theory* by Mas-Colell et al. said

> We now take $\alpha(x)$ as our utility function; that is, we assign a utility value $u(x)=\alpha(x)$ to every $x$. This utility level is also depicted in Figure 3.C.1. We need to check two properties of this function: that it represents the preference $\succsim$ (i.e., that $\alpha(x)\geq\alpha(y)$ if and only if $x\succsim y$) and that it is a continuous function.
>
> That $\alpha(x)$ represents preferences follows from its construction. Formally, suppose first that $\alpha(x)\geq\alpha(y)$. **By monotonicity, this implies that $\mathbf{\alpha(x)e\succsim\alpha(y)e}$.** Since $x\sim\alpha(x)e$ and $y\sim\alpha(y)e$, we have $x\succsim y$. Suppose, on the other hand, that $x\succsim y$. Then $\alpha(x)e\sim x\succsim y\sim\alpha(y)e$; and so **by monotonicity, we must have $\mathbf{\alpha(x)\geq\alpha(y)}$**. Hence, $\alpha(x)\geq\alpha(y)\iff x\succsim y$.

I have two questions about this second paragraph (as highlighted in bold).

1. I don't think monotonicity is sufficient to draw the conclusion. We also need continuity, right? See [this post](https://economics.stackexchange.com/questions/12403/continuous-rational-and-monotone-preference-relation-implies-x-succsim0?rq=1) and [this post](https://math.stackexchange.com/questions/4914969/prove-for-a-monotone-continuous-and-rational-preference-relation-succsim-on) for a related issue as well as the following lemma:

> **Lemma**$\quad$ *For a monotone, continuous, and rational preference relation $\succsim$ on $X=\mathbb{R}^L\_+$, $y\geq x$ implies $y\succsim x$.*

2. "...by monotonicity, we must have $\mathbf{\alpha(x)\geq\alpha(y)}$." Does this follow from the contrapositive of the definition of monotonicity?

> **Definition**$\quad$ The preference relation $\succsim$ on $X$ is *monotone* if $x\in X$ and $y\gg x$ implies $y\succ x$.

of which the contrapositive is

> The preference relation $\succsim$ on $X$ is *monotone* if $x\succsim y$ implies $x\geq y$.

---

**Reference** *Microeconomic Theory* by Andreu Mas-Colell, username_2 D. Whinston, and Jerry R. Green.

**Tags:** microeconomics, mathematical-economics, utility, preferences, theory

> **Comments:**
> **username_1**: Could you include an explicit reference to the book (authors and title) by editing your post?  
> **username_0**: @username_1 Yes, I added a reference.  
> **username_2**: Your contrapositive is not correct. It can be the case that neither $y\gg x$ nor $x\geq y$ holds. For example, if $y=(0,1)$ and $x=(0,0)$.  
> **username_0**: @username_2 So do we need to prove it by contradiction: Assume to the contrary that $\alpha(y)>\alpha(x)$. Then $\alpha(y)e\gg\alpha(x)e$ and so $y\sim\alpha(y)e\succ\alpha(x)e\sim x$, a contradiction.  
> **username_2**: Yes, that should work.  
> **username_0**: @username_2 Thank you very much!  

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## 1 answers

### Answer by username_3 on 2024-05-13. Score: 2 | Accepted answer

Def 3.B.2: The preference relation $\succsim$ on $X$ is monotone if $x\in X$ and $y\gg x$ implies $y\succ x$.

[1] If $\alpha(x)\geq \alpha(y)$, then there are two possibilities:

(i) $\alpha(x)>\alpha(y)$. In this case, $\alpha(x)e=(\alpha(x),\alpha(x),\ldots,\alpha(x))\gg (\alpha(y),\alpha(y),\ldots,\alpha(y))=\alpha(y)e$. So, by monotonicity, $\alpha(x)e\succ\alpha(y)e$. Therefore, $\alpha(x)e\succsim\alpha(y)e$.

(ii) $\alpha(x)=\alpha(y)$. In this case, $\alpha(x)e=(\alpha(x),\alpha(x),\ldots,\alpha(x))= (\alpha(y),\alpha(y),\ldots,\alpha(y))=\alpha(y)e$. So by completeness (reflexivity), $\alpha(x)e\succsim\alpha(y)e$

So, we don't need continuity to show that if $\alpha(x)\geq \alpha(y)$ holds then $\alpha(x)e\succsim\alpha(y)e$.

[2] If $\alpha(x)e\succsim \alpha(y)e$, then either $\alpha(x) > \alpha(y)$ or $\alpha(x) = \alpha(y)$. This is because if that is not the case and we have $\alpha(x) < \alpha(y)$ for some $x,y$, then by monotonicity, $\alpha(y)e\succ\alpha(x)e$, contradicting $\alpha(x)e\succsim \alpha(y)e$.

Contrapositive of monotonicity: The preference relation $\succsim$ on $X$ is monotone if $x\succsim y$ then either $x\notin X$ or $\neg y\gg x$.

> **Comments:**
> **username_0**: Thanks a lot! About question 1, I forgot that $\alpha(\cdot)\in\mathbb{R}_+$ instead of $\mathbb{R}^L_+$.  
> **username_0**: (Please correct me if I'm wrong.) After a second thought, I think your (ii) for question [1] depend on the ***lemma*** from the original question, and thus continuity. Completeness doesn't say anything about $\gg$ or $\leq$ relations.  
> **username_3**: It follows from completeness of $\succsim$ that $\succsim$ is reflexive, and therefore, $\forall x\in X$, $x\succsim x$ holds. Since $\alpha(x)=\alpha(y)$ implies that $\alpha(x)e=\alpha(y)e$ (i.e. both are the same bundle), so by reflexivity, $\alpha(x)e\succsim \alpha(x)e$ which is just another way of writing $\alpha(x)e\succsim \alpha(y)e$  

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