<ol>
<li><p>Requires continuity which is already assumed.</p>
<blockquote>
<p><em>Proof.</em> Consider an open ball <span class="math-container">$B_0$</span> around <span class="math-container">$0$</span> with radius <span class="math-container">$\epsilon_0 &lt; x/2$</span> and an open ball <span class="math-container">$B_x$</span> around <span class="math-container">$x$</span> with radius <span class="math-container">$\epsilon_x &lt; x/2$</span>. Pick any <span class="math-container">$z_x \in B_x$</span> and a <span class="math-container">$z_0 \in B_0$</span> and observe that <span class="math-container">$z_x &gt;&gt; z_0 \implies z_x \succ z_0$</span>. However, if <span class="math-container">$0 \succ x$</span>, then continuity requires <span class="math-container">$z_0 \succ z_x$</span> which is not possible.</p>
<p>Thus, <span class="math-container">$x \succeq 0$</span> for all <span class="math-container">$x \in X$</span> when monotonicity and continuity are both assumed.</p>
</blockquote>
</li>
<li><p>You may. Either way, it's correct.</p>
</li>
<li><p>Correct. However, a shorter proof would have been to construct <span class="math-container">$\alpha$</span> (as you did) without unnecessarily turning it into a <em>proof by contradiction</em>.</p>
<p>For non-emptiness of <span class="math-container">$A^{-}$</span>, pick <span class="math-container">$\alpha = 0$</span>. MWG's proof asusmes continuity, so we have <span class="math-container">$x \succeq 0 \ \forall \ x \in X$</span>.</p>
</li>
<li><p>The part you have shown (proven) looks correct to me.</p>
</li>
</ol>

CC BY-SA 4.0
2024-04-06T11:50:04.857Z
2024-04-06T11:50:04.857Z
uninterestedacademic
<p>I have difficulties understanding the first step of the proof of Proposition 3.C.1 in MWG.</p>
<blockquote>
<p><strong>Proposition 3.C.1</strong><span class="math-container">$\quad$</span> <em>Suppose that the rational preference relation <span class="math-container">$\succsim$</span> on <span class="math-container">$X$</span> is continuous. Then there is a continuous utility function <span class="math-container">$u(x)$</span> that represents <span class="math-container">$\succsim$</span>.</em></p>
</blockquote>
<blockquote>
<p><strong>Proof</strong><span class="math-container">$\quad$</span> For the case of <span class="math-container">$X=\mathbb{R}^L_+$</span> and a monotone preference relation, there is a relatively simple and intuitive proof that we present here with the help of Figure 3.C.1. <a href="https://i.sstatic.net/fPJmO.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/fPJmO.png" alt="enter image description here" /></a></p>
<p>Denote the diagonal ray in <span class="math-container">$\mathbb{R}^L_+$</span> (the locus of vectors with all <span class="math-container">$L$</span> components equal) by <span class="math-container">$Z$</span>. Let <span class="math-container">$e$</span> designate the <span class="math-container">$L$</span> vector whose elements are all equal to <span class="math-container">$1$</span>. Then <span class="math-container">$\alpha e\in Z$</span> for all nonnegative scalars <span class="math-container">$\alpha\geq0$</span>.</p>
<p><strong>Note that for every <span class="math-container">$x \in \mathbb{R}^L_+$</span>, monotonicity implies that <span class="math-container">$x\succsim0$</span>.</strong> <strong>Also note that for any <span class="math-container">$\overline{\alpha}$</span> such that <span class="math-container">$\overline{\alpha}e \gg x$</span> (as drawn in the figure), we have <span class="math-container">$\overline{\alpha}e \succsim x$</span>.</strong> Monotonicity and continuity can then be shown to imply that there is a unique value <span class="math-container">$\alpha(x)\in[0,\overline{\alpha}]$</span> such that <span class="math-container">$\alpha(x)e \sim x$</span>.</p>
<p>Formally, this can be shown as follows: By continuity, the upper and lower contour sets of <span class="math-container">$x$</span> are closed. <strong>Hence, the set <span class="math-container">$A^+ = \{\alpha\in\mathbb{R}_+:\alpha e \succsim x\}$</span> and <span class="math-container">$A^- = \{\alpha\in\mathbb{R}_+:x \succsim \alpha e\}$</span> are nonempty and closed.</strong> Note that by completeness of <span class="math-container">$\succsim$</span>, <span class="math-container">$\mathbb{R}_+ \subset (A^+ \bigcup A^-)$</span>. The nonemptiness and cloedness of <span class="math-container">$A^+$</span> and <span class="math-container">$A^-$</span>, along with the fact that <span class="math-container">$\mathbb{R}_+$</span> is connected, imply that <span class="math-container">$A^+ \bigcap A^- \neq \emptyset$</span>. Thus, there exists a scalar <span class="math-container">$\alpha$</span> such that <span class="math-container">$\alpha e \sim x$</span>. Furthermore, by monotonicity, <span class="math-container">$\alpha_1 e \succ \alpha_2 e$</span> whenever <span class="math-container">$\alpha_1 &gt; \alpha_2$</span>. Hence, there can be at most one scalar satisfying <span class="math-container">$\alpha e \sim x$</span>. This scalar is <span class="math-container">$\alpha(x)$</span>.</p>
</blockquote>
<h2>My Question</h2>
<p>I highlighted the sentences where I have trouble understanding:</p>
<ol>
<li>&quot;For every <span class="math-container">$x \in \mathbb{R}^L_+$</span>, monotonicity implies that <span class="math-container">$x\succsim0$</span>&quot;:</li>
</ol>
<blockquote>
<p>Although, it intuitively make sense, it does not quite follow the definition of monotonicity. The book defines monotonicity as follows:
<span class="math-container">\begin{align*}
\text{The preference relation $\succsim$ on $X$ is monotone if $x \in X$ and $y \gg x$ implies $y \succ x$.}
\end{align*}</span>
It doesn't say anything aout the case when <span class="math-container">$y \geq x$</span> but not <span class="math-container">$y \gg x$</span>. (For a related question, please see <a href="https://economics.stackexchange.com/questions/58052/understanding-the-definition-of-monotone">here</a>.) I couldn't understand why this statement is true.</p>
</blockquote>
<ol start="2">
<li>&quot;For any <span class="math-container">$\overline{\alpha}$</span> such that <span class="math-container">$\overline{\alpha}e \gg x$</span> (as drawn in the figure), we have <span class="math-container">$\overline{\alpha}e \succsim x$</span>&quot;:</li>
</ol>
<blockquote>
<p>This is a minor issue. Should we conclude <span class="math-container">$\overline{\alpha}e \succ x$</span>?</p>
</blockquote>
<ol start="3">
<li>&quot;The set <span class="math-container">$A^+ = \{\alpha\in\mathbb{R}_+:\alpha e \succsim x\}$</span> and <span class="math-container">$A^- = \{\alpha\in\mathbb{R}_+:x \succsim \alpha e\}$</span> are nonempty and closed.&quot;</li>
</ol>
<blockquote>
<p>I tried to prove the nonemptiness of these two sets, but couldn't figure it all out. Here is my attempt for the nonemptiness of <span class="math-container">$A^+$</span>:</p>
<blockquote>
<p>Assume to the contrary that <span class="math-container">$A^+=\emptyset$</span>. Then <span class="math-container">$x \succ \alpha e$</span> for all <span class="math-container">$\alpha \in \mathbb{R}_+$</span>. Denote <span class="math-container">$x=(x_1,\dots,x_L)$</span>. Let <span class="math-container">$\alpha'=\max\{x_1,\dots,x_L\}+1$</span>. Then <span class="math-container">$\alpha'e \succ x$</span>. This is a contradiction, so <span class="math-container">$A^+\neq\emptyset$</span>.</p>
</blockquote>
<p><em><strong>I have diffitulty proving the nonemptiness of <span class="math-container">$A^-$</span>. Someone please help me out:</strong></em></p>
<blockquote>
<p>Assume to the contrary that <span class="math-container">$A^-=\emptyset$</span>. Then <span class="math-container">$\alpha e \succ x$</span> for all <span class="math-container">$\alpha \in \mathbb{R}_+$</span>. This is certainly false if <span class="math-container">$x\gg0$</span>, because <span class="math-container">$x\succ\alpha''e$</span> where <span class="math-container">$\alpha''&lt;\min\{x_1,\dots,x_L\}$</span>. If <span class="math-container">$x=0$</span>, then <span class="math-container">$x\sim\alpha e$</span> for <span class="math-container">$\alpha=0$</span>, which shows the assumption is false. However, I cannot show a contradiction when <span class="math-container">$x\geq0$</span>. Because under its definition of monotonicity, it is perfectly possible for <span class="math-container">$0 \succ x$</span> (<span class="math-container">$\alpha=0$</span>) for <span class="math-container">$x\geq0$</span> but not <span class="math-container">$x\gg0$</span>.</p>
</blockquote>
</blockquote>
<ol start="4">
<li>Finally, this is my proof of the closedness of <span class="math-container">$A^+$</span> and <span class="math-container">$A^-$</span>. I would appreciate it if someone could help me check if it is correct.</li>
</ol>
<blockquote>
<p>Let <span class="math-container">$\alpha$</span> be a limit point of <span class="math-container">$A^+$</span>. Then there exists a sequence <span class="math-container">$\{\alpha_n\}$</span> in <span class="math-container">$A^+$</span> such that <span class="math-container">$\alpha = \lim_{n\to\infty}\alpha_n$</span> and <span class="math-container">$\alpha_n e \succsim x$</span> for all <span class="math-container">$n$</span>. Consider the sequence of pairs <span class="math-container">$\{(x,a_n e)\}_{n=1}^{\infty}$</span> with <span class="math-container">$\alpha_n e \succsim x$</span>, <span class="math-container">$x=\lim_{n\to\infty} x$</span>, and <span class="math-container">$\alpha e = \lim_{n\to\infty}\alpha_n e$</span>. By the continuity of <span class="math-container">$\succsim$</span>, we have <span class="math-container">$\alpha e \succsim x$</span>, so that <span class="math-container">$\alpha \in A^+$</span>. Hence, every limit point of <span class="math-container">$A^+$</span> is a point of <span class="math-container">$A^+$</span>, which means <span class="math-container">$A^+$</span> is closed. A similar argument would show that <span class="math-container">$A^-$</span> is closed.</p>
</blockquote>
<hr />
<p><strong>Update</strong></p>
<p>For question 1, it seems that the book was wrong. It should have said <em><strong>monotonicity and continuity imply that <span class="math-container">$x \succsim 0$</span></strong></em>.</p>

CC BY-SA 4.0
2024-04-06T14:45:12.950Z
A good related post: https://economics.stackexchange.com/questions/12403/continuous-rational-and-monotone-preference-relation-implies-x-succsim0?rq=1
Beerus
<microeconomics><mathematical-economics><preferences><decision-theory><choice-theory>
Question About Proof of Proposition 3.C.1 in MWG - Step 1
Beerus
# Question About Proof of Proposition 3.C.1 in MWG - Step 1

(Asked by: username_0 on 2024-04-06)

I have difficulties understanding the first step of the proof of Proposition 3.C.1 in MWG.

> **Proposition 3.C.1**$\quad$ *Suppose that the rational preference relation $\succsim$ on $X$ is continuous. Then there is a continuous utility function $u(x)$ that represents $\succsim$.*

> **Proof**$\quad$ For the case of $X=\mathbb{R}^L\_+$ and a monotone preference relation, there is a relatively simple and intuitive proof that we present here with the help of Figure 3.C.1. [![enter image description here](https://i.sstatic.net/fPJmO.png)](https://i.sstatic.net/fPJmO.png)
>
> Denote the diagonal ray in $\mathbb{R}^L\_+$ (the locus of vectors with all $L$ components equal) by $Z$. Let $e$ designate the $L$ vector whose elements are all equal to $1$. Then $\alpha e\in Z$ for all nonnegative scalars $\alpha\geq0$.
>
> **Note that for every $x \in \mathbb{R}^L\_+$, monotonicity implies that $x\succsim0$.** **Also note that for any $\overline{\alpha}$ such that $\overline{\alpha}e \gg x$ (as drawn in the figure), we have $\overline{\alpha}e \succsim x$.** Monotonicity and continuity can then be shown to imply that there is a unique value $\alpha(x)\in[0,\overline{\alpha}]$ such that $\alpha(x)e \sim x$.
>
> Formally, this can be shown as follows: By continuity, the upper and lower contour sets of $x$ are closed. **Hence, the set $A^+ = \{\alpha\in\mathbb{R}\_+:\alpha e \succsim x\}$ and $A^- = \{\alpha\in\mathbb{R}\_+:x \succsim \alpha e\}$ are nonempty and closed.** Note that by completeness of $\succsim$, $\mathbb{R}\_+ \subset (A^+ \bigcup A^-)$. The nonemptiness and cloedness of $A^+$ and $A^-$, along with the fact that $\mathbb{R}\_+$ is connected, imply that $A^+ \bigcap A^- \neq \emptyset$. Thus, there exists a scalar $\alpha$ such that $\alpha e \sim x$. Furthermore, by monotonicity, $\alpha\_1 e \succ \alpha\_2 e$ whenever $\alpha\_1 > \alpha\_2$. Hence, there can be at most one scalar satisfying $\alpha e \sim x$. This scalar is $\alpha(x)$.

## My Question

I highlighted the sentences where I have trouble understanding:

1. "For every $x \in \mathbb{R}^L\_+$, monotonicity implies that $x\succsim0$":

> Although, it intuitively make sense, it does not quite follow the definition of monotonicity. The book defines monotonicity as follows:
> \begin{align\*}
> \text{The preference relation $\succsim$ on $X$ is monotone if $x \in X$ and $y \gg x$ implies $y \succ x$.}
> \end{align\*}
> It doesn't say anything aout the case when $y \geq x$ but not $y \gg x$. (For a related question, please see [here](https://economics.stackexchange.com/questions/58052/understanding-the-definition-of-monotone).) I couldn't understand why this statement is true.

2. "For any $\overline{\alpha}$ such that $\overline{\alpha}e \gg x$ (as drawn in the figure), we have $\overline{\alpha}e \succsim x$":

> This is a minor issue. Should we conclude $\overline{\alpha}e \succ x$?

3. "The set $A^+ = \{\alpha\in\mathbb{R}\_+:\alpha e \succsim x\}$ and $A^- = \{\alpha\in\mathbb{R}\_+:x \succsim \alpha e\}$ are nonempty and closed."

> I tried to prove the nonemptiness of these two sets, but couldn't figure it all out. Here is my attempt for the nonemptiness of $A^+$:
>
> > Assume to the contrary that $A^+=\emptyset$. Then $x \succ \alpha e$ for all $\alpha \in \mathbb{R}\_+$. Denote $x=(x\_1,\dots,x\_L)$. Let $\alpha'=\max\{x\_1,\dots,x\_L\}+1$. Then $\alpha'e \succ x$. This is a contradiction, so $A^+\neq\emptyset$.
>
> ***I have diffitulty proving the nonemptiness of $A^-$. Someone please help me out:***
>
> > Assume to the contrary that $A^-=\emptyset$. Then $\alpha e \succ x$ for all $\alpha \in \mathbb{R}\_+$. This is certainly false if $x\gg0$, because $x\succ\alpha''e$ where $\alpha''<\min\{x\_1,\dots,x\_L\}$. If $x=0$, then $x\sim\alpha e$ for $\alpha=0$, which shows the assumption is false. However, I cannot show a contradiction when $x\geq0$. Because under its definition of monotonicity, it is perfectly possible for $0 \succ x$ ($\alpha=0$) for $x\geq0$ but not $x\gg0$.

4. Finally, this is my proof of the closedness of $A^+$ and $A^-$. I would appreciate it if someone could help me check if it is correct.

> Let $\alpha$ be a limit point of $A^+$. Then there exists a sequence $\{\alpha\_n\}$ in $A^+$ such that $\alpha = \lim\_{n\to\infty}\alpha\_n$ and $\alpha\_n e \succsim x$ for all $n$. Consider the sequence of pairs $\{(x,a\_n e)\}\_{n=1}^{\infty}$ with $\alpha\_n e \succsim x$, $x=\lim\_{n\to\infty} x$, and $\alpha e = \lim\_{n\to\infty}\alpha\_n e$. By the continuity of $\succsim$, we have $\alpha e \succsim x$, so that $\alpha \in A^+$. Hence, every limit point of $A^+$ is a point of $A^+$, which means $A^+$ is closed. A similar argument would show that $A^-$ is closed.

---

**Update**

For question 1, it seems that the book was wrong. It should have said ***monotonicity and continuity imply that $x \succsim 0$***.

**Tags:** microeconomics, mathematical-economics, preferences, decision-theory, choice-theory
> **Comments:**
> **username_0**: A good related post: https://economics.stackexchange.com/questions/12403/continuous-rational-and-monotone-preference-relation-implies-x-succsim0?rq=1  

---

## 1 answers

### Answer by username_1 on 2024-04-06. Score: 2 | Accepted answer

1. Requires continuity which is already assumed.

   > *Proof.* Consider an open ball $B\_0$ around $0$ with radius $\epsilon\_0 < x/2$ and an open ball $B\_x$ around $x$ with radius $\epsilon\_x < x/2$. Pick any $z\_x \in B\_x$ and a $z\_0 \in B\_0$ and observe that $z\_x >> z\_0 \implies z\_x \succ z\_0$. However, if $0 \succ x$, then continuity requires $z\_0 \succ z\_x$ which is not possible.
   >
   > Thus, $x \succeq 0$ for all $x \in X$ when monotonicity and continuity are both assumed.
2. You may. Either way, it's correct.
3. Correct. However, a shorter proof would have been to construct $\alpha$ (as you did) without unnecessarily turning it into a *proof by contradiction*.

   For non-emptiness of $A^{-}$, pick $\alpha = 0$. MWG's proof asusmes continuity, so we have $x \succeq 0 \ \forall \ x \in X$.
4. The part you have shown (proven) looks correct to me.


---

