<p>This is a small issue, but it confuses me.</p>
<p>In the proof of Proposition 3.C.1 of <em>Microeconomic Theory</em> by Mas-Colell et al, the book said</p>
<blockquote>
<p>... <span class="math-container">$\mathbb{R}_+\subset (A^+\bigcup A^-)$</span>. The nonemptiness and closedness 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>.</p>
</blockquote>
<p>My question is that I don't think <span class="math-container">$\mathbb{R}_+$</span> be a <em><strong>subset</strong></em> of <span class="math-container">$(A^+\bigcup A^-)$</span> would be sufficient to draw the conclusion.  Shouldn't the book have, instead, stated <span class="math-container">$\mathbb{R}_+ = (A^+\bigcup A^-)$</span>?</p>
<hr />
<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>Note that for every <span class="math-container">$x \in \mathbb{R}^L_+$</span>, monotonicity implies that <span class="math-container">$x\succsim0$</span>. Also note that for any <span class="math-container">$\bar{\alpha}$</span> such that <span class="math-container">$\bar{\alpha}e \gg x$</span> (as drawn in the figure), we have <span class="math-container">$\bar{\alpha}e \succsim x$</span>. Monotonicity and continuity can then be shown to imply that there is a unique value <span class="math-container">$\alpha(x)\in[0,\bar{\alpha}]$</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. 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">$\mathbf{\succsim}$</span>, <span class="math-container">$\mathbf{\mathbb{R}_+ \subset (A^+ \bigcup A^-)}$</span>. The nonemptiness and cloedness of <span class="math-container">$\mathbf{A^+}$</span> and <span class="math-container">$\mathbf{A^-}$</span>, along with the fact that <span class="math-container">$\mathbf{\mathbb{R}_+}$</span> is connected, imply that <span class="math-container">$\mathbf{A^+ \bigcap A^- \neq \emptyset}$</span>.</strong> 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>
<hr />
<p><strong>Reference</strong> <em>Microeconomic Theory</em> by Andreu Mas-Colell, Michael D. Whinston, and Jerry R. Green.</p>

CC BY-SA 4.0
2024-05-12T21:41:22.830Z
Are you familiar with the notion of connectedness in general topological spaces?
Michael Greinecker
CC BY-SA 4.0
2024-05-12T23:21:32.907Z
@MichaelGreinecker Honestly no. I learned the concept of connectedness from baby Rudin, which says the following: Two subsets $A$ and $B$ of a metric space $X$ are said to be *separated* if both $A\bigcap \overline{B}$ and $\overline{A}\bigcap B$ are empty. A set $E\subset X$ is said to be *connected* if $E$ is *not* a union of two nonempty separated sets.
Champa
CC BY-SA 4.0
2024-05-12T23:29:49.033Z
Okay, one can prove that a set is connected if it is not a subset of the disjoint union of two closed sets (or open sets, here equivalent) that both intersect the set. Then the argument in MWG goes through. Of course, in that context both sets are subsets of $\mathbb{R}_+$, so equality holds. Either argument works. There is no mistake here in MWG.
Michael Greinecker
CC BY-SA 4.0
2024-05-13T00:14:41.350Z
@MichaelGreinecker I'm still confused. Consider this example: Let $A=[0,1]$, $B=[2,3]$, and $C=A=[0,1]$. Clearly, $A$ and $B$ are nonempty and closed,  $C$ is connected, and $C\subset A\bigcup B$. But $A\bigcap B=\emptyset$. This is why I think MWG's  argument is wrong.
Champa
CC BY-SA 4.0
2024-05-13T00:17:22.960Z
But here, it is not the case that both $A$ and $B$ intersect $C$.
Michael Greinecker
CC BY-SA 4.0
2024-05-13T00:23:52.523Z
Yes, you are right. But that is essentially what MWG said. So I think what I wanted to say is that MWG is not rigorous here. They should perhaps at least mention something like $A^+\bigcap\mathbb{R}_+\neq\emptyset$ and $A^-\bigcap\mathbb{R}_+\neq\emptyset$. (Or they could change $\subset$ to $=$.) Because if we only take what they said, we cannot draw the conclusion that $A^+\bigcap A^-\neq\emptyset$.
Champa
CC BY-SA 4.0
2024-05-13T00:26:42.957Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/153118/discussion-between-champa-and-michael-greinecker).
Champa
CC BY-SA 4.0
2024-05-13T06:53:51.133Z
Could you include an explicit reference to the book (authors and title) by editing your post?
Richard Hardy
CC BY-SA 4.0
2024-05-13T12:56:51.197Z
@RichardHardy Yes, I added a reference.
Champa
<microeconomics><mathematical-economics><utility><preferences><theory>
Question About the Step Involving the Connectedness of $\mathbb{R}_+$ in the Proof of Proposition 3.C.1 from MWG
Champa
# Question About the Step Involving the Connectedness of $\mathbb{R}_+$ in the Proof of Proposition 3.C.1 from MWG

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

This is a small issue, but it confuses me.

In the proof of Proposition 3.C.1 of *Microeconomic Theory* by Mas-Colell et al, the book said

> ... $\mathbb{R}\_+\subset (A^+\bigcup A^-)$. The nonemptiness and closedness of $A^+$ and $A^-$, along with the fact that $\mathbb{R}^+$ is connected, imply that $A^+\bigcap A^-\neq\emptyset$.

My question is that I don't think $\mathbb{R}\_+$ be a ***subset*** of $(A^+\bigcup A^-)$ would be sufficient to draw the conclusion. Shouldn't the book have, instead, stated $\mathbb{R}\_+ = (A^+\bigcup A^-)$?

---

> **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 $\bar{\alpha}$ such that $\bar{\alpha}e \gg x$ (as drawn in the figure), we have $\bar{\alpha}e \succsim x$. Monotonicity and continuity can then be shown to imply that there is a unique value $\alpha(x)\in[0,\bar{\alpha}]$ 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 $\mathbf{\succsim}$, $\mathbf{\mathbb{R}\_+ \subset (A^+ \bigcup A^-)}$. The nonemptiness and cloedness of $\mathbf{A^+}$ and $\mathbf{A^-}$, along with the fact that $\mathbf{\mathbb{R}\_+}$ is connected, imply that $\mathbf{A^+ \bigcap A^- \neq \emptyset}$.** 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 $\alpha(x)$.

---

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

**Tags:** microeconomics, mathematical-economics, utility, preferences, theory
> **Comments:**
> **username_1**: Are you familiar with the notion of connectedness in general topological spaces?  
> **username_0**: @MichaelGreinecker Honestly no. I learned the concept of connectedness from baby Rudin, which says the following: Two subsets $A$ and $B$ of a metric space $X$ are said to be *separated* if both $A\bigcap \overline{B}$ and $\overline{A}\bigcap B$ are empty. A set $E\subset X$ is said to be *connected* if $E$ is *not* a union of two nonempty separated sets.  
> **username_1**: Okay, one can prove that a set is connected if it is not a subset of the disjoint union of two closed sets (or open sets, here equivalent) that both intersect the set. Then the argument in MWG goes through. Of course, in that context both sets are subsets of $\mathbb{R}_+$, so equality holds. Either argument works. There is no mistake here in MWG.  
> **username_0**: @MichaelGreinecker I'm still confused. Consider this example: Let $A=[0,1]$, $B=[2,3]$, and $C=A=[0,1]$. Clearly, $A$ and $B$ are nonempty and closed,  $C$ is connected, and $C\subset A\bigcup B$. But $A\bigcap B=\emptyset$. This is why I think MWG's  argument is wrong.  
> **username_1**: But here, it is not the case that both $A$ and $B$ intersect $C$.  
> **username_0**: Yes, you are right. But that is essentially what MWG said. So I think what I wanted to say is that MWG is not rigorous here. They should perhaps at least mention something like $A^+\bigcap\mathbb{R}_+\neq\emptyset$ and $A^-\bigcap\mathbb{R}_+\neq\emptyset$. (Or they could change $\subset$ to $=$.) Because if we only take what they said, we cannot draw the conclusion that $A^+\bigcap A^-\neq\emptyset$.  
> **username_0**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/153118/discussion-between-champa-and-michael-greinecker).  
> **username_2**: Could you include an explicit reference to the book (authors and title) by editing your post?  
> **username_0**: @RichardHardy Yes, I added a reference.  

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## 0 answers

