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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

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.

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  • $\begingroup$ Are you familiar with the notion of connectedness in general topological spaces? $\endgroup$
    – Michael Greinecker
    Commented May 12 at 21:41
  • $\begingroup$ @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. $\endgroup$
    – Champa
    Commented May 12 at 23:21
  • $\begingroup$ 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. $\endgroup$
    – Michael Greinecker
    Commented May 12 at 23:29
  • $\begingroup$ @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. $\endgroup$
    – Champa
    Commented May 13 at 0:14
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    $\begingroup$ Could you include an explicit reference to the book (authors and title) by editing your post? $\endgroup$
    – Richard Hardy
    Commented May 13 at 6:53

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