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

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.) I couldn't understand why this statement is true.

  1. "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 $\overline{\alpha}e \succ x$?

  1. "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 $x\gg0$.

  1. 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.


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


1 Answer 1

  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.


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