# Question About Proof of Proposition 3.C.1 in MWG - Step 1

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.

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.

Update

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

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.