# Question About the Step Proving $\alpha(x)$ Represents Preferences in the Proof of Proposition 3.C.1 from MWG

Continue from this question, the book Microeconomic Theory by Mas-Colell et al. said

We now take $$\alpha(x)$$ as our utility function; that is, we assign a utility value $$u(x)=\alpha(x)$$ to every $$x$$. This utility level is also depicted in Figure 3.C.1. We need to check two properties of this function: that it represents the preference $$\succsim$$ (i.e., that $$\alpha(x)\geq\alpha(y)$$ if and only if $$x\succsim y$$) and that it is a continuous function.

That $$\alpha(x)$$ represents preferences follows from its construction. Formally, suppose first that $$\alpha(x)\geq\alpha(y)$$. By monotonicity, this implies that $$\mathbf{\alpha(x)e\succsim\alpha(y)e}$$. Since $$x\sim\alpha(x)e$$ and $$y\sim\alpha(y)e$$, we have $$x\succsim y$$. Suppose, on the other hand, that $$x\succsim y$$. Then $$\alpha(x)e\sim x\succsim y\sim\alpha(y)e$$; and so by monotonicity, we must have $$\mathbf{\alpha(x)\geq\alpha(y)}$$. Hence, $$\alpha(x)\geq\alpha(y)\iff x\succsim y$$.

1. I don't think monotonicity is sufficient to draw the conclusion. We also need continuity, right? See this post and this post for a related issue as well as the following lemma:

Lemma$$\quad$$ For a monotone, continuous, and rational preference relation $$\succsim$$ on $$X=\mathbb{R}^L_+$$, $$y\geq x$$ implies $$y\succsim x$$.

1. "...by monotonicity, we must have $$\mathbf{\alpha(x)\geq\alpha(y)}$$." Does this follow from the contrapositive of the definition of monotonicity?

Definition$$\quad$$ The preference relation $$\succsim$$ on $$X$$ is monotone if $$x\in X$$ and $$y\gg x$$ implies $$y\succ x$$.

of which the contrapositive is

The preference relation $$\succsim$$ on $$X$$ is monotone if $$x\succsim y$$ implies $$x\geq y$$.

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

• Could you include an explicit reference to the book (authors and title) by editing your post? Commented May 13 at 6:53
• @RichardHardy Yes, I added a reference. Commented May 13 at 12:58
• Your contrapositive is not correct. It can be the case that neither $y\gg x$ nor $x\geq y$ holds. For example, if $y=(0,1)$ and $x=(0,0)$. Commented May 13 at 13:29
• @MichaelGreinecker So do we need to prove it by contradiction: Assume to the contrary that $\alpha(y)>\alpha(x)$. Then $\alpha(y)e\gg\alpha(x)e$ and so $y\sim\alpha(y)e\succ\alpha(x)e\sim x$, a contradiction. Commented May 13 at 14:21
• Yes, that should work. Commented May 13 at 17:13

Def 3.B.2: The preference relation $$\succsim$$ on $$X$$ is monotone if $$x\in X$$ and $$y\gg x$$ implies $$y\succ x$$.

[1] If $$\alpha(x)\geq \alpha(y)$$, then there are two possibilities:

(i) $$\alpha(x)>\alpha(y)$$. In this case, $$\alpha(x)e=(\alpha(x),\alpha(x),\ldots,\alpha(x))\gg (\alpha(y),\alpha(y),\ldots,\alpha(y))=\alpha(y)e$$. So, by monotonicity, $$\alpha(x)e\succ\alpha(y)e$$. Therefore, $$\alpha(x)e\succsim\alpha(y)e$$.

(ii) $$\alpha(x)=\alpha(y)$$. In this case, $$\alpha(x)e=(\alpha(x),\alpha(x),\ldots,\alpha(x))= (\alpha(y),\alpha(y),\ldots,\alpha(y))=\alpha(y)e$$. So by completeness (reflexivity), $$\alpha(x)e\succsim\alpha(y)e$$

So, we don't need continuity to show that if $$\alpha(x)\geq \alpha(y)$$ holds then $$\alpha(x)e\succsim\alpha(y)e$$.

[2] If $$\alpha(x)e\succsim \alpha(y)e$$, then either $$\alpha(x) > \alpha(y)$$ or $$\alpha(x) = \alpha(y)$$. This is because if that is not the case and we have $$\alpha(x) < \alpha(y)$$ for some $$x,y$$, then by monotonicity, $$\alpha(y)e\succ\alpha(x)e$$, contradicting $$\alpha(x)e\succsim \alpha(y)e$$.

Contrapositive of monotonicity: The preference relation $$\succsim$$ on $$X$$ is monotone if $$x\succsim y$$ then either $$x\notin X$$ or $$\neg y\gg x$$.

• Thanks a lot! About question 1, I forgot that $\alpha(\cdot)\in\mathbb{R}_+$ instead of $\mathbb{R}^L_+$. Commented May 13 at 17:19
• (Please correct me if I'm wrong.) After a second thought, I think your (ii) for question [1] depend on the lemma from the original question, and thus continuity. Completeness doesn't say anything about $\gg$ or $\leq$ relations. Commented May 13 at 21:47
• It follows from completeness of $\succsim$ that $\succsim$ is reflexive, and therefore, $\forall x\in X$, $x\succsim x$ holds. Since $\alpha(x)=\alpha(y)$ implies that $\alpha(x)e=\alpha(y)e$ (i.e. both are the same bundle), so by reflexivity, $\alpha(x)e\succsim \alpha(x)e$ which is just another way of writing $\alpha(x)e\succsim \alpha(y)e$
– Amit
Commented May 14 at 3:48