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

Question

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, $\alpha(x)\geq\alpha(y)\iff x\succsim y$.

I have two questions about this second paragraph (as highlighted in bold).

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 $y\succsim x$.

2. "...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 $x\geq y$.

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

Answer 1

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 $\neg y\gg x$.