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In Microeconomic Theory by Mas-Colell, Whinston, and Green, the definition of monotone preference relations is given as follows:

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

So suppose that $\succsim$ is monotone. I was wondering if it is possible for two bundles $x$ and $y$ to be such that $y \geq x$ but $x \succ y$?

I'm not sure about this, but I think that the above statement is not possible; that is, I think:

If $\succsim$ is monotone, then $y \succsim x$ whenever $y \geq x$.

However, I couldn't prove it. Could someone please help me out? Thanks a lot in advance!


Note that $x \geq y$ means $x_n \geq y_n$ for all $n=1,\dots,N$ and that $x \gg y$ means $x_n > y_n$ for all $n=1,\dots,N$.

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  • $\begingroup$ I think this works. Why can't my preferences be defined by a preference for less? That is, $U(x) > U(y)$ wherever $x<y$ such that $x\succ y$ whenever $y > x$. I was never under the impression that it was necessarily an increasing relation. Preference relations can be monotonically increasing or decreasing. Not sure how to go about directly proving this though, I would just suppose the preference relation I defined and then prove it by contradiction. $\endgroup$
    – Brennan
    Commented Apr 6 at 1:31
  • $\begingroup$ @Brennan But that's bizarre. If you take a look at my first question in this post "economics.stackexchange.com/questions/58053/… =", it should indicate that "$y \succsim x$ whenever $y \geq x$". $\endgroup$
    – Beerus
    Commented Apr 6 at 2:51
  • $\begingroup$ I don't think it's that bizarre. My preference relation is monotonically increasing in less suffering. Or it's monotonically decreasing in suffering. I dont see how monotonicity is violated if I just say I prefer less of something in a monotonic way. $\endgroup$
    – Brennan
    Commented Apr 6 at 16:00

1 Answer 1

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Throughout my answer, I assume that $x \succ y$ is read as "$x$ is strictly preferred to $y$". Refer to footnote (a) if that is not the case.

I was wondering if it is possible for two bundles $x$ and $y$ to be such that $y \geq x$ but $x \succ y$?

  1. Counterexample: $X = \{(1,1), (2,2), (3,1)\}$ such that $(2,2) \succ (1,1) \succ (3,1)$. Completeness, transitivity and monotonicity are satisfied. Now pick $x = (3,1)$ and $y = (1,1)$.

  2. With the assumption that $X \cong \mathbb{R}^m \ (m \in \mathbb{N})$ and continuity of $\succeq_X$, we have $y \geq x \implies y \succeq x$.

    Proof. Suppose $y \geq x$ for some $x,y \in X$ and assume for the sake of contradiction that $x \succ y$. Define $Z := \{z \in X : z >> y\}$. Consider any two open balls $B_x$ and $B_y$ and observe that $z_y \in B_y \cap Z$ must satisfy $x \succ z_y$ (due to continuity and our assumption that $x \succ y$) and $z_y \succ x$ (due to monotonicity) simultaneously. Contradiction!


(a) If $x \succ y$ is read as "$x$ is weakly preferred to $y$", consider $U(X) = 1$ and see that $(y \geq x) \land (x \succ y)$ holds for all $x,y \in X$ satisfying $y \geq x$.

(b) Definition of continuity of preferences: A preference relation $\succeq$ on $X$ is continuous if whenever $a \succ b$, $\exists$ open balls $B_a$ and $B_b$ such that for all $x \in B_a$ and $y \in B_b$, $x \succ y$.

(c) Note: I would appreciate an answer that has a weaker condition on $X$ compared to $X$ being isomorphic to $\mathbb{R}^m$ $(m \in \mathbb{N})$. Having said that, the above proof works for $X = \mathbb{R}^m_{\geq 0}$ $(m \in \mathbb{N})$ which makes more sense from the economic point.

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  • $\begingroup$ Correctme if I'm wrong. For your definition of continuity, I feel it should be: $a \succ b$ whenever there are open balls $B_a$ and $B_b$ such that $x \succ y$ for all $x \in B_a$ and $y \in B_b$. $\endgroup$
    – Beerus
    Commented Apr 6 at 14:51
  • $\begingroup$ @Beerus No, your definition is incorrect. Counterexample: Pick $X = \mathbb{R}^2, a = (2,2), b = (1,1)$. For $\epsilon = 0.1$, there are open balls $B(a,\epsilon)$ and $B(b,\epsilon)$ such that $x \succ y$ for all $x \in B(a,\epsilon)$ and $y \in B(y,\epsilon)$, and also $a \succ b$. However, lexicographic preferences not continuous. I took the definition from Ariel Rubinstein's Lecture Notes in Micro Theory, Lecture 2 (Utility), Page 16, Definition C1. $\endgroup$ Commented Apr 6 at 16:58
  • $\begingroup$ Oh, I see. Yes you are right! $\endgroup$
    – Beerus
    Commented Apr 6 at 17:34

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