<p>Throughout my answer, I assume that <span class="math-container">$x \succ y$</span> is read as &quot;<span class="math-container">$x$</span> is <strong>strictly</strong> preferred to <span class="math-container">$y$</span>&quot;. Refer to footnote (a) if that is not the case.</p>
<blockquote>
<p>I was wondering if it is possible for two bundles <span class="math-container">$x$</span> and <span class="math-container">$y$</span> to be such that <span class="math-container">$y \geq x$</span> but <span class="math-container">$x \succ y$</span>?</p>
</blockquote>
<ol>
<li><p>Counterexample: <span class="math-container">$X = \{(1,1), (2,2), (3,1)\}$</span> such that <span class="math-container">$(2,2) \succ
    (1,1) \succ (3,1)$</span>. Completeness, transitivity and monotonicity are
satisfied. Now pick <span class="math-container">$x = (3,1)$</span> and <span class="math-container">$y = (1,1)$</span>.</p>
</li>
<li><p><strong>With the assumption that <span class="math-container">$X \cong \mathbb{R}^m \ (m \in \mathbb{N})$</span> and continuity of <span class="math-container">$\succeq_X$</span>, we have <span class="math-container">$y \geq x
\implies y \succeq x$</span>.</strong></p>
<p><em><strong>Proof.</strong></em> Suppose <span class="math-container">$y \geq x$</span> for some <span class="math-container">$x,y \in X$</span> and assume for the sake of contradiction that <span class="math-container">$x \succ y$</span>. Define <span class="math-container">$Z := \{z \in X : z &gt;&gt; y\}$</span>. Consider any two open balls <span class="math-container">$B_x$</span> and <span class="math-container">$B_y$</span> and observe that <span class="math-container">$z_y \in B_y \cap Z$</span> must satisfy <span class="math-container">$x \succ z_y$</span> (due to continuity and our assumption that <span class="math-container">$x \succ y$</span>) and <span class="math-container">$z_y \succ x$</span> (due to monotonicity) simultaneously. Contradiction!</p>
</li>
</ol>
<hr />
<p><sub>(a) If <span class="math-container">$x \succ y$</span> is read as &quot;<span class="math-container">$x$</span> is <strong>weakly</strong> preferred to <span class="math-container">$y$</span>&quot;, consider <span class="math-container">$U(X) = 1$</span> and see that <span class="math-container">$(y \geq x) \land (x \succ y)$</span> holds for all <span class="math-container">$x,y \in X$</span> satisfying <span class="math-container">$y \geq x$</span>.</sub></p>
<p><sub>(b) <strong>Definition of continuity of preferences:</strong> A preference relation <span class="math-container">$\succeq$</span> on <span class="math-container">$X$</span> is continuous if whenever <span class="math-container">$a \succ b$</span>, <span class="math-container">$\exists$</span> open balls <span class="math-container">$B_a$</span> and <span class="math-container">$B_b$</span> such that for all <span class="math-container">$x \in B_a$</span> and <span class="math-container">$y \in B_b$</span>, <span class="math-container">$x \succ y$</span>.</sub></p>
<p><sub>(c) <strong>Note:</strong> I would appreciate an answer that has a weaker condition on <span class="math-container">$X$</span> compared to <span class="math-container">$X$</span> being isomorphic to <span class="math-container">$\mathbb{R}^m$</span> <span class="math-container">$(m \in \mathbb{N})$</span>. Having said that, the above proof works for <span class="math-container">$X = \mathbb{R}^m_{\geq 0}$</span> <span class="math-container">$(m \in \mathbb{N})$</span> which makes more sense from the economic point.</sub></p>

CC BY-SA 4.0
2024-04-06T14:51:47.420Z
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 $y \in B_b$*.
Beerus
CC BY-SA 4.0
2024-04-06T16:58:05.173Z
@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](https://economics.stackexchange.com/questions/56370/lexicographic-preference-relation-not-continuous). I took the definition from Ariel Rubinstein's Lecture Notes in Micro Theory, Lecture 2 (Utility), Page 16, Definition C1.
uninterestedacademic
CC BY-SA 4.0
2024-04-06T17:34:30.990Z
Oh, I see. Yes you are right!
Beerus
CC BY-SA 4.0
2024-04-06T04:09:51.730Z
2024-04-06T09:12:42.043Z
2024-04-06T09:12:42.043Z
uninterestedacademic
<p>In <em>Microeconomic Theory</em> by Mas-Colell, Whinston, and Green, the definition of monotone preference relations is given as follows:</p>
<blockquote>
<p><strong>Definition 3.B.2</strong><span class="math-container">$\quad$</span> The preference relation <span class="math-container">$\succsim$</span> on <span class="math-container">$X$</span> is <em>monotone</em> if <span class="math-container">$x \in X$</span> and <span class="math-container">$y \gg x$</span> implies <span class="math-container">$y \succ x$</span>.</p>
</blockquote>
<p>So suppose that <span class="math-container">$\succsim$</span> is monotone. I was wondering if it is possible for two bundles <span class="math-container">$x$</span> and <span class="math-container">$y$</span> to be such that <span class="math-container">$y \geq x$</span> but <span class="math-container">$x \succ y$</span>?</p>
<p>I'm not sure about this, but I think that the above statement is not possible; that is, I think:</p>
<blockquote>
<p>If <span class="math-container">$\succsim$</span> is monotone, then <span class="math-container">$y \succsim x$</span> whenever <span class="math-container">$y \geq x$</span>.</p>
</blockquote>
<p>However, I couldn't prove it. Could someone please help me out? Thanks a lot in advance!</p>
<hr />
<p>Note that <span class="math-container">$x \geq y$</span> means <span class="math-container">$x_n \geq y_n$</span> for all <span class="math-container">$n=1,\dots,N$</span> and that <span class="math-container">$x \gg y$</span> means <span class="math-container">$x_n &gt; y_n$</span> for all <span class="math-container">$n=1,\dots,N$</span>.</p>

CC BY-SA 4.0
2024-04-06T01:31:23.210Z
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.
Brennan
CC BY-SA 4.0
2024-04-06T02:51:30.563Z
@Brennan But that's bizarre. If you take a look at my first question in this post "https://economics.stackexchange.com/questions/58053/question-about-proof-of-proposition-3-c-1-in-mwg-step-1?noredirect=1&lq=1 =", it should indicate that "$y \succsim x$ whenever $y \geq x$".
Beerus
CC BY-SA 4.0
2024-04-06T16:00:11.527Z
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.
Brennan
<microeconomics><mathematical-economics><preferences><decision-theory><choice-theory>
Understanding the definition of monotone
Beerus
# Understanding the definition of monotone

(Asked by: username_0 on 2024-04-05)

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 $n=1,\dots,N$.

**Tags:** microeconomics, mathematical-economics, preferences, decision-theory, choice-theory
> **Comments:**
> **username_1**: 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.  
> **username_0**: @Brennan But that's bizarre. If you take a look at my first question in this post "https://economics.stackexchange.com/questions/58053/question-about-proof-of-proposition-3-c-1-in-mwg-step-1?noredirect=1&lq=1 =", it should indicate that "$y \succsim x$ whenever $y \geq x$".  
> **username_1**: 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.  

---

## 1 answers

### Answer by username_2 on 2024-04-06. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_0**: 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 $y \in B_b$*.  
> **username_2**: @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](https://economics.stackexchange.com/questions/56370/lexicographic-preference-relation-not-continuous). I took the definition from Ariel Rubinstein's Lecture Notes in Micro Theory, Lecture 2 (Utility), Page 16, Definition C1.  
> **username_0**: Oh, I see. Yes you are right!  

---

