<p>Assume that, towards a contradiction, that both <span class="math-container">$\succeq$</span> and <span class="math-container">$\succeq^\ast$</span> rationalise the choice function and that they are different.</p>
<p>The fact that <span class="math-container">$\succeq$</span> and <span class="math-container">$\succeq^\ast$</span> are different means that there should exist options <span class="math-container">$x, y$</span> such that <span class="math-container">$\succeq$</span> and <span class="math-container">$\succeq^\ast$</span> disagree on the preference over <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. So, for example, <span class="math-container">$x \succeq y$</span> and <span class="math-container">$y \succ^\ast x$</span>.</p>
<p>Then as <span class="math-container">$\succeq$</span> rationalises <span class="math-container">$C$</span>, it must be that
<span class="math-container">$$
x \in C(\{x,y\}). \tag{1}
$$</span>
On the other hand, as <span class="math-container">$y \succ^\ast x$</span> and <span class="math-container">$\succeq^\ast$</span> also rationalises <span class="math-container">$C$</span>, it must be that <span class="math-container">$\{y\} = C(\{x,y\})$</span> which implies that
<span class="math-container">$$
x \notin C(\{x,y\}). \tag{2}
$$</span>
We see that <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> contradict each other.</p>
<p>Other cases lead to similar contradictions.</p>

CC BY-SA 4.0
2021-10-09T12:38:19.673Z
Wonderful. This contradiction approach makes so much more sense. Any idea on what the author was trying to get at by his way of doing it?
Rumi
CC BY-SA 4.0
2021-10-09T12:40:18.503Z
@Rumi As your choice over any binary set $\{x,y\}$ determines whether $x$ is better than $y$, $y$ is better than $x$ or the two are indifferent, we have that all preferences  over two alternatives are determined by the binary choice sets. This gives a unique preference order over all alternatives.
tdm
CC BY-SA 4.0
2021-10-09T12:52:39.970Z
I got the preference order part down but I was having a hard time imagining how it would be unique. Thanks for the contradiction it really helps!
Rumi
CC BY-SA 4.0
2021-10-09T12:27:54.100Z
2021-10-09T12:27:54.100Z
tdm
<p>Going through a proof in Mas Colell and I am not understanding how (iii) shows uniqueness of the rationalizing preference relation. I understand that well <span class="math-container">$\beta$</span> is the power set so it contains all pairwise elements, and <span class="math-container">$C(.)$</span> gives us the preference for these pairs, but how is this implying uniqueness of the rational preference relation.</p>
<p>In other words, how does this rule out the possibility that no other rational preference relation exists? Could we start off by assuming some other rational preference relation exists and reach a contradiction, which would imply uniqueness?</p>
<p><a href="https://i.sstatic.net/OZtEU.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/OZtEU.png" alt="Mas Collel" /></a></p>

CC BY-SA 4.0
2021-10-09T12:53:40.447Z
Hi please do not include pictures of the text in a question. It is against our rules laid out in the help center to post pictures of texts or equations
1muflon1
<microeconomics><preferences><proof>
How do you establish uniqueness of a rational preference relation?
Rumi
# How do you establish uniqueness of a rational preference relation?

(Asked by: username_0 on 2021-10-09)

Going through a proof in Mas Colell and I am not understanding how (iii) shows uniqueness of the rationalizing preference relation. I understand that well $\beta$ is the power set so it contains all pairwise elements, and $C(.)$ gives us the preference for these pairs, but how is this implying uniqueness of the rational preference relation.

In other words, how does this rule out the possibility that no other rational preference relation exists? Could we start off by assuming some other rational preference relation exists and reach a contradiction, which would imply uniqueness?

[![Mas Collel](https://i.sstatic.net/OZtEU.png)](https://i.sstatic.net/OZtEU.png)

**Tags:** microeconomics, preferences, proof
> **Comments:**
> **username_1**: Hi please do not include pictures of the text in a question. It is against our rules laid out in the help center to post pictures of texts or equations  

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## 1 answers

### Answer by username_2 on 2021-10-09. Score: 3 | Accepted answer

Assume that, towards a contradiction, that both $\succeq$ and $\succeq^\ast$ rationalise the choice function and that they are different.

The fact that $\succeq$ and $\succeq^\ast$ are different means that there should exist options $x, y$ such that $\succeq$ and $\succeq^\ast$ disagree on the preference over $x$ and $y$. So, for example, $x \succeq y$ and $y \succ^\ast x$.

Then as $\succeq$ rationalises $C$, it must be that
$$
x \in C(\{x,y\}). \tag{1}
$$
On the other hand, as $y \succ^\ast x$ and $\succeq^\ast$ also rationalises $C$, it must be that $\{y\} = C(\{x,y\})$ which implies that
$$
x \notin C(\{x,y\}). \tag{2}
$$
We see that $(1)$ and $(2)$ contradict each other.

Other cases lead to similar contradictions.

> **Comments:**
> **username_0**: Wonderful. This contradiction approach makes so much more sense. Any idea on what the author was trying to get at by his way of doing it?  
> **username_2**: @Rumi As your choice over any binary set $\{x,y\}$ determines whether $x$ is better than $y$, $y$ is better than $x$ or the two are indifferent, we have that all preferences  over two alternatives are determined by the binary choice sets. This gives a unique preference order over all alternatives.  
> **username_0**: I got the preference order part down but I was having a hard time imagining how it would be unique. Thanks for the contradiction it really helps!  

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