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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

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  • $\begingroup$ 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 $\endgroup$
    – 1muflon1
    Oct 9, 2021 at 12:53

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Rumi
    Oct 9, 2021 at 12:38
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    $\begingroup$ @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. $\endgroup$
    – tdm
    Oct 9, 2021 at 12:40
  • $\begingroup$ 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! $\endgroup$
    – Rumi
    Oct 9, 2021 at 12:52

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