<p>Here is an attempt at a direct proof:</p>

<p>Let all the assumptions hold, and further assume that $x \succcurlyeq^* y$. Then for some $B \supseteq \{x,y\}$, $C^*(B) \supseteq \{x\}$ (by definition of $\succcurlyeq^*$), which further implies $x \succcurlyeq z$ for all $z \in B$, and therefore $x \succcurlyeq y$ (by definition of $C^*$). (This shows that $\succcurlyeq^* \subseteq \succcurlyeq$; the other direction follows from looking at doubleton sets $B = \{x,y\}$).</p>

<p>What is left to show is that $C^*$ satisfies the weak axiom. So take any $B' \supseteq \{x,y\}$, such that $y \in C^*(B')$. If no such set exists, then $y$ is not revealed preferred to $x$ and we are done. If such a set exists then by definition $y \succcurlyeq z$ for all $z \in B'$. We already know $x \succcurlyeq y$ so by transativity $x \succcurlyeq z$ for all $z \in B'$ and hence $x \in C^*(B')$. Again,  $y$ is not revealed preferred to $x$ and we are done.</p>

<p>Notice, while we used transitivity, we did not use completeness. This hints at the fact that the statement can actually be generalized to pre-orders (reflexive and transitive), by allowing $C^*$ to be empty. This point is somewhat lost by appealing to the second result.</p>

CC BY-SA 3.0
2016-07-01T16:48:06.997Z
2016-07-02T02:05:57.077Z
2016-07-02T02:05:57.077Z
201p
<blockquote>
  <p>If $\succsim$ is rational, then $B \mapsto C^*(B, \succsim)$ satisfies the weak axiom, and $\succsim=\succsim^*$</p>
</blockquote>

<p>Previously in the same theorem actually, they proved the following:</p>

<blockquote>
  <p>If $C$ is a choice rule satisfying the weak axiom, then $\succsim^*$ is rational, and for all $B \in \mathcal{P}(X), C(B) = C^*(B, \succsim^*).$</p>
</blockquote>

<p>Note: $C^*(B) = C^*(B, \succsim) = \{ x \in B: \forall y \in B, x \succsim y \}$, and the weak axiom is referring to the weak axiom of revealed preference which a choice rule satisfies if $[x \succsim^* y ] \Rightarrow \neg (\exists B)[y \succ_B x]$</p>

<p>My try of the proof: </p>

<p>First suppose $B \mapsto C^*(B, \succsim)$ does not satisfy the weak axiom, then $\succsim^*$ is not rational by the earlier part of the theorem, so if we can prove $\succsim = \succsim^*$ we have a contradiction which will prove the statement.</p>

<p>As $\succsim$ is rational, then there is a utility function $u: X \to \mathbb{R}$ that represents $\succsim$, then $\{x \in B: \forall y \in B, u(x) \geq u(y) \} = C^*(B, \succsim)$ but then $\exists B$ s.t. $x,y \in B$ and since $B \mapsto C^*(B, \succsim)$ then $u(x) \in C^*(B, \succsim)$ so $x \in C(B)$ hence $\succsim = \succsim^*$</p>

CC BY-SA 3.0
2016-07-01T19:05:59.063Z
Pardon, what does $\mapsto$ denote and does $\mathcal{P}(X)$ mean power set of $X$?
Kitsune Cavalry
CC BY-SA 3.0
2016-07-01T20:03:55.920Z
$B \mapsto C^*(B, \succsim) $ denotes the function $C^*: B \to C^*(B, \succsim)$.  And yes used mathcal P, $\mathcal{P}(X)$, to denote the powerset of $X$ :)
Sunhwa
<mathematical-economics><preferences>
If $\succsim$ is rational, then $B \mapsto C^*(B, \succsim)$ satisfies the weak axiom, and $\succsim=\succsim^*$
Sunhwa
# If $\succsim$ is rational, then $B \mapsto C^*(B, \succsim)$ satisfies the weak axiom, and $\succsim=\succsim^*$

(Asked by: username_0 on 2016-05-05)

> If $\succsim$ is rational, then $B \mapsto C^\*(B, \succsim)$ satisfies the weak axiom, and $\succsim=\succsim^\*$

Previously in the same theorem actually, they proved the following:

> If $C$ is a choice rule satisfying the weak axiom, then $\succsim^\*$ is rational, and for all $B \in \mathcal{P}(X), C(B) = C^\*(B, \succsim^\*).$

Note: $C^\*(B) = C^\*(B, \succsim) = \{ x \in B: \forall y \in B, x \succsim y \}$, and the weak axiom is referring to the weak axiom of revealed preference which a choice rule satisfies if $[x \succsim^\* y ] \Rightarrow \neg (\exists B)[y \succ\_B x]$

My try of the proof:

First suppose $B \mapsto C^\*(B, \succsim)$ does not satisfy the weak axiom, then $\succsim^\*$ is not rational by the earlier part of the theorem, so if we can prove $\succsim = \succsim^\*$ we have a contradiction which will prove the statement.

As $\succsim$ is rational, then there is a utility function $u: X \to \mathbb{R}$ that represents $\succsim$, then $\{x \in B: \forall y \in B, u(x) \geq u(y) \} = C^\*(B, \succsim)$ but then $\exists B$ s.t. $x,y \in B$ and since $B \mapsto C^\*(B, \succsim)$ then $u(x) \in C^\*(B, \succsim)$ so $x \in C(B)$ hence $\succsim = \succsim^\*$

**Tags:** mathematical-economics, preferences

> **Comments:**
> **username_1**: Pardon, what does $\mapsto$ denote and does $\mathcal{P}(X)$ mean power set of $X$?  
> **username_0**: $B \mapsto C^*(B, \succsim) $ denotes the function $C^*: B \to C^*(B, \succsim)$.  And yes used mathcal P, $\mathcal{P}(X)$, to denote the powerset of $X$ :)  

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## 1 answers

### Answer by username_2 on 2016-07-01. Score: 2 | Accepted answer

Here is an attempt at a direct proof:

Let all the assumptions hold, and further assume that $x \succcurlyeq^\* y$. Then for some $B \supseteq \{x,y\}$, $C^\*(B) \supseteq \{x\}$ (by definition of $\succcurlyeq^\*$), which further implies $x \succcurlyeq z$ for all $z \in B$, and therefore $x \succcurlyeq y$ (by definition of $C^\*$). (This shows that $\succcurlyeq^\* \subseteq \succcurlyeq$; the other direction follows from looking at doubleton sets $B = \{x,y\}$).

What is left to show is that $C^\*$ satisfies the weak axiom. So take any $B' \supseteq \{x,y\}$, such that $y \in C^\*(B')$. If no such set exists, then $y$ is not revealed preferred to $x$ and we are done. If such a set exists then by definition $y \succcurlyeq z$ for all $z \in B'$. We already know $x \succcurlyeq y$ so by transativity $x \succcurlyeq z$ for all $z \in B'$ and hence $x \in C^\*(B')$. Again, $y$ is not revealed preferred to $x$ and we are done.

Notice, while we used transitivity, we did not use completeness. This hints at the fact that the statement can actually be generalized to pre-orders (reflexive and transitive), by allowing $C^\*$ to be empty. This point is somewhat lost by appealing to the second result.


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