<p>The English translation of <span class="math-container">$c_u(A)=\{x\in A:u(x)\ge u(y) \text{ for all }y\in A\}$</span> is </p>

<blockquote>
  <p><span class="math-container">$c_u(A)$</span> is a subset of elements in the set <span class="math-container">$A$</span> that satisfies the following condition: if an element is in <span class="math-container">$c_u(A)$</span> then <strong>this element must generate a utility no lower than any other element in the set <span class="math-container">$A$</span></strong>. </p>
</blockquote>

<p>In other words, <span class="math-container">$c_u(A)$</span> contains the decision maker's most-preferred element(s) in <span class="math-container">$A$</span>. </p>

<p>Note that "<span class="math-container">$x$</span>" in the definition is <em>any</em> element in <span class="math-container">$A$</span> that satisfies the bolded condition above, and "<span class="math-container">$y$</span>" is <em>any</em> element in <span class="math-container">$A$</span> (without further restrictions). I emphasize the word <em>any</em> to highlight the fact that <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are merely placeholders and therefore do not refer to specific elements in either set. In fact, we may well have defined <span class="math-container">$c_u(A)$</span> as <span class="math-container">$\{y\in A: u(y)\ge u(x)  \text{ for all }x\in A\}$</span> and the interpretation/English translation will be exactly the same (i.e. verbatim) as in the quote-block above. </p>

<hr>

<p>Let's work through an example. Let there be three possible alternatives, denoted as follows:
<span class="math-container">\begin{equation}
a_1=\text{one \$5 bill}, \qquad 
a_2=\text{one \$10 bill}, \qquad 
a_3=\text{two \$5 bills}. 
\end{equation}</span>
Further suppose that the decision maker cares only about the total amount of money in each alternative, and thus 
<span class="math-container">\begin{equation}
u(a_2)=u(a_3)&gt;u(a_1).
\end{equation}</span></p>

<h2>Example 1.</h2>

<p>Let <span class="math-container">$A=\{a_1,a_2,a_3\}$</span> and let's forget about <span class="math-container">$B$</span> for now. </p>

<p>"<strong>Suppose <span class="math-container">$x,y\in A$</span> and <span class="math-container">$x\in c_u(A)$</span>.</strong>" The first part of the sentence (before "and") suggests that both <span class="math-container">$x$</span> and <span class="math-container">$y$</span> (again, placeholders here) can be either <span class="math-container">$a_1$</span>, <span class="math-container">$a_2$</span>, or <span class="math-container">$a_3$</span>, but the second part of the sentence (after "and") rules out <span class="math-container">$x=a_1$</span> since <span class="math-container">$a_1$</span> is not one of the most preferred alternatives.</p>

<p>"<strong>Then <span class="math-container">$u(x)\ge u(y)$</span>.</strong>" This follows from <span class="math-container">$x\in c_u(A)$</span>, since <span class="math-container">$x$</span> is either <span class="math-container">$a_2$</span> or <span class="math-container">$a_3$</span> and <span class="math-container">$y$</span> is either <span class="math-container">$a_1$</span>, <span class="math-container">$a_2$</span>, or <span class="math-container">$a_3$</span>.</p>

<p>"<strong>If <span class="math-container">$y\notin c_u(A)$</span>, then <span class="math-container">$u(z)&gt;u(y)$</span> for some <span class="math-container">$z\in A$</span>.</strong>" The "if" part of this sentence restricts <span class="math-container">$y$</span> to be only <span class="math-container">$a_1$</span>, because if <span class="math-container">$y$</span> were either <span class="math-container">$a_2$</span> or <span class="math-container">$a_3$</span>, it would have been included in the set <span class="math-container">$c_u(A)$</span>. Consequently, since <span class="math-container">$y$</span> is not one of the most preferred alternatives, some element in <span class="math-container">$A$</span> must be strictly preferred to <span class="math-container">$y$</span>, and we know that this "some element" must be either <span class="math-container">$a_2$</span> or <span class="math-container">$a_3$</span>. [The textbook uses the variable <span class="math-container">$z$</span> instead of <span class="math-container">$x$</span> because it wants to allow for the possibility that <span class="math-container">$z\in A$</span> but <span class="math-container">$z\notin B$</span>. Otherwise, it would have used <span class="math-container">$x$</span> instead.] </p>

<p>The rest of the proof should be straightforward from here on. </p>

<h2>Example 2.</h2>

<p>Let <span class="math-container">$A=\{a_1,a_2\}$</span>. I'll leave you to verify that in this example, the textbook's proof works in the same way without loss of generality. </p>

CC BY-SA 4.0
2019-03-30T17:57:51.183Z
Your examples make things much clearer for me. However, how do we know $x \in c_u (B)$? Is $A$ in $x \in c_u (A)$ in the first sentence of the proof just a placeholder for whatever set we are looking at? In other words, once we arrive at the last sentence, we immediately know that $x \in c_u (B)$ because the first sentence said that $x \in c_u (A)$? If not, I am not sure how we know based on the sentences preceding the last sentence that $x \in c_u (B)$.
Ryan da Silva
CC BY-SA 4.0
2019-03-31T06:07:44.473Z
@RyandaSilva: The proof does not claim that $x\in c_u(B)$; it simply claims that $y\notin c_u(B)$.
Herr K.
CC BY-SA 4.0
2019-03-31T06:12:20.317Z
@RyandaSilva: By definition, $c_u(A)$ contains the most preferred elements in set $A$, and $c_u(B)$ contains the most preferred elements in set $B$. We're given 3 conditions regarding two elements $x,y$ that simultaneously belong to both $A$ and $B$: i) $x$ and $y$ are in both sets $A$ and $B$; ii) $x\in c_u(A)$, namely, $x$ is one of the most preferred elements in $A$; and iii) $y\notin c_u(A)$, namely, $y$ is not one of the most preferred elements in $A$. The last two conditions imply that $x$ is strictly preferred to $y$.
Herr K.
CC BY-SA 4.0
2019-03-31T06:15:26.903Z
When the choice context is defined by $A$, $x$ would be chosen over $y$. When the context changes to $B$, choice coherence simply requires that the relative ranking between $x$ and $y$ remain unchanged, i.e. $y$ cannot be preferred to $x$. In other words, $y$ cannot be one of the best elements in $B$ when $x$ is also present in $B$, namely $y\notin c_u(B)$. Whether $x$ is still one of the best alternatives is not specified.
Herr K.
CC BY-SA 4.0
2019-03-12T19:31:08.000Z
2019-03-12T19:31:08.000Z
Herr K.
<p>If there is no <span class="math-container">$z \in A$</span> such that
<span class="math-container">$u(z) &gt; u(y)$</span> (perhaps because no other <span class="math-container">$z$</span> exists, perhaps for other reasons)
then by definition of <span class="math-container">$c_u$</span> we know that <span class="math-container">$y \in c_u(A)$</span>. This would contradict the assumption of the third sentence, that <span class="math-container">$y \notin c_u(A)$</span>.</p>

<blockquote>
  <p>If, moreover, <span class="math-container">$y \notin c_u(A)$</span>...</p>
</blockquote>

CC BY-SA 4.0
2019-03-11T09:36:58.017Z
What in the definition of $c_u$ implies that $y \in c_u(A)$ if there is no $z \in A$ such that $u(z) > u(y)$? It seems to me that we know is that $x \in c_u(A)$.
Ryan da Silva
CC BY-SA 4.0
2019-03-11T10:21:33.180Z
Is $x$ special in any way though, or is it just notation that could also be $\alpha$ or,  say,  $y$?
Giskard
CC BY-SA 4.0
2019-03-11T12:18:22.700Z
I am not sure I understand your question. He says that $x$ represents a certain commodity in the first sentence of (b). From then on, it should still represent that commodity. If he had initially chosen $\alpha$ instead of $x$, he would then have to use $\alpha$ to continue to represent that commodity.
Ryan da Silva
CC BY-SA 4.0
2019-03-11T12:51:42.147Z
He does not say that... please read the statement more carefully or read up on set notation.
Giskard
CC BY-SA 4.0
2019-03-11T16:27:04.383Z
@RyandaSilva: The statement does not imply that there is necessarily a third alternative that is distinct from $x$ and $y$. I think your difficulty is in interpreting "... for some $z\in A$" as "... for some $z\in A$ **and $z\ne x$**. This is unwarranted. The correct interpretation should be "... for some $z\in A$ **and $z$ may or may not be the same as $x$**".
Herr K.
CC BY-SA 4.0
2019-03-11T22:32:32.217Z
@HerrK. If $z$ is the same as $x$, then $u(z) > u(y)$ would contradict the previous sentence and the definition of the utility function. Furthermore, if $z$ is not the same as $x$, then I still don't see how we could rule out that the individual is indifferent between $z$ and $y$.
Ryan da Silva
CC BY-SA 4.0
2019-03-11T23:37:45.803Z
@RyandaSilva: 1) $u(x)>u(y)$ is not a contradiction to $u(x)\ge u(y)$. 2) The variable $y$ in the definition is merely *a placeholder*, not a specific alternative. 3) Indifference between $z$ and $y$ is ruled out because of the supposition that $y\notin c_u(A)$ and that $c_u(A)$ is not empty (it contains at least $x$).
Herr K.
CC BY-SA 4.0
2019-03-12T03:25:08.837Z
@HerrK. 3) Why doesn't the part of the choice function that states that $u(x) \geq u(y)$ for all $y \in A$ imply that the individual could be indifferent between $x$ and some $y \in A$ even if $y \notin c_u(A)$ and $x \in c_u(A)$?
Ryan da Silva
CC BY-SA 4.0
2019-03-13T05:17:56.670Z
@RyandaSilva because that would contradict the definition of $c_u$.
Theoretical Economist
CC BY-SA 4.0
2019-03-11T05:58:47.963Z
2019-03-11T05:58:47.963Z
Giskard
<p>In the first chapter of <a href="http://assets.press.princeton.edu/chapters/s9890.pdf" rel="nofollow noreferrer">Kreps (2013)</a>, there is a proof that the choice function satisfies choice coherence. Kreps writes:</p>

<p><a href="https://i.sstatic.net/vVgqb.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/vVgqb.png" alt="enter image description here"></a></p>

<p>I do not understand how the third sentence of (b) logically follows from the previous two sentences. How do we necessarily know that there is a third commodity, z, in A? There might just be two commodities. Furthermore, how do we necessarily know that u(z) > u(y)? Why can't the consumer possibly be indifferent between z and y?</p>

<microeconomics><utility><preferences>
Proof of Choice Coherence in Kreps (2013)
Ryan da Silva
# Proof of Choice Coherence in Kreps (2013)

(Asked by: username_0 on 2019-03-11)

In the first chapter of [Kreps (2013)](http://assets.press.princeton.edu/chapters/s9890.pdf), there is a proof that the choice function satisfies choice coherence. Kreps writes:

[![enter image description here](https://i.sstatic.net/vVgqb.png)](https://i.sstatic.net/vVgqb.png)

I do not understand how the third sentence of (b) logically follows from the previous two sentences. How do we necessarily know that there is a third commodity, z, in A? There might just be two commodities. Furthermore, how do we necessarily know that u(z) > u(y)? Why can't the consumer possibly be indifferent between z and y?

**Tags:** microeconomics, utility, preferences


---

## 2 answers

### Answer by username_1 on 2019-03-12. Score: 1

The English translation of $c\_u(A)=\{x\in A:u(x)\ge u(y) \text{ for all }y\in A\}$ is

> $c\_u(A)$ is a subset of elements in the set $A$ that satisfies the following condition: if an element is in $c\_u(A)$ then **this element must generate a utility no lower than any other element in the set $A$**.

In other words, $c\_u(A)$ contains the decision maker's most-preferred element(s) in $A$.

Note that "$x$" in the definition is *any* element in $A$ that satisfies the bolded condition above, and "$y$" is *any* element in $A$ (without further restrictions). I emphasize the word *any* to highlight the fact that $x$ and $y$ are merely placeholders and therefore do not refer to specific elements in either set. In fact, we may well have defined $c\_u(A)$ as $\{y\in A: u(y)\ge u(x) \text{ for all }x\in A\}$ and the interpretation/English translation will be exactly the same (i.e. verbatim) as in the quote-block above.

---

Let's work through an example. Let there be three possible alternatives, denoted as follows:
\begin{equation}
a\_1=\text{one \$5 bill}, \qquad
a\_2=\text{one \$10 bill}, \qquad
a\_3=\text{two \$5 bills}.
\end{equation}
Further suppose that the decision maker cares only about the total amount of money in each alternative, and thus
\begin{equation}
u(a\_2)=u(a\_3)>u(a\_1).
\end{equation}

## Example 1.

Let $A=\{a\_1,a\_2,a\_3\}$ and let's forget about $B$ for now.

"**Suppose $x,y\in A$ and $x\in c\_u(A)$.**" The first part of the sentence (before "and") suggests that both $x$ and $y$ (again, placeholders here) can be either $a\_1$, $a\_2$, or $a\_3$, but the second part of the sentence (after "and") rules out $x=a\_1$ since $a\_1$ is not one of the most preferred alternatives.

"**Then $u(x)\ge u(y)$.**" This follows from $x\in c\_u(A)$, since $x$ is either $a\_2$ or $a\_3$ and $y$ is either $a\_1$, $a\_2$, or $a\_3$.

"**If $y\notin c\_u(A)$, then $u(z)>u(y)$ for some $z\in A$.**" The "if" part of this sentence restricts $y$ to be only $a\_1$, because if $y$ were either $a\_2$ or $a\_3$, it would have been included in the set $c\_u(A)$. Consequently, since $y$ is not one of the most preferred alternatives, some element in $A$ must be strictly preferred to $y$, and we know that this "some element" must be either $a\_2$ or $a\_3$. [The textbook uses the variable $z$ instead of $x$ because it wants to allow for the possibility that $z\in A$ but $z\notin B$. Otherwise, it would have used $x$ instead.]

The rest of the proof should be straightforward from here on.

## Example 2.

Let $A=\{a\_1,a\_2\}$. I'll leave you to verify that in this example, the textbook's proof works in the same way without loss of generality.

> **Comments:**
> **username_0**: Your examples make things much clearer for me. However, how do we know $x \in c_u (B)$? Is $A$ in $x \in c_u (A)$ in the first sentence of the proof just a placeholder for whatever set we are looking at? In other words, once we arrive at the last sentence, we immediately know that $x \in c_u (B)$ because the first sentence said that $x \in c_u (A)$? If not, I am not sure how we know based on the sentences preceding the last sentence that $x \in c_u (B)$.  
> **username_1**: @RyandaSilva: The proof does not claim that $x\in c_u(B)$; it simply claims that $y\notin c_u(B)$.  
> **username_1**: @RyandaSilva: By definition, $c_u(A)$ contains the most preferred elements in set $A$, and $c_u(B)$ contains the most preferred elements in set $B$. We're given 3 conditions regarding two elements $x,y$ that simultaneously belong to both $A$ and $B$: i) $x$ and $y$ are in both sets $A$ and $B$; ii) $x\in c_u(A)$, namely, $x$ is one of the most preferred elements in $A$; and iii) $y\notin c_u(A)$, namely, $y$ is not one of the most preferred elements in $A$. The last two conditions imply that $x$ is strictly preferred to $y$.  
> **username_1**: When the choice context is defined by $A$, $x$ would be chosen over $y$. When the context changes to $B$, choice coherence simply requires that the relative ranking between $x$ and $y$ remain unchanged, i.e. $y$ cannot be preferred to $x$. In other words, $y$ cannot be one of the best elements in $B$ when $x$ is also present in $B$, namely $y\notin c_u(B)$. Whether $x$ is still one of the best alternatives is not specified.  

---

### Answer by username_2 on 2019-03-11. Score: 1

If there is no $z \in A$ such that
$u(z) > u(y)$ (perhaps because no other $z$ exists, perhaps for other reasons)
then by definition of $c\_u$ we know that $y \in c\_u(A)$. This would contradict the assumption of the third sentence, that $y \notin c\_u(A)$.

> If, moreover, $y \notin c\_u(A)$...

> **Comments:**
> **username_0**: What in the definition of $c_u$ implies that $y \in c_u(A)$ if there is no $z \in A$ such that $u(z) > u(y)$? It seems to me that we know is that $x \in c_u(A)$.  
> **username_2**: Is $x$ special in any way though, or is it just notation that could also be $\alpha$ or,  say,  $y$?  
> **username_0**: I am not sure I understand your question. He says that $x$ represents a certain commodity in the first sentence of (b). From then on, it should still represent that commodity. If he had initially chosen $\alpha$ instead of $x$, he would then have to use $\alpha$ to continue to represent that commodity.  
> **username_2**: He does not say that... please read the statement more carefully or read up on set notation.  
> **username_1**: @RyandaSilva: The statement does not imply that there is necessarily a third alternative that is distinct from $x$ and $y$. I think your difficulty is in interpreting "... for some $z\in A$" as "... for some $z\in A$ **and $z\ne x$**. This is unwarranted. The correct interpretation should be "... for some $z\in A$ **and $z$ may or may not be the same as $x$**".  
> **username_0**: @HerrK. If $z$ is the same as $x$, then $u(z) > u(y)$ would contradict the previous sentence and the definition of the utility function. Furthermore, if $z$ is not the same as $x$, then I still don't see how we could rule out that the individual is indifferent between $z$ and $y$.  
> **username_1**: @RyandaSilva: 1) $u(x)>u(y)$ is not a contradiction to $u(x)\ge u(y)$. 2) The variable $y$ in the definition is merely *a placeholder*, not a specific alternative. 3) Indifference between $z$ and $y$ is ruled out because of the supposition that $y\notin c_u(A)$ and that $c_u(A)$ is not empty (it contains at least $x$).  
> **username_0**: @HerrK. 3) Why doesn't the part of the choice function that states that $u(x) \geq u(y)$ for all $y \in A$ imply that the individual could be indifferent between $x$ and some $y \in A$ even if $y \notin c_u(A)$ and $x \in c_u(A)$?  
> **username_3**: @RyandaSilva because that would contradict the definition of $c_u$.  

---

