In the first chapter of Kreps (2013), there is a proof that the choice function satisfies choice coherence. Kreps writes:

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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?


2 Answers 2


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, $y \notin c_u(A)$...

  • $\begingroup$ 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 $x \in c_u(A)$. $\endgroup$ Mar 11, 2019 at 9:36
  • $\begingroup$ Is $x$ special in any way though, or is it just notation that could also be $\alpha$ or, say, $y$? $\endgroup$
    – Giskard
    Mar 11, 2019 at 10:21
  • $\begingroup$ 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. $\endgroup$ Mar 11, 2019 at 12:18
  • 1
    $\begingroup$ He does not say that... please read the statement more carefully or read up on set notation. $\endgroup$
    – Giskard
    Mar 11, 2019 at 12:51
  • 2
    $\begingroup$ @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 $x$). $\endgroup$
    – Herr K.
    Mar 11, 2019 at 23:37

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.

  • $\begingroup$ 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 $x \in c_u (B)$. $\endgroup$ Mar 30, 2019 at 17:57
  • $\begingroup$ @RyandaSilva: The proof does not claim that $x\in c_u(B)$; it simply claims that $y\notin c_u(B)$. $\endgroup$
    – Herr K.
    Mar 31, 2019 at 6:07
  • $\begingroup$ @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 $y$. $\endgroup$
    – Herr K.
    Mar 31, 2019 at 6:12
  • $\begingroup$ 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. $\endgroup$
    – Herr K.
    Mar 31, 2019 at 6:15

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