# Proof of Choice Coherence in Kreps (2013)

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

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

• 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 Mar 11 '19 at 9:36
• Is $x$ special in any way though, or is it just notation that could also be $\alpha$ or, say, $y$? – Giskard Mar 11 '19 at 10:21
• 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 Mar 11 '19 at 12:18
• He does not say that... please read the statement more carefully or read up on set notation. – Giskard Mar 11 '19 at 12:51
• @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. Mar 11 '19 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.

• 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 Mar 30 '19 at 17:57
• @RyandaSilva: The proof does not claim that $x\in c_u(B)$; it simply claims that $y\notin c_u(B)$. – Herr K. Mar 31 '19 at 6:07
• @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. Mar 31 '19 at 6:12
• 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. Mar 31 '19 at 6:15