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I am a little confused about the statement of 1.D.5 in MWG, which I will reproduce here for convenience. I have "solved" the problem, I just don't understand something particular.

$\textbf{(1.D.5)}$ Let $X = \{x,y,z\}$ and $\mathscr{B} = \{ \{x,y\},\{y,z\},\{z,x\} \}$. Suppose that choice is now stochastic in the sense that, for every $B \in \mathscr{B}$, $C(B)$ is a frequency distribution over alternatives in $B$. For example, if $B = \{x,y\}$, we write $C(B) = (C_x(B), C_y(B))...$

So, my issue is that it seems that $C : \mathscr{B} \to \mathbb{R}_+^2$ is not well-defined, and would be more appropriately defined as a function $A (\subseteq X^2) \to \mathbb{R}_+^2$. What am I missing?

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  • $\begingroup$ On second thought, I suppose any function is well-defined if you explicitly enumerate each input/output pair. But the problem doesn’t do that. $\endgroup$
    – Seh-kai
    Commented Sep 5, 2020 at 11:29
  • $\begingroup$ Isn't $\mathscr B\subset X^2$? The book also specifies the range of the $C$ function as the set of non-negative 2-vectors whose elements sum to $1$. $\endgroup$
    – Herr K.
    Commented Sep 5, 2020 at 17:17
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    $\begingroup$ I believe $\mathscr{B}$ is a subset of the powerset of $X$; i.e. all subsets of $X$. My issue is that I don't know what $C(\{ z,x \})$ is supposed to be. Is it $= (C_z(B),C_x(B))? $ Or is it $= (C_x(B),C_z(B))$? The outputs of $C$ are ordered based on the elements of $B$, but since $B$ is a set, $B$ carries no ordering. $\endgroup$
    – Seh-kai
    Commented Sep 5, 2020 at 19:29
  • $\begingroup$ You're right about $B$ having no order. I missed that part. $\endgroup$
    – Herr K.
    Commented Sep 5, 2020 at 20:55

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