# Is possible that, in a choice structure $(B,C)$, $C(b) = \emptyset$, for some $b \in B$

It is a very simple question but with some implications. I just start reading Mas-Colell and it's not clear for me if it is possible a choice structure, $$(B,C(.))$$, where $$\exists b \in B, C(b)= \emptyset$$.

For example: $$B=\{\{x,y,z\},\{x,y\},\{x,z\},\{y,z\},\{x\},\{y\},\{z\}\}$$

$$C(\{x,y,z\}) = \{z\}$$

$$C(\{x,y\}) = \emptyset$$

$$C(\{y,z\}) = \{z\}$$

$$C(\{x,z\}) = \{z\}$$

$$C(\{x\}) = \emptyset$$

$$C(\{y\}) = \emptyset$$

$$C(\{z\}) = \{z\}$$

is a valid choice structure?

Mas-Colell defines choice structure and choice rules in section 1.C, page 10 on oxford press 1995 edition.

I first though about it because I think it would have some interesting implications, but it was false. Anyway, the question remains. Thanks!

• Can you provide more context such as a page number of MWG? – Bayesian Jan 13 at 10:05
• If you want to model "no choice" from some menu of items $\bar X$ you can just let $X = \bar X \cup \{o\}$ where the element $o$ is an outside option and then let $o$ be a member of all subsets in $B$, then you formally live up to requirement in MWG that sets in $B$ are non-empty and still allow for something like "no-choice" which seems to be the purpose of allowing for $C(b) = \emptyset$. – Jesper Hybel Jan 13 at 10:31
• @Bayesian sorry! I will edit, MWG defines choice structure and choice rule in the section 1.C, page 10. – 49328481 Jan 13 at 23:36