# $\alpha$ condition and choice structure

I am trying to prove a statement that if $c_R$ is a choice structure, then it must satisfy Sen's $\alpha$ condition.

However, if $X=\{a,b,c \}$, I can construct some choice structure like $c(\{a\})=a,c(\{b\})=b, c(\{c\})=c, c(\{a,b\})=a,c(\{b,c\})=b,c(\{c,a\})=c,c(\{a,b,c\})=a$.

Such choice structure clearly violates the $\alpha$ (shrinking) condition.

So can someone point out where do I make a mistake? Or is the statement assuming the relation $R$ to be rational?

Thanks!

## 1 Answer

A choice structure can be whatever it wants to be. A choice structure that is rationalizable by a preference relation must (among other things) satisfy Sen's $\alpha$.

So if you want to show that a rationalizable choice structure satisfies $\alpha$ then you can show that any that satisfies WARP will also satisfy $\alpha$.

I recommend Kreps or the UPitt notes for more info on the relationship between choice structures and preference.

http://www.pitt.edu/~luca/ECON2100/lecture_02.pdf

• So do you mean a choice structure satisfies the $\alpha$ condition only if we can find a rational preference $R$ s.t. $c_R=c$?
– ask
Commented Sep 11, 2016 at 17:59
• Wait. "only if" means sufficient right? Then no as $\alpha$ is necessary, but not sufficient.
– VCG
Commented Sep 11, 2016 at 18:03