0

I'm trying to prove that path independence implies that Sen's alpha holds. Can someone guide me on how i can approach this proof?

Specifically Sen's alpha essentially states, that if for a choice function, I am the champion of the world, then I must also be a champion of the country I'm from.

Asaf Karagila
  • 393,674
Saad Hirani
  • 1

1 Answers1

1

Suppose PI is satisfied and take any A,B such that $B \subseteq A$. Take any $ x \in B $ and suppose that $ x \in C(A) $, but, towards a contradiction, that $ x \notin C(B) $.

Then:

$ x \in C(A)=C((A\setminus B) \cup B)=C(C(A\setminus B) \cup C(B))$

Where the second equality is given by PI. This means that $ x \in C(A \setminus B) \cup C(B) $ but since we had that $ x \notin C(B) $ then it must be that $ x \in C(A \setminus B)$ which implies that $ x \in A \wedge x \notin B $, a contradiction.

This concludes the proof and so we get alpha.

Arturo
  • 11