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.
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.