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 November 9th, 2017, 02:35 PM #1 Newbie   Joined: Nov 2017 From: Texas Posts: 2 Thanks: 0 Is this proof correct? Hi, the problem is this: A choice function satisﬁes condition α if whenever x = C(A) and x ∈ B ⊂ A, it follows that x = C(B) as well. Axiom 1: If x ≠ C(A) then C(A) = C(A\{x}) And I need to show whether condition α implies Axiom 1. C(A) is choice function. Here is a definition of choice function (just in case you need it): "Let X be a ﬁnite set of alternatives. Let P(X) be the set of all non-empty subsets of X. We call the elements of P(X) as menus. A choice function is a map C : P(X) → X such that C(A) ∈ A, for all A ∈ P(X). It selects a single element from each menu. A choice correspondence is a map C : P(X) →P(X) such that C(A) ⊆ A, for all A ∈ P(X). From each menu, at least one alternative is chosen." And I found that Axiom 1 does not imply condition α. Here is a counter example: We have X = {x, y, z}. Take A = {x, y, z} and y ∈ C({x, y, z}) y ∈ C({x, y}), z ∈ C({x, z}) and {y, z} ∈ C({y, z}). Here we have a contradiction because C({x, y, z}) ≠ C({y, z}). Thank you. Last edited by skipjack; November 10th, 2017 at 03:23 AM. Reason: to change "condition 1 to condition α" and amend Axiom 1
 November 9th, 2017, 03:57 PM #2 Newbie   Joined: Nov 2017 From: Texas Posts: 2 Thanks: 0 Sorry for the double post, but I can't find any option to edit the post. I have an error: it must say condition α rather than condition 1. So Condition 1 is Condition α. Last edited by skipjack; November 9th, 2017 at 04:01 PM.
 November 9th, 2017, 04:02 PM #3 Global Moderator   Joined: Dec 2006 Posts: 19,723 Thanks: 1807 I've edited it to change the 1 to α for you. Thanks from greg1313 and neto333
 November 10th, 2017, 03:25 AM #4 Global Moderator   Joined: Dec 2006 Posts: 19,723 Thanks: 1807 I've also edited Axiom 1 for neto333 to change the first "=" to "≠".

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