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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: 20,636 Thanks: 2080 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: 20,636 Thanks: 2080 I've also edited Axiom 1 for neto333 to change the first "=" to "≠". Tags correct, proof Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post RanberSingh Number Theory 2 April 27th, 2017 08:06 AM sKebess Number Theory 9 April 9th, 2017 08:27 AM Magnitude Calculus 3 February 16th, 2017 10:28 AM 123qwerty Calculus 4 May 10th, 2016 08:00 AM Gunuu Applied Math 5 September 21st, 2008 07:17 AM

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