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November 9th, 2017, 03:35 PM   #1
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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 04:23 AM. Reason: to change "condition 1 to condition α" and amend Axiom 1
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November 9th, 2017, 04:57 PM   #2
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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 05:01 PM.
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November 9th, 2017, 05:02 PM   #3
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I've edited it to change the 1 to α for you.
Thanks from greg1313 and neto333
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November 10th, 2017, 04:25 AM   #4
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I've also edited Axiom 1 for neto333 to change the first "=" to "≠".
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