# Is this proof correct?

i'm new at this, so i´m really sorry sorry if i do something wrong. 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 if 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 nonempty 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 implies 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.

• Which condition is Condition 1? – Herr K. Nov 10 '17 at 0:45
• im sorry. I meant condition α. Ty for letting me know that error. – neto333 Nov 10 '17 at 0:53
• Something about Axiom 1 doesn't look right. Suppose $A=\{x,y,z\}$. Your Axiom 1 says $x=C(A)\;\Rightarrow\;C(A)=C(A\setminus\{x\})$, which means $x=C(\{y,z\})$?? – Herr K. Nov 10 '17 at 1:47
• Another mistake. Sorry i should read my post more carefully next time, Axiom 1 should say $x≠C(A)$ then $C(A) = C(A\{x})$ .Sorry im having some trouble trying to writhe in code. – neto333 Nov 10 '17 at 1:59

Your "counterexample" also violates condition $\alpha$: $y=C(\{x,y,z\})$ but $y\ne C(\{y,z\})$. So it's not really a counterexample.
• Sorry, i think you are right. I completely forgot what i was trying to proof. We have that ${y, z}∈C({ y, z})$ so why this violates condition α – neto333 Nov 10 '17 at 2:49
• @neto333: $\{y\}\ne\{y,z\}$. Note that the definitions use the "$=$" sign. – Herr K. Nov 10 '17 at 3:15