<p>Your "counterexample" also violates condition $\alpha$: $y=C(\{x,y,z\})$ but $y\ne C(\{y,z\})$. So it's not really a counterexample.</p>

CC BY-SA 3.0
2017-11-10T02:39:28.837Z
@neto333: Your question says you want to "show if condition α implies Axiom 1", not the other way around. Also, your example does not satisfy Axiom 1 either.
Herr K.
CC BY-SA 3.0
2017-11-10T02:49:32.423Z
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
CC BY-SA 3.0
2017-11-10T03:15:54.723Z
@neto333: $\{y\}\ne\{y,z\}$. Note that the definitions use the "$=$" sign.
Herr K.
CC BY-SA 3.0
2017-11-10T03:18:41.717Z
Ok, i think your answer is the answer. Just one more question before i mark this post as answered. I was not able to find any counter example (maybe im wrong) so can we say condition α implies Axiom 1? Or im i missing a counter example?
neto333
CC BY-SA 3.0
2017-11-10T03:23:25.580Z
@neto333: Yes, the implication can be proved mathematically.
Herr K.
CC BY-SA 3.0
2017-11-10T02:25:50.043Z
2017-11-10T02:25:50.043Z
Herr K.
<p>i'm new at this, so i´m really sorry sorry if i do something wrong.
The problem is this:
A choice function satisﬁes <strong>condition α</strong> 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.</p>

<p>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"</p>

<p>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. </p>

CC BY-SA 3.0
2017-11-10T00:45:54.187Z
Which condition is Condition 1?
Herr K.
CC BY-SA 3.0
2017-11-10T00:53:54.967Z
im sorry. I meant condition α. Ty for letting me know that error.
neto333
CC BY-SA 3.0
2017-11-10T01:47:57.650Z
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.
CC BY-SA 3.0
2017-11-10T01:59:05.077Z
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
<microeconomics><choice-theory>
Is this proof correct?
neto333
# Is this proof correct?

(Asked by: username_0 on 2017-11-09)

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.

**Tags:** microeconomics, choice-theory

> **Comments:**
> **username_1**: Which condition is Condition 1?  
> **username_0**: im sorry. I meant condition α. Ty for letting me know that error.  
> **username_1**: 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 $x=C(\{y,z\})$??  
> **username_0**: 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.  

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## 1 answers

### Answer by username_1 on 2017-11-10. Score: 2 | Accepted answer

Your "counterexample" also violates condition $\alpha$: $y=C(\{x,y,z\})$ but $y\ne C(\{y,z\})$. So it's not really a counterexample.

> **Comments:**
> **username_1**: @username_0: Your question says you want to "show if condition α implies Axiom 1", not the other way around. Also, your example does not satisfy Axiom 1 either.  
> **username_0**: 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  
> **username_1**: @username_0: $\{y\}\ne\{y,z\}$. Note that the definitions use the "$=$" sign.  
> **username_0**: Ok, i think your answer is the answer. Just one more question before i mark this post as answered. I was not able to find any counter example (maybe im wrong) so can we say condition α implies Axiom 1? Or im i missing a counter example?  
> **username_1**: @username_0: Yes, the implication can be proved mathematically.  

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