A preference is a relation P on the elements of V

[toughcookie723]

Given a Set V, a preference is a relation P on the elements of V that satisfies the following:

For any a,b in V, either aPb or bPa, if aPb and bPa, then a=b. Finally P is transitive.

Prove:

Preferences are NOT equivalence relations.

--We know that it is not an equivalence relation because it fails symmetry.



This is what I have for proof:

Suppose preferences are equivalence relations.
Suppose V={a, b, c, ....)

By definition of a preference, if aPb and bPa, then a = b

Since V is a set and by definition of sets, for any a, b in V, a doesn't equal b.

-> (a,b) doesn't equal (b, a). --> not symmetric.

Since preferences fail symmetry, they are not equivalent relations.

Is this correct thinking of the proof? THANK YOU for your help/guidance.

 

[pka]

Given a Set V, a preference is a relation P on the elements of V that satisfies the following:
For any a,b in V, either aPb or bPa, if aPb and bPa, then a=b. Finally P is transitive.
Prove:
Preferences are NOT equivalence relations.
--We know that it is not an equivalence relation because it fails symmetry.
This is what I have for proof:
Suppose preferences are equivalence relations.
Suppose V={a, b, c, ....)
By definition of a preference, if aPb and bPa, then a = b
Since V is a set and by definition of sets, for any a, b in V, a doesn't equal b.
-> (a,b) doesn't equal (b, a). --> not symmetric.
Since preferences fail symmetry, they are not equivalent relations.
Is this correct thinking of the proof? THANK YOU for your help/guidance.

I have one quibble with this: What if V has only one element in it?
Then the statement is false.
So, if \(\displaystyle \|V\|>1\) it is true.

 

[toughcookie723]

You are absolutely right, I forgot to mention that! Thank you!