Jomo
Elite Member
 Joined
 Dec 30, 2014
 Messages
 3,529
Hi,
I did a proof that my mathematician friend said was wrong. He claims that I should show bijections. I am sure that he is correct but I do not see why my trivial proof is wrong.
The statement of the theorem along with my proof is below.
Theorem 2: Let U be some universal set. Define a relation on \(\displaystyle \mathcal{P}(U)\) by \(\displaystyle A\, \sim\, B\) iff \(\displaystyle A\) and \(\displaystyle B\) have the same cardinality; that is, iff there is a bijection \(\displaystyle f\, :\, A\, \rightarrow\, B.\) Then \(\displaystyle \sim\) is an equivalence relation on \(\displaystyle \mathcal{P}(U).\)
Proof:
Reflexive: \(\displaystyle A\, \sim\, A\) since \(\displaystyle A\) and \(\displaystyle A\) have the same cardinality.
Symmetric: Suppose \(\displaystyle A\, \sim\, B.\) Then \(\displaystyle A\) and \(\displaystyle B\) have the same cardinality, so \(\displaystyle B\) and \(\displaystyle A\) have the same cardinality. Then \(\displaystyle B\, \sim\, A.\)
Transitive: Suppose \(\displaystyle A\, \sim\, B\) and \(\displaystyle B\, \sim\, C.\) Then \(\displaystyle A\) and \(\displaystyle B\) have the same cardinality, and \(\displaystyle B\) and \(\displaystyle C\) have the same cardinality. Since a set has only one cardinality, it follows that \(\displaystyle A\) and \(\displaystyle C\) must have the same cardinality, and \(\displaystyle A\, \sim\, C.\)
Since the relation \(\displaystyle \sim\) is reflexive, symmetric, and transitive, it is an equivalence relation.
Thanks for your time!
I did a proof that my mathematician friend said was wrong. He claims that I should show bijections. I am sure that he is correct but I do not see why my trivial proof is wrong.
The statement of the theorem along with my proof is below.
Theorem 2: Let U be some universal set. Define a relation on \(\displaystyle \mathcal{P}(U)\) by \(\displaystyle A\, \sim\, B\) iff \(\displaystyle A\) and \(\displaystyle B\) have the same cardinality; that is, iff there is a bijection \(\displaystyle f\, :\, A\, \rightarrow\, B.\) Then \(\displaystyle \sim\) is an equivalence relation on \(\displaystyle \mathcal{P}(U).\)
Proof:
Reflexive: \(\displaystyle A\, \sim\, A\) since \(\displaystyle A\) and \(\displaystyle A\) have the same cardinality.
Symmetric: Suppose \(\displaystyle A\, \sim\, B.\) Then \(\displaystyle A\) and \(\displaystyle B\) have the same cardinality, so \(\displaystyle B\) and \(\displaystyle A\) have the same cardinality. Then \(\displaystyle B\, \sim\, A.\)
Transitive: Suppose \(\displaystyle A\, \sim\, B\) and \(\displaystyle B\, \sim\, C.\) Then \(\displaystyle A\) and \(\displaystyle B\) have the same cardinality, and \(\displaystyle B\) and \(\displaystyle C\) have the same cardinality. Since a set has only one cardinality, it follows that \(\displaystyle A\) and \(\displaystyle C\) must have the same cardinality, and \(\displaystyle A\, \sim\, C.\)
Since the relation \(\displaystyle \sim\) is reflexive, symmetric, and transitive, it is an equivalence relation.
Thanks for your time!
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