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Are the following 3 statements true and does the cantor-bernstein theorem follow

  1. Nov 23, 2012 #1
    1.There exists an injection from A to B ⇔ A ≤ B
    2.There exists an injection from B to A ⇔ B ≤ A
    3.If A ≤ B and B ≤ A, then A = B

    Does this prove the Cantor Bernstein theorem? Which says that if 1 and 2 then there exists a Bijection between A and B (A = B)

    And if it does, why is there a different, longer proof for it?
     
  2. jcsd
  3. Nov 23, 2012 #2
    No. 1) and 2) are definitions of ≤ and ≥ for cardinalities of sets. 3) is a nontrivial consequence for which you have provided no argument at all.
     
  4. Nov 23, 2012 #3
    Statement 3 IS the Cantor-Schroeder-Berstein theorem: "If the cardinality of A is less than or equal to the cardinality of B, and the cardinality of B is less than or equal to the cardinality of A, then the cardinality of A is equal to the cardinality of B." You can also state it as "If there is an injection from A to B, and there is an injection from B to A, then there is a bijection from A to B." As Norweigan said, it requires a nontrivial argument to prove this theorem.

    EDIT: See the easy-to-understand proof here.
     
    Last edited: Nov 23, 2012
  5. Nov 23, 2012 #4
    I am reading that proof now but where is the flaw in my reasoning?
    A ≤B and B ≤ A is like saying A = B or A is strictly less than B and B is strictly less than A, which is a contradiction, so A must = B.
     
  6. Nov 23, 2012 #5
    If A and B were numbers, then yes it would be trivially true that if A≤B and B≤A then A would equal B. But A and B are sets, and what we mean by A≤B is that "there exists an injection from A to B". We don't know beforehand whether "less than or equal to" for sets, which has to do with existence of an injection, has the same properties as "less than or equal to" for numbers. We have to prove it. So you can't use your familiar properties of numbers, like the fact that two things can't be strictly less than each other.
     
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