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Bijection between AuB and A with A infinite set and B enumerable set.

  1. Aug 26, 2013 #1
    1. The problem statement, all variables and given/known data.
    Let A and B be disjoint sets, A infinite and B enumerable. Prove that there exists a bijection between AuB and A.

    3. The attempt at a solution.

    I have an idea of how to prove this statement, but I got stuck in the middle, so here is what I've done:
    There are just two cases to consider:

    1)A is enumerable or 2)A is uncountable.

    I think that the first case is pretty simple, since if A and B are both enumerable and disjoint sets, then, AuB is also enumerable. By definition of enumerable, there exist f and g both bijective functions from N to A, and from AuB to N respectively. If I define h=fog, this function is bijective because it is composition of bijective functions and it goes from the set AuB to the set A. In conclusion, h is the function that satisfies the required conditions.

    I am having some difficulty in constructing the desired function when A is uncountable. If A is uncountable, then it has a proper subset which is countable. Lets call that set S. Using properties of cardinality, I can prove that A-S is still uncountable. And the set A is exactly the union of A-S and S.
    The idea I had to construct the bijection was defining the function in this way:
    f restricted to the set B is the composition hog where h is a bijective function with N as its domain and S as its codomain and g is a bijective function with B as its domain and N as its codomain. Now I need to define f restricted to the set A, and here is where I get stuck. I mean, can I prove that there is a bijection from A to its proper uncountable subset A-S?. If this was true, then it would we very easy to prove that f is a bijection from AuB to the set A=(A-S)uS.
    Or maybe there is an easier way to solve the problem, without having to consider separated cases but this was all I could think of.
     
  2. jcsd
  3. Aug 27, 2013 #2

    mfb

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    Staff: Mentor

    Why do you make that so complicated? There is a bijection between A-S and A-S, and the remaining part of the bijection is equivalent to the case you solved before.
     
  4. Aug 27, 2013 #3
    You are absolutely right, it was quite simple but I couldn't see it. Thanks!
     
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