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Proof involving numerical equivalence of sets

  1. May 30, 2014 #1
    1. The problem statement, all variables and given/known data

    Show that for a set A[itex]\subset[/itex]N, which is numerically equivalent to N=Z+, and the set B = A [itex]\cup[/itex]{0}, it holds that A and B are numerically equivalent, i.e., that A [itex]\approx[/itex]B

    Hint: Recall the definition of A≈B and use the fact that A is numerically equivalent to N. Note that 0 [itex]\notin[/itex] N.


    2. Relevant equations



    3. The attempt at a solution

    I really have little clue of how to complete this proof, this is sort of a wild guess, any help appreciated:

    It is given that A≈N. This means [itex]\exists[/itex]f:A→N such that f is a bijection. Therefore, f:A→N such that Im[f] = N and [itex]\forall[/itex]x1, x2[itex]\in[/itex]A, x1[itex]\neq[/itex]x2→f(x1)[itex]\neq[/itex]f(x2). Because A [itex]\subset[/itex]N, and A≈A by definition, then there is a function g:A→A such that g is a bijection. This describes the function g(n) = n for [itex]\forall[/itex]n[itex]\in[/itex]A. Then we can define a function h(n) = g(n-1). Because B = A[itex]\cup[/itex]{0}, g:A→B is a bijective function. This is true b/c Im[g] = B and f(x1) [itex]\neq[/itex]f(x2)→g(x1-1)[itex]\neq[/itex]g(x2-1). Thus, we have found a bijection g:A→B, and therefore A[itex]\approx[/itex]B. This concludes the proof.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 30, 2014 #2

    haruspex

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    What are the domain and range of h? For a given n in the domain of h, how do you know n-1 is in the domain of g?
    Thinking in terms of the function f was a good start, but I don't see where you made use of it. Try combining f with the n-1 idea.
     
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