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Group Theory Sets and Mappings

  1. Mar 16, 2009 #1

    ZTV

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    1. The problem statement, all variables and given/known data

    Prove that f: S -> T is one-to-one if and only if f(AnB) = f(A) n f(B)
    for every pair of subsets A and B of S

    2. Relevant equations

    See above

    3. The attempt at a solution

    Part 1:

    Starting with the assumption f(AnB) = f(A) n f(B)

    Let f(a) = f(b) [I'm going to try and prove that a=b for one-to-one]

    f(a) is an element of f(A)
    f(b) is an element of f(B)
    f(a) = f (b)
    So
    f(a) is an element of f(A) n (B)
    From assumption
    f(a) and f(b) are elements of f(A) n f(B) so also elements of f (AnB)

    so a, b are elements of AnB

    Can't work out how to show they are equal though..

    2. Assume f is one-to-one

    So f(a)=f(b) and a=b

    If a=b, a is an element of AnB
    so f(a) is an element of f(AnB)

    Likewise, if f(a)=f(b) then f(a) is an element of f(A) n f(B)

    Don't know where to go from here.
     
  2. jcsd
  3. Mar 16, 2009 #2

    CompuChip

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    For the first one, note that the assumption holds for any sets A and B. So you can choose them conveniently (hint: sets can consist of just one point).

    For the second one, I suggest showing two inclusions f(A n B) c f(A) n f(B) and f(A) n f(B) c f(A n B).
    One of them is trivial (if y is the image of a point x in A n B, then x is in A so it's the image of a point in A and x is in B so it's the image of a point in B, so it's definitely in f(A) and f(B)). For the other one you will need the injectivity, just write out definitions.
     
  4. Mar 16, 2009 #3

    ZTV

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    Thanks... for the second one I'd considered looking at inclusions but couldn't work out how to implement it.

    Don't understand what you mean about choosing convenient sets though, can you explain?
     
  5. Mar 16, 2009 #4

    CompuChip

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    Well, you have assumed (I hope, otherwise do it now :smile:) that f(AnB) = f(A) n f(B) for any subsets A and B of S, because that's what it said in the statement.

    So to conclude that a = b, you can choose them as you like: if the assumption holds for any A and B, then in particular they hold for some A and B that you choose. For example, you are allowed to take A = B, A empty, B = S, A = S \ B, or whatever you think you need to prove that a = b.
    My hint to you is: a single point x of S is also a subset {x} c S.
     
  6. Mar 16, 2009 #5

    ZTV

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    Is for every A and B the same as any A and B?
     
  7. Mar 16, 2009 #6

    CompuChip

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    Yep, they both mean [itex]\forall A \subset S, \, \forall B \subset S[/itex].
     
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