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Does this make sense in set theory?

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data
    Let X be any set, f a function. Let f:X->Y

    Does f(A) make sense for A in X?

    I know f^(-1)(B) makes sense for B in Y.


    3. The attempt at a solution
    I can't see why not
     
  2. jcsd
  3. Sep 16, 2007 #2

    HallsofIvy

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    Yes, of course it does. f(A) is defined as {y| y= f(x) for some x in A}. In other words it is all values in y that you can actually "get to" using f.

    f-1(B)= {x| f(x) is in B}.

    Can you answer this: if f(x)= x2 and A= [-2, 1], what is f(A)?
     
  4. Sep 16, 2007 #3
    Wouldn't you say for all x in A?

    Since f is bijective, f(A)=[1,4]
     
  5. Sep 16, 2007 #4
    You could do that, but why waste the words when there's no confusion?

    And x^2 is *not* bijective. Think again...
     
  6. Sep 16, 2007 #5

    HallsofIvy

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    No, I wouldn't! 4 is in f(A) for the example below because 4= f(-2). But it certainly is NOT true that 4= f(x) for ALL x in A!

    Since f is clearly not bijective (it in neither injective nor surjective) that is not correct. To make one obvious point 0 is in [-2,1] so f(0)= 0 must be in f(A). It is NOT in [1, 4].
     
  7. Sep 16, 2007 #6
    My bad, I should have drawn a graph. f(A)=[0,4]

    With regards to 'f(A) is defined as {y| y= f(x) for some x in A}' What if I view y as a variable? So it is intepreted as all y in the codomain of f such that y=f(x) for all x in A.
     
  8. Sep 16, 2007 #7

    Hurkyl

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    It depends on how you're doing your set-building.

    If we use comprehension, we are thinking about the elements of Y in the image of A.
    [tex]f(A) = \{ \, y \in Y \mid \exists x \in A : f(x) = y \, \}[/tex]

    If we use replacement, we are thinking about replacing every element of A with its image.
    [tex]f(A) = \{ \, f(x) \mid x \in A \, \}[/tex]
     
    Last edited: Sep 16, 2007
  9. Sep 16, 2007 #8

    HallsofIvy

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    in {y| y= f(x) for some x in A}, y is a member of a set, not a variable.
     
  10. Sep 16, 2007 #9
    RIght. So I was thinking of 'replacement language' when intepreting HallsIvy's 'comprehension language' which obviously is not allowed hence the confusion.
     
  11. Sep 16, 2007 #10

    Hurkyl

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    What do you mean by "is not allowed"? :confused:
     
  12. Sep 16, 2007 #11
    Adding y in replacement language
    [tex]f(A) = \{ \, y=f(x) \mid x \in A \, \}[/tex]

    and intepreting that y as the same y in comprehension language.
     
  13. Sep 16, 2007 #12

    Hurkyl

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    This certainly isn't grammatically correct. You have a free variable y, and the expression doesn't match any set-builder notation I know.

    Anyways, it strikes me that I should have made a more positive response, and I'll do that now.

    The axiom of comprehension (a.k.a. axiom of subsets) says that if you have some set S and some predicate P, there exists a set containing exactly those elements of S that satisfy P. A standard way to denote that set is with the syntax
    [tex]\{ \, s \in S \mid P(s) \, \}.[/tex]
    Sometimes, when S can be inferred from the context, you will see the shorthand notation
    [tex]\{ \, s \mid P(s) \, \}.[/tex]

    So, the expression
    [tex]f(A) = \{ \, y \in Y \mid \exists x \in A : f(x) = y \, \}[/tex]
    is certainly a grammatically correct mathematical statement.
     
  14. Sep 17, 2007 #13

    HallsofIvy

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    Ouch! "Replacement Language"? "Comprehesion Language"? This is getting too deep for me. Whatever happened to good old set theory?
     
  15. Sep 17, 2007 #14
    Left with Russell when he got his barber to shave him.
     
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