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Does there exist a function s.t

  1. Apr 21, 2012 #1
    I'm reviewing some problems for a coming up test and was assigned some practice problems and wondering if you guys could say if you agree with my answers or not.

    Given sets A and B does there exist a function s.t
    [itex] \forall (b \in B) (|f^{-1}(\{b\})| < ∞)[/itex]

    f: Z to R - No elements in R > Z, thus set B will always contain a subset that does not get mapped via the inverse function.

    f: R to Z - No, there exists no inverse function that satisfies this
  2. jcsd
  3. Apr 21, 2012 #2

    I seen what you said before moderators deleted it and yea R is the reals and Z the integers.

    For the second statement you said to look at the floor function an inverse function does not exist for this.

    I firmly believe it's a no and no as the number of elements in the range are greater then the number of elements of in domain, which would imply no inverse function even exists for (2) and vice versa for (1).
  4. Apr 21, 2012 #3

    D H

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    The moderators did not delete SteveL27's post. He deleted it himself, probably because he didn't like his answer.
  5. Apr 21, 2012 #4
    Correct, I deleted my own answer. There were some subtleties involving ambiguities in the OP's question (or my own misunderstanding of the question) and I either had to make my answer a lot better or delete it. I took the lazy way out :-)
    Last edited: Apr 21, 2012
  6. Apr 21, 2012 #5

    But, I am right and wrong at same time, the notation in this context I believe does not mean an inverse function, but just the inverse. :X
  7. Apr 21, 2012 #6
    LOL. You were too quick for me :-)

    Yes, inverse of a subset of the range. Not an inverse function. Also I was thinking the absolute values of the members of the inverse image had to be finite ... my mistake. You mean that each inverse image is a finite set. Your notation's fine, that was my error.

    But then for a function from Z to R: since there is no possible surjection from Z onto R, we are certainly not required to try to find one! Then the standard inclusion i:Z->R : n |-> n works fine. In that case the inverse image of any real number contains either zero or one element, depending on whether that real number happens to be an integer. So it satisfies your requirement. Yes?
    Last edited: Apr 21, 2012
  8. Apr 21, 2012 #7
    I am going to change my answers,

    Z to R the answer is yes, for example f(1) = 1, f(2) = 2 .......f(n) = n.

    Then f^-1({1}) = 1, f^-1({2}) = 2, f^-1({3.045346}) = null

    and |{null}| = 1 which is finite.

    So, yea you are definitely correct on that one(I was going to post it but you got here first :D ).

    R to Z the answer has to be no because no matter what interval you pick in R I can show there there more reals than integers that lye in that interval.

    But, yea thanks for help appreciate it. I was also confused by notation and the hand outs are a bit ambigious so had to look at a pdf document I found online that goes into detail explaining various properties of functions . Got to a part where it said this functions inverse is XXX but, it does not have an inverse function and then it all clicked.

    Thanks again
  9. Apr 21, 2012 #8
    You got it.

    The inverse notation is ambiguous so it's good to explicitly realize that.

    If we have f(x) = x^2, just to take an example, then we would say f(3) = 9.

    But then we will typically write

    [itex]f^{-1}(3) = \{-3, 3\}[/itex]

    when we should actually write

    [itex]f^{-1}(\{3\}) = \{-3, 3\}[/itex]

    In other words we intend to denote the inverse image of a subset of the range; but by convention (or "abuse of notation") we omit the curly braces. We do NOT intend to imply that [itex]f^{-1}[/itex] is the inverse function to f. Rather [itex]f^{-1}[/itex] is a function from subsets of the range to subsets of the domain. You always have to make sure you know exactly what they mean by [itex]f^{-1}[/itex] because sometimes they mean the the inverse function of f; and other times they mean the inverse image of a point or subset of the range.
  10. Apr 21, 2012 #9

    D H

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    Why not just have f-1 be the floor function? This still has f-1(f(x)) for all x in Z, and that's the definition of the inverse function.
  11. Apr 22, 2012 #10
    If f-1 was the floor function wouldn't this imply f does not exist(because the floor function has no inverse)

    f-1(x) = max {m in Z | m ≤ x }

    for example f-1(2.4) = 2; f-1(2.3) = 2

    then f(2) goes to 2.4 and 2.3 so it isn't a function.
  12. Apr 22, 2012 #11

    D H

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    When dealing with different sets A and B, just because f(g(x)) = x for all x in A does not necessarily mean that g(f(y)) = y for all y in B. The function g can be viewed as an inverse of f, but f is not necessarily an inverse of g.
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