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Inverse Images and Sets (union & intersection)

  1. Sep 7, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose f is a function with sets A and B.
    1. Show that:
    [tex]I_{f} \left(A \cap B\right) = I_{f} \left(A\right) \cap I_{f} \left(B\right)[/tex]
    Inverse Image of F (A intersects B) = Inverse Image of F (A) intersects Inverse Image of B.

    2. Show by giving a counter example that:
    [tex]f\left(A \cap B\right) \neq f\left(A\right) \cap f \left(B\right)[/tex]
    F (A intersects B) does not equal F (A) intersects F(B)


    2. Relevant equations

    Knowledge of Sets and Inverse Images

    3. The attempt at a solution
    1.
    Let c be an element of [tex]I_{f} \left(A \cap B\right)[/tex].
    By the definition of [tex]I_{f} \left(A \cap B\right) [/tex], there is a [tex]d\in(A \cap B)[/tex] so that [tex]I_{f}(d)=c[/tex].
    Since, [tex]d\in(A \cap B)[/tex], [tex] d \in A & d \in B[/tex]. Since [tex]d\inA, I_{f}(d)\in I_{f}(A)[/tex]. This follows alongside [tex]d\inB, I_{f}(d)\inI_{f}(B)[/tex].
    Since [tex]I_{f}(d)=c \in I_{f}(A) [/tex] and [tex]I_{f}(d)=c \in I_{f}(B), c = I_{f}(A)\capI_{f}(B)[/tex].

    Thoughts? Also would I need to show that the [tex]I_{f}(A)\capI_{f}(B) \in I_{f} \left(A \cap B\right)[/tex] to show true equality?

    2.
    [tex]f\left(A \cap B\right) \neq f\left(A\right) \cap f \left(B\right)[/tex]
    I'm thinking either the absolute value function or a square function of some sort would show that it is not equal. Though, I'm not sure how to proceed with depicting the counter example.

    Sincerely,

    NA
     
    Last edited: Sep 7, 2010
  2. jcsd
  3. Sep 7, 2010 #2
    Okay, so my counterexample for No. 2 is:

    f(x)=|x| {-3,...,-1} = A and {1,...,200} = B

    There is no intersection between A ∩ B. However, there is an intersection with f(A) ∩ f(B) that gives the set {1,3}.

    Thus, {null} != {1,3}.

    Figured out the wording that I was missing =) Just need help with i now!
     
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