1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
    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?

    [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.


    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook