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Proving image of intersection?

  1. Feb 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Let F be a relation from X to Y and let A and B be subsets of X. Then,

    [itex]F(A \cap B) \subseteq F(A) \cap F(B)[/itex]

    3. The attempt at a solution

    Let [itex]y \in F(A \cap B)[/itex]. Then, [itex]\exists x \in A \cap B[/itex], so [itex]\exists x \in A[/itex] and [itex]x \in B[/itex].

    Then, [itex]y \in F(A)[/itex] and [itex]y \in F(B)[/itex], so [itex]y \in F(A) \cap F(B)[/itex].

    Therefore, [itex]y \in F(A \cap B) \Rightarrow y \in F(A) \cap F(B)[/itex], and hence, [itex]F(A \cap B) \subseteq F(A) \cap F(B)[/itex].

    I'm having trouble showing that the right side is not a subset of the left. Thanks for any help.
     
  2. jcsd
  3. Feb 19, 2012 #2

    HallsofIvy

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    [itex]\exists x \in A\cap B[/itex] such that F(x)= y. You might want to say that!

    That's because it may not be true! That's the reason for the "[itex]\subseteq[/itex]" rather than just "[itex]\subset[/itex]".
     
  4. Feb 19, 2012 #3
    This is actually the second part of the problem I'm on. The previous one was about the image of a union being equal to the union of the images. I mentioned your suggestion in that part.

    I understand that. I have to show that it's NOT a subset of the left hand side. I just don't know how to do that using the kind of logic I used in proving the left side was a subset of the right.
     
  5. Feb 21, 2012 #4

    HallsofIvy

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    And I will say again that you can't prove that- it is not true- unless you mean proper subset.
     
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