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Analysis: proving inverses involving sets

  1. Nov 9, 2008 #1
    Let f: X → Y and A is a subset of Y and B is a subset of Y. Prove that:

    a) f⁻¹(A union B) = f⁻¹(A) union f⁻¹(B)
    b) f⁻¹(A intersetion B) = f⁻¹(A) intersection f⁻¹(B).

    I know that f⁻¹(A) = {x ε X : f(x) ε A}
    and so f⁻¹(B) {x ε X : f(x) ε B}

    but after that I really don't understand how to prove this.
     
    Last edited: Nov 9, 2008
  2. jcsd
  3. Nov 9, 2008 #2

    HallsofIvy

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    Any time you are asked to prove X= Y for sets X and Y you do two things:
    Prove that X is a subset of Y and then that Y is a subset of X.

    To prove X is a subset of Y, you start "if x is a member of X" and then use the definitions of X and Y to show that x must be a member of Y.

    For example, here, if x is in f-1(A union B) then f(x) is in A union B. That means either f(x) is in A or f(x) is in B. If f(x) is in A, then x is in f-1(A). If f(x) is in B, then x is in f-1(B). In either case, x is in f-1(A) union f-1(B). That proves that f-1(A union B) is a subset of f-1(A) union f-1(B). Now do it the other way: if x is in f-1(A) union f-1(B), then ...
     
  4. Nov 9, 2008 #3

    CompuChip

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    Welcome to PF, vikkisut.
    It's really just a matter of plugging in the definitions:
    [tex]f^{-1}(A \cup B) = \{ x \in X: f(x) \in A \cup B \} \stackrel{\star}{=} \{ x \in X: f(x) \in A \text{ or } f(x) \in B \} \stackrel{\star}{=} \{ x \in X: f(x) \in A \} \cup \{ x \in X: f(x) \in B \} = f^{-1}(A) \cup f^{-1}(B)[/tex]
    That is just a partial proof though; I have marked two of the identities with a [itex]\star[/itex], I suggest you try to prove these by a standard argument (let x be in one of them and show that it is in the other and vice versa).
     
  5. Nov 9, 2008 #4
    Sorry I don't quite understand what you're suggesting me to do there? Let x be in what exactly? Be part part of A, and then show that it is in B?
     
  6. Nov 9, 2008 #5

    CompuChip

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    No, let it be in
    [tex]\{ x \in X: f(x) \in A \cup B \}[/tex]
    and prove it to be in
    [tex]\{ x \in X: f(x) \in A \} \cup \{ x \in X: f(x) \in B \}[/tex]
    Also see HallsOfIvy's post, for more explanation.

    [edit]You should also prove the converse of course, letting it be in the latter and showing that it is in the former.[/edit]

    [edit]1500 posts by the way, almost went unnoticed :tongue:[/edit]
     
  7. Nov 9, 2008 #6
    oh okay, thank you very much :)
     
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