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Abstract/Discrete/Algebraic Mathematics.

  1. Aug 3, 2010 #1
    Show/Prove that if f: X → Y is one-to-one on X, and A subset of X, then f^-1(f(A))<= subset A.

    If you wouldn’t mind, please check whether I did it correctly. Thanks in advance.

    Suppose x Є A
    Then, f(x) Є f(A)
    By image function y =f(x)
    Thus, y = f(x) Є f(A)
    And by inverse image, f^-1(y) = f-1(f(x)) Є f^-1(f(A)) <= subset A

    Therefore: Є f^-1(f(A)) <= subset A.
    Last edited: Aug 3, 2010
  2. jcsd
  3. Aug 12, 2010 #2
    That's right.
  4. Aug 12, 2010 #3
    Your proof doesn't make sense.

    You want to prove that f-1(f(A)) ⊆ A.
    Thus, suppose x ∈ f-1(f(A)). You want to prove x ∈ A.
    Since x ∈ f-1(f(A)), this means f(x) ∈ f(A).
    Thus there is some x' ∈ A such that f(x) = f(x').
    Since f is one-to-one, we have x = x', and thus x ∈ A.

    This is really the only way to prove it.


    Aside: For any function f: X → Y and any subset A of X, we have A ⊆ f-1(f(A)). This is because for any x ∈ A, f(x) ∈ f(A), so x ∈ f-1(f(A)). Thus, if f is one-to-one, you in fact have f-1(f(A)) = A.

    Aside 2: If f: X → Y is a function such that f-1(f(A)) ⊆ A for every subset A of X, then f is one-to-one. Proof: If x, x' ∈ X, let A = {x}. If f(x') = f(x), then f(x') ∈ f(A), so y ∈ f-1(f(A)) ⊆ A. But then x' = x, because A only contains x. Thus:
    A function f: X → Y is one-to-one if and only if for every A ⊆ X, f-1(f(A)) ⊆ A.​
    Last edited: Aug 12, 2010
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