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Math Help - Proof Function (1-1, Restriction)

  1. #1
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    Proof Function (1-1, Restriction)

    Hi,
    I have to prove that a restriction of a one-to-one function is one-to-one.
    Intuitively this makes sense to me, but I don't see how to approach this proof formally. Any help would be appreciated.
    Thanks
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  2. #2
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    Quote Originally Posted by tbyou87 View Post
    Hi,
    I have to prove that a restriction of a one-to-one function is one-to-one.
    What does it mean a restriction of a one-to-one function?
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  3. #3
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    1-1 restriction function

    f: A -> B is 1-1 iff whenever f(x) = f(y), then x = y.

    Let f: A -> B and let D subset of A. The restriction of f to D is the function
    f|D = {(x, y): y = f(x) and x element D}.

    The theorem makes sense. For example, f: R -> R where f(x) = x is 1-1. If we restirct the domain to D subset of R F: D -> R where f(x) = x is still 1-1.

    I just don't know how to formally prove this.

    Thanks
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  4. #4
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    Quote Originally Posted by tbyou87 View Post
    f: A -> B is 1-1 iff whenever f(x) = f(y), then x = y.

    Let f: A -> B and let D subset of A. The restriction of f to D is the function
    f|D = {(x, y): y = f(x) and x element D}.

    The theorem makes sense. For example, f: R -> R where f(x) = x is 1-1. If we restirct the domain to D subset of R F: D -> R where f(x) = x is still 1-1.

    I just don't know how to formally prove this.

    Thanks
    Let A and B be non-empty sets.

    And f:A --> B be a one-to-one map.

    Let D be a non-trivial subset of A.

    Then g: D --- > B defined as g(x)=f(x).

    We need to show that g(a)=g(b) implies a=b.

    Let us assume that it fails, i.e. their this collapsing. Meaning we can find "a" and "b" distinct such that, f(a)=f(b). But then that means f(a)=f(b). But that cannot be because f is one-to-one. Thus, a contradiction.
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