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Math Help - one-one function

  1. #1
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    one-one function

    Definition of 0ne-one function:

    f:A\rightarrow B is one-one if  f(a)=f(b) implies a=b.


    Now let A=\varnothing and B=\varnothing.


    I know the logical statement g: a\notin \varnothing \Rightarrow b\notin \varnothing is true, since were P is true and Q is true, P \Rightarrow Q is true by vacuous proof, but it's not the definition of one-one function.

    Question: How can we prove g:\varnothing \rightarrow \varnothing  using the definition?

    g(blank)= g(blank) implies blank = blank or
    g(\qquad)=g(\qquad) implies \qquad\qquad\qquad = \qquad\qquad\qquad ?
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  2. #2
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    Quote Originally Posted by novice View Post
    Definition of 0ne-one function:

    f:A\rightarrow B is one-one if  f(a)=f(b) implies a=b.


    Now let A=\varnothing and B=\varnothing.


    I know the logical statement g: a\notin \varnothing \Rightarrow b\notin \varnothing is true, since were P is true and Q is true, P \Rightarrow Q is true by vacuous proof, but it's not the definition of one-one function.

    Question: How can we prove g:\varnothing \rightarrow \varnothing  using the definition?

    g(blank)= g(blank) implies blank = blank or
    g(\qquad)=g(\qquad) implies \qquad\qquad\qquad = \qquad\qquad\qquad ?
    I might be missing something, but my thought process would be: since there is no a or b for which g(a) or g(b) is defined, the statement

    g(a)=g(b) \implies a=b.

    is vacuously true because g(a)=g(b) can never be true.
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  3. #3
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    Quote Originally Posted by undefined View Post
    I might be missing something, but my thought process would be: since there is no a or b for which g(a) or g(b) is defined, the statement

    g(a)=g(b) \implies a=b.

    is vacuously true because g(a)=g(b) can never be true.
    If we look at the set of functions B^A, we see that the is a function since |B|^{|A|}=0^0=1

    The proof in my book shows this:

    Given A and B are empty sets.
    Let b\in B. A function f:A \rightarrow B is a one-one function for every a \in A.

    I am not convinced, and very annoyed by it since a\notin A and b\notin B.
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by novice View Post
    If we look at the set of functions B^A, we see that the is a function since |B|^{|A|}=0^0=1

    The proof in my book shows this:

    Given A and B are empty sets.
    Let b\in B. A function f:A \rightarrow B is a one-one function for every a \in A.

    I am not convinced, and very annoyed by it since a\notin A and b\notin B.
    Yeah I don't understand the statement

    Let b\in B.

    Isn't that like writing

    Let 0 = 1.

    ?
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