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Thread: one-one function

  1. #1
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    one-one function

    Definition of 0ne-one function:

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


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


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

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

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

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


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


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

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

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

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

    is vacuously true because $\displaystyle g(a)=g(b)$ 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 $\displaystyle a$ or $\displaystyle b$ for which $\displaystyle g(a)$ or $\displaystyle g(b)$ is defined, the statement

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

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

    The proof in my book shows this:

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

    I am not convinced, and very annoyed by it since $\displaystyle a\notin A$ and $\displaystyle b\notin B$.
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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 $\displaystyle B^A$, we see that the is a function since $\displaystyle |B|^{|A|}=0^0=1$

    The proof in my book shows this:

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

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

    Let $\displaystyle b\in B$.

    Isn't that like writing

    Let $\displaystyle 0 = 1$.

    ?
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