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Math Help - some play on empty sets

  1. #1
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    some play on empty sets

    Hello

    I just have some general question regarding functions. Consider the sets A and B. Now if A\neq \varnothing\mbox{ and }B=\varnothing, then the definition of the function f:A\to B fails but there is still a function g:B\to A, even though in both cases A\times B=\varnothing . So if we define ^AB as the set of functions from A to B, then if
    B=\varnothing\mbox{ and } A\neq \varnothing, then

    ^AB=\varnothing\mbox{  and  } ^BA=\{\varnothing\}

    is that right reasoning ? I need this result in some proof.
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  2. #2
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    Re: some play on empty sets

    Quote Originally Posted by issacnewton View Post
    I just have some general question regarding functions. Consider the sets A and B. Now if A\neq \varnothing\mbox{ and }B=\varnothing, then the definition of the function f:A\to B fails but there is still a function g:B\to A, even though in both cases A\times B=\varnothing . So if we define ^AB as the set of functions from A to B, then if B=\varnothing\mbox{ and } A\neq \varnothing, then
    ^AB=\varnothing\mbox{  and  } ^BA=\{\varnothing\}
    is that right reasoning ? I need this result in some proof.
    This question can start an argument. The emptyset is problematic.
    That said, I will give a quote directly from Halmos.
    If each of X~\&~Y is a set and Y^X is the set of all functions X\to Y then the quote
    " (i) Y^{\emptyset} has exactly one element, namely \emptyset, whether Y is empty or not, and (ii) if X is not empty, then \emptyset^X is empty."
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  3. #3
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    Re: some play on empty sets

    Plato, thats what I am saying too, am I not ?
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  4. #4
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    Re: some play on empty sets

    Quote Originally Posted by issacnewton View Post
    Plato, thats what I am saying too, am I not ?
    I think so. But that notation is confusing.
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  5. #5
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    Re: some play on empty sets

    Thats what Daniel (Velleman) uses in his book. Just following him.......
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