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Proof of mapping to and from null set

  1. Sep 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Using the precise denition of a function and a little logic, show that, for every set [itex]Y[/itex] , there is exactly one function [itex]f [/itex] from [itex] \emptyset [/itex] to [itex]Y[/itex] . When is [itex]f[/itex] injective? Surjective?

    Let [itex] X[/itex] be a set. Show that there are either no functions from [itex]X[/itex] to [itex] \emptyset [/itex]
    or exactly one such function, depending on whether [itex] X \neq \emptyset [/itex] or [itex] X= \emptyset [/itex]

    2. Relevant equations



    3. The attempt at a solution

    So far all I can think of is that the definition of a function I think is: [tex] f:X \rightarrow Y, f=\{(x,f(x)), such\ that\ \forall x \in X, \ \exists ! y\in Y : f(x)=y \}[/tex]

    Or something like that. I'm having a hard time trying to write a proper definition of a function, especially because I'm fairly sure my teacher wants something to do with ordered pairs, (as that's how functions are graphed and he mentioned in class that functions should be defined by their graphs...)

    Anyway, regarding the null set: All I can think is that the null set is also a subset of Y, and therefore the one function that maps from the null set, is the one that also maps to the null set? And that the function is surjective and injective only when [itex] Y= \emptyset [/itex]. But then I'm very confused about how you can map a non-elements to non-elements, so perhaps I'm wrong, in which case, I have no idea where to go with this problem.

    Thanks for any help!
     
  2. jcsd
  3. Sep 7, 2012 #2

    Dick

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    It is a little challenging to chase these definitions around. But a function is a set of ordered pairs. An empty set of ordered pairs is technically a function. When does that work?
     
  4. Sep 8, 2012 #3
    Okay, so far I follow.

    And then... What?? An empty set of ordered pairs is still just an empty set, right? So the null set is a function? I'm sorry. I don't understand.
     
  5. Sep 8, 2012 #4

    HallsofIvy

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    Yes, that's a perfectly valid definition. It is a symbolic way of saying that a function is a set of ordered pair such that no two distinct pairs have the same first member. Recall that "[itex]\forall x \in X[/itex]" can also be interpreted as "if [itex]x\in X[/itex]". And what do you know about propositions of the for "if A then B" when B is false? What if A is false?

     
  6. Sep 8, 2012 #5

    Dick

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    Yes. It's called the empty function. http://en.wikipedia.org/wiki/Empty_function
     
  7. Sep 9, 2012 #6
    Hmm... Okay. So I read the wiki, and then googled the empty function, but I'm still conceptually struggling a little... I have:

    [tex] f:X \to Y, f=\{(x,f(x)), such\ that\ \forall x \in X, \ \exists ! y\in Y : f(x)=y \}[/tex] thus for all functions [tex] f \subseteq X \times Y : (x_1, y_1)= (x_1, y_2) \Leftrightarrow y_1=y_2 [/tex] So, let [itex] \ X=\emptyset , f:\emptyset \to Y [/itex] then [itex] (x_1, y_1) = (x_1, y_2), y_1\neq y_2[/itex] never occurs because there are no [itex] x \in \emptyset[/itex]. Therefore [itex] f:\emptyset \to Y[/itex] satisfies the definition of a function.

    Does that work? Then I would argue there is only one such function because from the definition of a function:
    [itex] \ f \subseteq X \times Y. [/itex] Then, when [itex] \ X=\emptyset \ then\ f \subseteq \emptyset \times Y = \emptyset [/itex] thus f is unique?

    I think I'm close, but I'm not sure if my logic is totally sound. Also, I'm still getting used to using various symbols and terminology, so correct me if I've made any mistakes. Thanks again!
     
  8. Sep 9, 2012 #7

    vela

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    Perhaps I'm missing something obvious, but your definition looks messed up to me. f is a subset of XxY, right? So f(x) necessarily has to be an element of Y. Yet your definition is saying that f(x) is not an element in Y for all possible values of x.

    Yes, this is right. The definition of a function is satisfied vacuously.

    Yeah, this looks good, though you might want to wait for a mathematician to chime in.

     
  9. Sep 9, 2012 #8

    Dick

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    I think [itex]\exists ! y[/itex] is supposed to mean, 'there exists a unique y'. And it looks pretty good to me too.
     
  10. Sep 9, 2012 #9

    vela

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    Ah, I've never seen that notation.
     
  11. Sep 9, 2012 #10
    Yeah, sorry. The ! meant unique. My teacher uses it, so I did as well.

    Also, regarding when the function is surjective or injective. I believe it is only surjective when Y is also the empty set, because then there is nothing in Y to map to, thus technically each element of Y IS mapped to. Same for injective, because no element will ever be mapped to more than once. However, now I'm confused as to whether the function is bijective? (which isn't part of the question but now I'm not sure...)

    And for the second part, I said that because there is no [itex] y \in \emptyset[/itex] then the definition of a function, [itex] f=\{(x,f(x)), such\ that\ \forall x \in X, \ \exists ! y\in Y : f(x)=y \} [/itex] cannot hold, because no y's exist. Unless X is also the null set, in which case, the above argument holds.
     
    Last edited: Sep 10, 2012
  12. Sep 10, 2012 #11

    Dick

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    I think you got that right. A function from the empty set to the empty set is bijective. Why not? And if f:X->Y and X is nonempty and Y empty is then there are no functions. It all seems ok to me.
     
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