Results 1 to 5 of 5

Math Help - Is emptyset a function?

  1. #1
    Junior Member
    Joined
    Feb 2010
    From
    New Jersey
    Posts
    73

    Is emptyset a function?

    My discrete book is defining a function, f, as a special type of relationship in which if both (a,b) \in f and (a,c) \in f, then b=c (and a relation is defined as a set of ordered pairs).

    So, is the empty set not a function because it doesn't have any ordered pairs, or is it a function because it does not violate the definition of a function?

    For each of the following relations, please answer these questions:
    (1) Is it a function? If not, explain why.
    (2) If yes, what are it's domain and range?
    (3) Is the function one-to-one? If not, explain why.
    (4) If yes, what is the inverse function?

    a,b,c,d,e,... I already did
    f. f=\emptyset
    (1) Yes (trivially), because there are no ordered pairs in f, it does not violate the definition of function.
    (2) dom f = im f = \emptyset
    (3) Yes (trivially), since the definition of one-to-one is not violated
    (4) f^{-1}=\emptyset

    Above is how I wrote up my homework (but it's not due until Thu.), but I'm not confident it's the right answer.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Feb 2010
    Posts
    466
    Thanks
    4
    Quote Originally Posted by MSUMathStdnt View Post
    My discrete book is defining a function, f, as a special type of relationship in which if both (a,b) \in f and (a,c) \in f, then b=c (and a relation is defined as a set of ordered pairs).

    So, is the empty set not a function because it doesn't have any ordered pairs, or is it a function because it does not violate the definition of a function?
    I wouldn't put it as "not violating" but rather as "does fulfill."

    When we say "S is a set of ordered pairs" we mean "For all x, if x in S, then x is an ordered pair." Well, the emptyset fulfills that requirement. For all x, if x is in the empty set then x is an ordered pair. So, in that very specific sense (which is the only sense that matters toward this question), yes, the empty set is a set of ordered pairs.

    Quote Originally Posted by MSUMathStdnt View Post
    Above is how I wrote up my homework (but it's not due until Thu.), but I'm not confident it's the right answer.
    All correct, except I would modify (1) as I mentioned above.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2010
    From
    New Jersey
    Posts
    73
    Quote Originally Posted by MoeBlee View Post
    I wouldn't put it as "not violating" but rather as "does fulfill."

    When we say "S is a set of ordered pairs" we mean "For all x, if x in S, then x is an ordered pair." Well, the emptyset fulfills that requirement. For all x, if x is in the empty set then x is an ordered pair. So, in that very specific sense (which is the only sense that matters toward this question), yes, the empty set is a set of ordered pairs.

    All correct, except I would modify (1) as I mentioned above.
    I understand what you're saying. But I still don't see how to word it (although I'll probably get full credit as long as I've got the idea right). How does this sound:

    (1) Yes (trivially). There are no ordered pairs in f, therefore; for all ordered pairs in f, there are none that have the same first value and a different second value.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    This reminds me of a thread I saw recently.
    http://mymathforum.com/viewtopic.php?f=22&t=18683
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Feb 2010
    Posts
    466
    Thanks
    4
    But you missed mentioning that every member of the empty set is an ordered pair.

    I'll do it in English [where '0' stands for the empty set]:

    For all x, if x is in 0 then, x is an ordered pair. So 0 is a relation. And for all x, y, z, if <x y> and <x z> are in 0 then y=z. So 0 is a relation that is moreover a function.

    In symbols:

    Ax(x in 0 -> x is an ordered pair).
    So 0 is a relation.
    Axyz((<x y> in 0 & <x z> in 0>) -> y=z).
    So 0 is a function.

    /

    If you want to get more detailed, you can mention that

    Ax(x in 0 -> x is an ordered pair)

    Axyz((<x y> in 0 & <x z> in 0>) -> y=z)

    are true because the antecedent in each is false.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 20
    Last Post: November 27th 2012, 06:28 AM
  2. Replies: 0
    Last Post: October 19th 2011, 05:49 AM
  3. Replies: 4
    Last Post: October 27th 2010, 06:41 AM
  4. Replies: 3
    Last Post: September 14th 2010, 03:46 PM
  5. Replies: 2
    Last Post: September 2nd 2010, 11:28 AM

Search Tags


/mathhelpforum @mathhelpforum