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Homework Help: Question about Empty Set?

  1. Apr 13, 2010 #1
    Do all sets include Empty set by default?

    Like I have { {a}, {b}, {c} }, its equal to { emptySet, {a}, {b}, {c} }???

    See I have this question it goes something like.

    The power set of {a, b} is {emptySet, {a}, {b}, {a,b}}

    So for all A, B that belongs to the power set {a,b}, A R B where A is a subset of B

    Which means

    I get

    {a} -> {a}
    {b} -> {b}
    {a} -> {a, b}
    {b} -> {a, b}
    {a, b} -> {a, b}
    emptySet -> emptySet

    So I know that this relation has no greatest element.

    But my answer also states this relation least element is the emptySet? Why is this so? is does not relate to anyone other than itself, unless all sets contains emptySet by default? Or am I doing something wrong here?
  2. jcsd
  3. Apr 13, 2010 #2
    All sets have the empty set has a subset. Thus, for every set A, the empty set is in the power set of A. It is not true, however, that the empty set is an element of all sets.
  4. Apr 13, 2010 #3
    Erm... did you mean all sets have the empty set AS a subset?

    Because I don't get what you are saying.
  5. Apr 13, 2010 #4
    Yes, that is what he meant. Do you understand then?
  6. Apr 13, 2010 #5
    Erm... basically you guys are saying that the all sets has an empty set as a subset but an empty set does not exists in every set?

    That just doesn't make sense? Its there in every set, but its not there in all sets?
  7. Apr 13, 2010 #6
    The empty set is a subset of every set. It's not an element of every set.
  8. Apr 13, 2010 #7
    Maybe your confusion originates from the assumption that subsets of a set can only contain elements of that set? Well, admitted, that does sound very logical :)

    But you have to look at the definition of subset.


    B is a subset of A <=> for each element x in B, x is an element of A

    As you can see, if you fill in the empty set as B, you see that the condition is satisfied, because every element of the empty set is also in A.

    EDIT: and to understand that not every set has to have the empty set, define C = {all sets that are not the empty set}. C surely doesn't hold the empty set. Of course, maybe D = {1,2} would've been an easier example, but still...
  9. Apr 13, 2010 #8
    You are confusing the notion of subset with the notion of being a member of a set. When you say that the empty set "exists in" a set, that is equivalent to saying that the empty set is one of the elements or members of the set.
  10. Apr 14, 2010 #9
    I wonder if C is well-defined since many paradoxes arise when considering such large collections. In particular, I don't think C works well with the fact that the power set of a set is always larger than the set. Edit: nevermind, the power set of C of course contains the empty set, which is not an element of C.
    Last edited: Apr 14, 2010
  11. Apr 14, 2010 #10
    I think that only arises when there is some form of self-reference. And C is not empty so I think we're in the clear.
  12. Apr 14, 2010 #11
    As far as i'm concerned, zpconn is right-empty set is a subset of every single set, but it's not an element of it. Hence it's inproper to wirte {1,2}={empty set, 1, 2}, but as it's fine to say {1,2}+emptyset={1,2}, emtyset is a subset of {1,2}, and so on...
  13. Apr 14, 2010 #12
    This idea confused me too when I first met it ;-)

    The empty set is considered a subset of all sets, but it can't be an element of all sets. If it was, it would be an element of itself, but that would contradict its own definition as a set with no elements. So we need to distinguish between the concept of a subset and an element. (An element is also called a member.) A set is said to contain its elements. The elements are said to be in the set or belong to the set.

    For example, the set {1,2} is a subset of the set {1,2,3}, but {1,2} isn't an element of {1,2,3}. The set {1,2,3} has only three elements, 1, 2 and 3. Each of its elements is a single number. None of its elements is a set containing more than one number.

    We could define a set {1,2,3,{1,7}}. This set has four elements, the numbers 1, 2 and 3, and the set {1,2}. Subsets of this set are {1}, {2}, {3}, {1,2}, {1,3}, {1,{1,7}}, {2,3}, {2,{1,7}}, {3,{1,7}}, {1,2,3}, {1,2,{1,7}}, {2,3,{1,7}}, but not {2,7} because the number 7 on its own isn't an element of {1,2,3,{1,7}}, only the set containing both numbers 1 and 7.
  14. Apr 14, 2010 #13


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    If it existed, then the power set of C U {{}} is a subset of C U {{}} which is a contradiction.

    It's easy to show the existence of C implies "self-reference" -- the standard proof that the class of all sets is not itself a set applies to C as well. If C were to exist, the typical interesting set is the subset of C consisting of those sets that do not contain themselves.
  15. Apr 14, 2010 #14
    Hm, interesting. C doesn't exist.

    I get the second half of your post, but what do you mean by "If it existed, then the power set of C U {{}} is a subset of C U {{}} which is a contradiction."? I don't see the contradiction.
  16. Apr 14, 2010 #15
    I believe it is a contradiction with the fact that there does not exist a surjective function from A onto it's power set
  17. Apr 14, 2010 #16
    Well, I suppose my guess was right that it was related to the power set axiom...
  18. Apr 14, 2010 #17
    Oh of course.

    But one more thing: how come you have to prove that a set can exist? Bertrand Russell didn't have to prove that A = {all the sets that do not contain themselves} existed, he could just assume it did -- at least, that's what he did, isn't it? At what point did the theory of sets decide you had to check?
  19. Apr 14, 2010 #18


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    Cantor's "naïve" set theory had unrestricted comprehension -- any logical predicate you could write down corresponded to a set. Therefore, in that theory, the class A that you wrote is, in fact, a set. Russell's paradox uses A to prove that "naïve" set theory was inconsistent

    In typical modern set theories, Russell's argument is no longer a paradox, but instead a proof by contradiction that there is not a set of all sets.
  20. Apr 15, 2010 #19
  21. Apr 15, 2010 #20
    Hurkyl, that's very interesting. It sounds logical that you can proof a set cannot exist because it leads to a contradiction, but of course, Cantor's "naïve" -as you call it- set theory where you can define any set, also feels very logical, as it feels as if sets exist as soon as you define them. Seems like a pickle point, and subtle, surely. Where should I go to find out more about this kind of things? What is the name of the study concerning these matters (talking about types of logic and its axioms)? Is it a typical graduate course? Can you advice any books I can throw myself on in the holidays?
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