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Very Easy Set Theory Question

  1. Aug 17, 2006 #1
    I want to make sure I understand the meaning of membership and subset.

    For example, if I have a set x, then is x a member/subset of the set

    S = {{y},x}

    I came to the conclusion that x is a member of the set S because S contains x as an element, and x is also a subset of S because S contains the set x and x is a subset of x. This doesn't sound right, I've used the same argument for two different things?? Hmmm...

    But if

    S = {y,{x}}

    then x would NOT be a member of S because the set S does not contain the set x as an element (but it does contain the set {x} as an element and {x} \neq x). And x is also NOT a subset of S for reasons I cannot think of.
    Last edited: Aug 17, 2006
  2. jcsd
  3. Aug 18, 2006 #2
    Also, if I have the power set P(x), is x a member of P(x)?

    If the set x is to be a member of the power set P(x) then P(x) must contain the set x. For example, let x = {a,b} then P(x) = {{},{a},{b},{a,b}}. And since if I remove the outer brackets I have {}, {a}, {b}, and {a,b} as members, and there is {a,b}! Since this will happen with whatever set x I start with, x will always be a member of P(x).

    Is x a subset of P(x)?

    I figured that since P(x) consists of all subsets of x and since x is a subset of itself, then x is a subset of P(x). Is it true that for every set x, x is always a subset of the power set P(x)?
  4. Aug 18, 2006 #3
    Yes x is a member of S, but x is not a subset of S, however {x} is a subeset of S.

    This is correct.
  5. Aug 18, 2006 #4
    I take it that x is a set? Then yes x belongs to P(x).

    P(x) is the SET of all subsets of x, so yes x is a subset of x and hence it will belong to P(x), but remember the conditions on a subset

    A is a subset of B if for every element a in A a is also in B.

    You should notice that none of the elements of x will belong to the power set of x and hence x is not a subset of P(X).
  6. Aug 18, 2006 #5
    Right. So {x} is a subset of S because every element of {x} is in S (namely the only element of {x} is x, and x is in S). However, not every element of x (say z) is in S because S contains only two distinct sets (namely x and {y}) and z is not one of these.
  7. Aug 18, 2006 #6
    Also that x is not necessarily a set.
  8. Aug 18, 2006 #7


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    That's not strictly true; if x is a transitive set, then it will be a subset of its power set. (if and only if)
  9. Aug 18, 2006 #8
    Thanks I didn't know that, is that the only case where a set will be a subset of its powerset?
  10. Aug 18, 2006 #9
    Yes, I will always be assuming that x is a set.
  11. Aug 18, 2006 #10
    But what if my set was not P(x) but instead P({x})!?

    P({x}) is the set of all subsets of {x}. But {x} is a set of one element, namely x, so P({x}) = {{},{x}} does it not? and x is not an member (but {x} would be!?).

    Is x a subset of P({x})? No, would be my answer, because we cannot guarantee that a member of x is in P({x}) which contains only the empty set and the set containing the set x.
  12. Aug 18, 2006 #11
    Yes that would be correct again.
  13. Aug 18, 2006 #12


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    I don't know if it will help, but (I think) if you assume the axiom of foundation, you can actually represent sets as trees. For example, the set {x, {y}} can be represented as:

    Code (Text):
      / \
     /   \
    x     *
    In this representation, an asterisk denotes the top of set, and all of the children of an asterisk are the elements of that set. (If x and y were themselves sets, you could substitute their trees into the above diagram)

    I say "I think", because I'm pretty sure that the axiom of foundation guarantees every set can be drawn like this, but not 100% sure.
  14. Aug 18, 2006 #13


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    Technicalities again. The answer is yes if x is the empty set, or the set containing the empty set. (Or, if we reject the axiom of foundation, and allow x = {{x}} or x = {{}, {x}})
    Last edited: Aug 18, 2006
  15. Aug 18, 2006 #14
    Perhaps this will help: I will always assume that x is non-empty, x is not an member of itself, and y is not x.
  16. Aug 18, 2006 #15
    The Axiom of Foundation has not been covered in lectures yet, so Im pretty sure Im not meant to use it. However, your idea may come in handy never-the-less...I like it.
  17. Aug 18, 2006 #16


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    Well, allowing recursion, and sets with transfinite depth is a sort of trappy thing, and not really needed for doing ordinary mathematics. So for most purposes, you're going to want to assume foundation. I mainly just want you to be aware of its necessity. :smile:
  18. Aug 20, 2006 #17
    What about U P(x), that is the union of the power set?

    I figured that since the union is "like" bracket removal, x will not be a member. For example, take x = {1,2}. Then

    U P(x) = U {{},{1},{2},{1,2}} = {} U {1} U {2} U {1,2} = ?? not sure what goes here.
    Last edited: Aug 20, 2006
  19. Aug 20, 2006 #18
    I was thinking, (assuming that x is not a member of itself and is non-empty), x is not a member of x BUT x is a member of {x}. So, would x be both a member AND an element of the set {{x},x}?
  20. Aug 20, 2006 #19


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    Note that, in your particular case, U P(x) = {1, 2} = x. Can you prove the general case?

    Yes... but member and element are synonyms.

    I think you meant to ask:

    Would {x} be an element and a subset of {{x},x}?

    And the answer would be yes.
  21. Aug 20, 2006 #20
    ...you read my mind! ;)
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