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Question re: sets
How can a set contain itself as a member?

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Question re: sets

graymath
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Posted 08/17/09 - 06:46 AM:	#21
Why is there a problem with "the set of all sets"? I accept that it is a purely conceptual construct, but I cannot see an obvious problem with it. The set A contains all sets A is a set Therefore set A contains A. Unless there is some definition of set which does not allow for the membership to other sets, I don't see any contradiction. Considering, for example, that "The Set of All Mammals", is a member of "The Set of All Sets Containing Members of the Animal Kingdom", I don't see any problem with sets being members of other sets. I believe the source of the confusion may lie in the choice of language (sometimes) used with set theory. The notion that a set "contains" items, seems to suggest the existence of some spatial (whether conceptual or otherwise) domain in which these items exist. Given this, it seems absurd to suggest a set can contain itself as an item, since the item contained is supposed to be equivalent to the containing item - which, I accept, involves a contradiction. But membership to a set implies the existence of a property common among all the sets' members. It does not provide a complete definition of the object. For example, "dog" belongs to the set of all mammals, but it also belongs to the set of all four legged animals. The properties of "dog" used to determine its membership are different for each set. Membership to a set rarely provides an exhaustive definiton of the item concerned. As Russell showed, there is, however, a contradiction involved in "the set of all sets which are not members of themselves" (ie - if it is a member of itself, then it isn't, if it isn't, then it is).

Elotemuygrande
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Posted 08/22/09 - 04:49 PM:	#22
swstephe wrote: From my background, I don't think it is usual set theory to define a set as containing other sets. Set members are usually atomic, or the empty set. Maybe there are some theories that attempt this, but you run into conflicts. Out of practice former math major here. I would say is normal for sets to contain other sets. Russell once gave a definition of what a natural number is in terms of a set of sets that have one to one mappings between their members for example. Using that definition you could look at the set of natural numbers as a set of sets containing sets

Elotemuygrande
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Posted 08/22/09 - 04:57 PM:	#23
swstephe wrote: Had to look it up -- but it seems the criticism is shared, that members of sets should all be constructable objects. But the criticism isn't proven, (since it depends on some loosening of terms). All responses to Russell's paradox involve restricting the member set. Even Russell, himself, restricts "theory of types" to hierarchical, (rather than functional), definitions. Even so, when you talk about a set being a member of itself, the set has to be an empty set, or they aren't the same set anymore. I'm not sure I agree with the aspect about the empty set, at least according to the view I was presented: {} - contains no members { {} } - contains one member, the empty set Also, the power set(set of all subsets of a set) of a set will have cardinality 2^n. In the case of the empty set this would be 2^0=1.

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Posted 08/23/09 - 11:51 PM:	#24
Arkady wrote: Question: as someone who is nearly totally new to set theory, I would like the following conundrum resolved: A set can contain itself as a member, correct? Not in ZFC. So, consider set [A], which contains itself, along with a number of other sets as members: A = [A B C D] Is this not logically equivalent to the proposition: "All A's are A's BUT not all A's are A's"? What you're trying to say is unclear.

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Posted 08/24/09 - 12:00 AM:	#25
Arkady wrote: I see your point. I shouldn't have said "All A's are A's." If "are" is taken to mean "equals" (which is what I intended), then that would mean that all members of A are equal to A itself, which I definitely don't mean. So, let me re-phrase my argument a bit. [A]=[A], as per the basic principle of self-identity: anything is equal to itself. Let's say: [A]=[1,2,3] [B]=[4,5,6] [C]=[7,8,9] [D]=[10,11,12] Now, [A] contains itself as a member, along with [B], [C], and [D]. So: [A]=[[A] [B] [C] [D]], which is equivalent to saying: [A]=[1,2,3,4,5,6,7,8,9,10,11,12] So, we have an [A] which contains 1,2,3, and a "different" A, which contains 1,2,3,4,5,6,7,8,9,10,11,12. But, this should be impossible, given that [A] should be equal to itself. So, either sets can't really contain themselves as members, or [A] is somehow not equal to itself. You are confused. First, you're using brackets incorrectly. Second, you define A as {1,2,3}, but then later define it as {A, B, C, D}. Third, you do not seem to even have an intuitive understanding of what a set is. If A is a set that has B as a member where B = {1,2,3}, it does not follow that 1, 2, and 3 are members of A. They are members of a member of A and members of the union of A, but this is something different.

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Posted 08/24/09 - 12:17 AM:	#26
Elotemuygrande wrote: swstephe wrote: Had to look it up -- but it seems the criticism is shared, that members of sets should all be constructable objects. But the criticism isn't proven, (since it depends on some loosening of terms). All responses to Russell's paradox involve restricting the member set. Even Russell, himself, restricts "theory of types" to hierarchical, (rather than functional), definitions. Even so, when you talk about a set being a member of itself, the set has to be an empty set, or they aren't the same set anymore. I'm not sure I agree with the aspect about the empty set, at least according to the view I was presented: {} - contains no members { {} } - contains one member, the empty set Also, the power set(set of all subsets of a set) of a set will have cardinality 2^n. In the case of the empty set this would be 2^0=1. Yes, you are correct. I will also mention that the definite article is used to speak of the empty set. That is, there is only one empty set. This is a consequence of extensionality (if A and B are empty sets, they have the same members, therefore, A = B).

realistcat
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Posted 08/26/09 - 03:49 PM:	#27
all of this depends on your theory of sets. there is a theory of sets that we might call "concrete group theory." suppose I see two cats curled up on the sofa, then there is group cats curled up on this sofa. This group exists only in virtue of the two cats that make it up and has no reality beyond those two cats, according to concrete group theory. We can define first-order groups as groups whose members are all concrete particulars. The cats and the sofa are also a first-order group. Because they have no reality beyond the particulars that make them up, there can be no empty set. If i say There are zero cats on the bed we can take this as equivalent to: the property cat on the bed is not instantiated. This is true because there is no actual state of affairs of a cat being on the bed. The only entity presupposed here is the property cat on the bed. First-order properties are properties of concrete particulars. so second-order sets are groups that include first-order properties or first-order properties and concrete particulars. there is here no idea of order within concrete sets. as long as the base entities in set A are the same as set B, it's the same set. I suppose one objection to the theory of concrete sets is that it can't support infinitistic mathematics, and that's true. there aren't enough concrete particulars. but note that in computerland there is also no infinitistic mathematics. Not strictly speaking. that's because all computer brains are finite. so calculators and computers have to resort to methods of approximation for infinitistic math.

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