Can a set be member of itself?

[traldi]

Sorry if I write something stupid, but I am just a student. I really want to understand why the rejection of Cantor´s Set Theory.

Considering the Cantor´s set concept, can a set be member of itself?

Although I suppose that the answer is yes, my intuition answer no, it can´t!

"A set is a collection into whole of definite, distinct objects of our intuition or our thought. The objects are called the elements of the set."

The elements of a Set have to be definite to Set exists.
When I imagine a Set as a element of itself, it have to be definite to my intuition. This Set is not definite to my intuition, cause I need first imagine his definite elements. It never ends! Thats why a set cant be member of itself.

Would someone explain where my affirmation is wrong? thanks.

[robert Ihnot]

Consider, is this a question of intuition or how set theory is usually defined? And the further question why set theory is so defined, is something you and I have not gone into.

Consider the Axiom of the Empty Set and the Axiom of the Power Set.

Also, take the matter of whether a number N is considered a divisor of itself.

[CompuChip]

Your question reminds me of Russell's paradox (http://en.wikipedia.org/wiki/Russell's_paradox)

[CRGreathouse]

Zermelo-Frankel set theory does not allow such sets due to the Axiom of Foundation/Regularity. There are non well-founded set theories, which typically replace the Axiom of Foundation with some statement that implies the negation of foundation. The Anti-Foundation Axiom comes to mind.

Perhaps the simplest example of a non well-founded set is
\Phi=\{\Phi\}=\{\{\{\cdots\}\}\}.

[HallsofIvy]

In naive set theory, yes a set can contain itself and then you get Russell's paradox. In more advanced set theory, a "set", by definition, cannot contains sets and so cannot contain itself.

[traldi]

Thanks a lot about all the replies. The last reply answer my first question (in my context).

So, a set can be member of itself in the naive set theory. This is the common sense.

I have an argument to say that it false:

Cantor -"A set is a collection into whole of definite, distinct objects of our intuition or our thought."

My argument - To a set be definite to my intuition, it have to be definite members to my intuition. One of set´s element is not definite yet, because it is myself. So I can´t close a Set. So a set can´t be member of itself.


Would someone write me why the argument is false or write me an argument that proof that a set can be member of itself in the naive set theory.

I am not intend to change nothing, just argument to help me understand.

[CRGreathouse]

So, a set can be member of itself in the naive set theory. This is the common sense.

I have an argument to say that it false:

In set theory, the concept of "set" is undefined. Cantor gave a description, not a definition. In naive set theory that's all you have to go on.

In axiomatic set theory, sets are defined by their properties (and apply to anything with those properties).

[Dragonfall]

ZF -FA +some form of anti-foundation axiom is consistent if ZF is. So there is no sense in which non-well-founded sets don't "exist". Read Peter Aczel's "Non-Well-Founded Sets" for more info.

I spent a semester studying non-well-founded sets. There is really nothing interesting about them. The big question is when do you consider two sets to be "equal" when non-well-founded sets are introduced? The answer is that there are infinitely many ways to do that, and they're all consistent if ZF is.

[LukeD]

In naive set theory, yes a set can contain itself and then you get Russell's paradox. In more advanced set theory, a "set", by definition, cannot contains sets and so cannot contain itself.

The problem with naive set theory that leads to Russel's paradox is not the non-well foundedness though, it's the naive usage of comprehension. In no form of any non-well founded set theory is it possible to form the "set" that is used in Russel's paradox, so there is no problem.

[mathwonk]

in some set theories, the basic object is called a "class", and a "set" is defined as a class which is "small" enough to be an element of another class.

this is briefly discussed in wikipedia. there may also be an appendix in kelley's general topology on this, but it has been over 40 years since i looked at it.

[matt grime]

In more advanced set theory, a "set", by definition, cannot contains sets and so cannot contain itself.

I'm sure Halls realized he mis-spoke here, but to clarify, it is certainly possible for a set to contain another set in 'advanced set theory'.

After all, how does one define the integers a la peano? One has the empty set label that 0, a set that contains the empty set, call that 1, then a set that contains 0 and 1, call that 2, etc.

It just is not possible (in the ones that I'm aware of people using) for it to contain itself as a member.

[DavidWhitbeck]

The power set would not actually be a set by that advanced definition. So what object would it be?

[Dragonfall]

What is a class that contains a proper class called? Type-2 class?

[CRGreathouse]

What is a class that contains a proper class called? Type-2 class?

In two-sorted theories, classes are never contained in anything (though some are identified with sets, which are contained in classes and/or sets). In type theory, there's no such thing as a proper class, just ever-higher types. A type 2 set contains type 1 sets (only), etc.

In theories with proper classes you can talk about collections of proper classes, but such collections are not objects of those theories.

[Doctoress SD]

This bothered me when we first studied sets. We were told that R is a subset of R. I was like, NO WAY and I argued about it. But apparently that's the way it is.

The discussion is as follows:

Hi there,
I am writing to query on something you said in a lecture.
You said R is a subset of R.
R being all numbers between - infinity and +infinity (non - discrete, or continuous)
Whereas R+ is all positive numbers to infinity including fractions etc. And R- are all negative numbers (Both integers and non integers). Also a set of all integers, and a set of all +ve integers and a set of all -ve integers and any other set you choose to make. Would be subsets of R but R can not be a subset or R because R is R. Does that make sense?

Thank you and I hope you let me know because that has confused me a little.

Regards,

...[Me]...

...The definition is as follows.

Let A and B be sets (i.e., collections of objects, the objects being referred to as the elements of the sets).
By definition, "A is a subset of B" means that every element of A is also an element of B.

An immediate consequence of the definition of subset is that any set A is a subset of itself, because every element of A is an element of A. In particular, R is a subset of R.

I hope it is clear now.

...[Lecturer]...


This went on.

SD

[uman]

"R is a subset of R" is clearly true, but does not mean the same thing as "R is a member of R", which is clearly false.

[Doctoress SD]

So then are you suggesting there is a difference between subset and part of? I do not believe it is as clear as you state.

[uman]

{1,2} is a subset of {1,2,3,4}. {1,2} is not a member of {1,2,3,4}; the only members of {1,2,3,4} are 1, 2, 3, and 4.

[CRGreathouse]

Being an element of X is an undefined notion. The four elements of {1, 2, {3, 4}, 5} are:
1
2
{3, 4}
5

Being a subset of X is defined on the basis of "element of". Y is a subset of X if every element of Y is an element of X. So the subsets of {1, 2, {3, 4}, 5} are
{1, 2, {3, 4}, 5}
{1, 2, {3, 4}}
{1, 2, 5}
{1, 2}
{1, {3, 4}, 5}
{1, {3, 4}}
{1, 5}
{1}
{2, {3, 4}, 5}
{2, {3, 4}}
{2, 5}
{2}
{{3, 4}, 5}
{{3, 4}}
{5}
{}

[Dragonfall]

Be careful when dealing with formalities, as 2 is both an element and a subset of {0, 1, 2}.

[CRGreathouse]

Be careful when dealing with formalities, as 2 is both an element and a subset of {0, 1, 2}.

No. 2 is an element of {0, 1, 2}, and {2} is a subset of {0, 1, 2}.

{2} is both an element and a subset of {0, 1, 2, {2}}, though.

[Hubert]

No. 2 is an element of {0, 1, 2}, and {2} is a subset of {0, 1, 2}.

{2} is both an element and a subset of {0, 1, 2, {2}}, though.

I think that he is referring to the standard definition of 2 as the set {0,1}.

[CRGreathouse]

I think that he is referring to the standard definition of 2 as the set {0,1}.

It's possible, in which case he would be both correct and confusing for people unfamiliar with that definition -- including, presumably, our OP.

But for all definitions of 0, 1, and 2 the things I wrote hold:

2 is an element of {0, 1, 2}
{2} is a subset of {0, 1, 2}
{2} is an element and a subset of {0, 1, 2, {2}}

[Doctoress SD]

Lol, and 2 is a member of all three.

[HallsofIvy]

This bothered me when we first studied sets. We were told that R is a subset of R. I was like, NO WAY and I argued about it. But apparently that's the way it is.
Any set is a "subset" of itself. It is not a proper subset. Are you clear on the difference between those two concepts?

"R is a subset of R" is clearly true, but does not mean the same thing as "R is a member of R", which is clearly false.

So then are you suggesting there is a difference between subset and part of? I do not believe it is as clear as you state.
He didn't say "part of", he said "member of". Both "subset of" and "member of" have specific mathematical definitions. "Part of" does not.

[Doctoress SD]

What is the mathmatical difference between a member of and a subset of?

[Dragonfall]

"A is an element of B" can't be explained any further.

"A is a subset of B" if every element of A is also an element of B.

[CompuChip]

You might also want to read this (http://en.wikipedia.org/wiki/Element_%28mathematics%29) and this (http://en.wikipedia.org/wiki/Subset) article carefully, to try to understand the distinction between a set and its elements.

[noisysignal]

What is the mathmatical difference between a member of and a subset of?

A set is a collection of entities/elements. Consider for instance, a basket B having 5 marbles (of different color). This is our set B. The 5 marbles here are the elements/members of the set B. Now, imagine that you put your hand inside and randomly pick out elements. What you have in your hand now is a *subset* of the set B. Clearly, you could have picked no marbles (the null set) or you could have picked out all 5 marbles, in which case you have chosen the entire set B as your subset.

Now, if we (erroneously) state that a set is a member of itself, it is like saying that the basket B having 5 marbles has a basket B having 5 marbles ...

This is an indefinite recursive declaration which has no meaning and hence is different from what we mean by a subset.

I hope you understood the difference

[Hurkyl]

We can represent sets in terms of a rooted tree derived from the membership relation. For example,


*
/|
/ |
1 2

The * in this diagram denotes the set whose elements are '1' and '2'.


*
/|\
/ | \
1 2 *
/|
/ |
1 2

The top * in this diagram denotes the set whose elements are '1', '2', and { the set whose elements are '1' and '2' }.


*
/|\
/ | \
1 2 *
/|\
/ | \
1 2 *
/|\
/ | \
1 2 .
.
.

This tree depicts the structure of a set whose elements are '1', '2', and itself.


There's a slight subtlety here: this tree is not well-founded1, meaning that it contains an infinite downward path. Normally, if well-founded sets are depicted by the same tree, then they are the same set. But when dealing with non-well-founded sets, this is no longer true: there could be many different sets that have this same structure.

(Common approaches to set theory explicitly forbid such non-well-founded sets)


1: I'm a little uncomfortable with dealing with non-well-founded sets, so I'm not perfectly confident in what I've written here. (in particular, I might be talking about a closely related property rather than well-foundedness itself)

[Doctoress SD]

A set is a collection of entities/elements. Consider for instance, a basket B having 5 marbles (of different color). This is our set B. The 5 marbles here are the elements/members of the set B. Now, imagine that you put your hand inside and randomly pick out elements. What you have in your hand now is a *subset* of the set B. Clearly, you could have picked no marbles (the null set) or you could have picked out all 5 marbles, in which case you have chosen the entire set B as your subset.

Now, if we (erroneously) state that a set is a member of itself, it is like saying that the basket B having 5 marbles has a basket B having 5 marbles ...

This is an indefinite recursive declaration which has no meaning and hence is different from what we mean by a subset.

I hope you understood the difference


Oh, that makes loads more sense. Thank you very much. There is a rather large difference then. Cool.

[maverick280857]

Oh, that makes loads more sense. Thank you very much. There is a rather large difference then. Cool.

(PS--Sorry for possible repetition...haven't seen all the OPs' posts yet.)

There is a small point here, which I think you appreciate but I'll mention it for the sake of completeness. A set is a subset of itself, but as Kartik has pointed out, its not a member of itself. So you need to distinguish between a subset and a member. (I personally prefer the term 'element' rather than 'member'). There are some interesting implications of this in mathematical logic.

[Doctoress SD]

(PS--Sorry for possible repetition...haven't seen all the OPs' posts yet.)

There is a small point here, which I think you appreciate but I'll mention it for the sake of completeness. A set is a subset of itself, but as Kartik has pointed out, its not a member of itself. So you need to distinguish between a subset and a member. (I personally prefer the term 'element' rather than 'member'). There are some interesting implications of this in mathematical logic.

Ok, now you got me more confused.

[CompuChip]

If this confuses you then the concepts of member and subset, and the difference between them, still doesn't make enough sense to you.
If A is a set, then "B is a subset" means: every element of B is also an element of A.
So "a set is a subset of itself" means by definition that every element of the set is an element of the set.

[Doctoress SD]

I know that.

[jambaugh]

See Finsler Set Theory. Finsler played with the idea of self containing sets. Basically you construct sets as nodes of directed networks.


To avoid Russel paradoxes you must be absolutely constructive in defining these sets. However you have more flexibility in that you start with directed nets and see if a node satisfies the requirements of being a Finsler Set.

[maverick280857]

Ok, now you got me more confused.

Sorry about that. I've been away a few days so I couldn't get a chance to reply. Anyway, I hope your confusion has been removed.

[DeaconJohn]

Consider, is this a question of intuition or how set theory is usually defined? And the further question why set theory is so defined, is something you and I have not gone into.

Consider the Axiom of the Empty Set and the Axiom of the Power Set.

Also, take the matter of whether a number N is considered a divisor of itself.

Robert,
Very thought provoking comment about the relationship of the "matter of whether a number N is considered a divisor of itself" and the status of self-referencing statements in mathematics.
Let's see if I can make a first step at formalizing your "matter." (Well put! by the way.)
Let S be the set of all natural numbers "n" such that "n divides n."
OK, that's a shot. If "n" is a divisor of itself (obviously just a matter of definition in this case), then S = N. (N = {1, 2, 3, ... } = the set of natural numbers.
Note that Russell's paradox works with the set of all sets that are NOT elements of themselves. It's not the "matter" of sets that are elements of themselves that causes the problem, it's the self-referencing nature of the matter.
If I remember correctly, the nature of self-referencing statements are the essence of Godel's proof of his theorem.
For a "popular" example, consider the statement p = "This statement is false." Ask the question whether p is true or false. If p is true then it is false. If p is false then it is true.
Hmmmmmmmmmmmmmmmmmmmm.
Awesome are the mysteries of the universe that we live in.
DeaconJohn

[dream runner]

Alright, it seems logical that in ZFC set theory, a set cannot be a member of itself. My question pertains to the definition of ordinals. According to Thomas Jech's edition of set theory, a set is ordinal if it is both transitive and well ordered by membership. I've been poking around trying to find an example of a set which is transitive and not well ordered by membership and only two possibilities seem to arise:

1: By definition, a set A is not well ordered by membership if there exists some subset B of A does not contain a least member. Thus every element b within B implies the existence of another element c within B which is also an element of b. This seems recursively to lead to an infinite set B.

2: A common counterexample I've seen is when proving that the class Ord of all ordinals is a proper class for else it would contain a member [alpha] which is an element of itself, and thus not well ordered. Furthermore, by our above discussion, no set exists such that it is a member of itself and thus no set exists containing a set which is a member of itself, as this flies in the face of the Axiom Schema of Seperation.

It seems that both cases pose a stark contradiction and thus imply that no such set exists. However, absurdity is not a requisite for truth, thus I would like to know, since all evidence seems to point towards the contrary, whether it is even possible for any set A to be transitive by the definition that every element of A is also a subset of A, and yet not be well ordered by membership.


P.S. My idea of a set here does not include any urelements, which seems rational, for consider the set

A = {b, {b}}
The only way this set can possibly be transitive is if b is a subset of A, and the only non set element which inherently a member and a subset is the null element, thus
A = {null, {null}} is transitive, and adding another element would require that element be the successor {{null}}. Much like a russian nesting doll.

[Hurkyl]

1: By definition, a set A is not well ordered by membership if there exists some subset B of A does not contain a least member. Thus every element b within B implies the existence of another element c within B which is also an element of b. This seems recursively to lead to an infinite set B.
You have an omission here -- you're implicitly assuming that membership is an order relation. However, that need not be true, even for transitive sets. For example, define:
A = {}
B = {A} = {{}}
C = {B} = {{{}}}
D = {A, B, C} = {{}, {{}}, {{{}}}}
D is a transitive set. However, membership does not satisfy the trichotomy law: all three of the following statements are false:
A is a member of C
C is a member of A
A = C

[dream runner]

Aha!!! Thank you so much. I was spending too much time studying ordinals, in which membership is a linear ordering before thinking of this question.

Sometimes thinking too deeply about another topic can lead me into a quagmire such as these.

[dream runner]

And I haven't gotten to the trichotomy axiom quite yet, and how it applies to binary relations. So I'm assuming in order for the membership relation to be considered as an order relation it must be trichotomous, which makes sense. Very interesting.