# Can a set be member of itself?

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.

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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
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CRGreathouse
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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
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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.

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
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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).

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.

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
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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
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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.

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

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

CRGreathouse
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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.

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:

Myself-10 November 2007 said:
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

"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.

{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
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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}
{}

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

CRGreathouse
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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.

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
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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}}

Lol, and 2 is a member of all three.

HallsofIvy