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  • If we define (i.e., by definition) a set as an abstract collection of at least one element, and
  • If we say that by nature (i.e., as an axiom) all sets contain one element that is the set itself at the same time

then

  • There is no empty set, and
  • All sets contain themselves, and
  • There is no set of all sets that do not contain themselves, i.e., there is no Russell's paradox.

As an example, the set of all cats would have a cat member that is, at the same time, the set of all cats. Also, the set of all numbers would have a number that is, at the same time, the set of all numbers.

There would be a casting test operator. With it you could ask "Is this cat, at the same time, the set of all cats?". Or you could answer questions like "Is that cat a member of this cat?" with an answer like "Yes, because this cat is, at the same time, the set of all cats". Another one: "Does the set of all cats is a member of itself?" with answer "Yes, because one of the cats is, at the same time, the set of all cats".

Do you see any contradiction in a theory sketch like this?

Do you see any contradiction if we add a universal set as an axiom? i.e.:

  • For all elements, there exists a unique set that contains all of them.

I have no degree in philosophy and I am also not a native english speaker, however, I have been humble while asking this question. Please, be humble while answering it.

Edit: I accepted waf9000 answer because it is true that with traditional equality axioms this theory sketch is contradictory. I will keep experimenting with new equality axioms outside this question. Thanks a lot for your time and patience.

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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Philosophy Meta, or in Philosophy Chat. Comments continuing discussion may be removed.
    – Geoffrey Thomas
    Feb 28, 2023 at 8:39
  • If you define sets in terms of "abstract groups," then how are you defining groups? I'm assuming you're not using the usual definition form abstract algebra.
    – Sandejo
    Feb 28, 2023 at 21:25
  • Fixed it. Thanks. Mar 1, 2023 at 12:23
  • While your edit did fix the issue with the word "group" already having another definition, you didn't fix the main issue, with the ambiguity in your definition of a set. In particular, what defines an "abstract collection?"
    – Sandejo
    Mar 2, 2023 at 7:26
  • I currently don't have one. I will try to be developing a more formal one in the future. It has the property of having one or more elements only once and with no order in particular. I will change "theory" with "theory sketch". Thanks. Mar 2, 2023 at 14:58

10 Answers 10

9

I think we can find a contradiction with ZFC immediately:

Let S = {A, B} with A =/= B.
WLOG, by your hypothesis, assume that A = S. (One of these elements needs to equal the whole set!)
Then, also consider the set T = {A}. We can construct this by using the Axiom of Specification to find a subset of S that doesn't contain B.
Again, by your hypothesis, we have A = T; by transitivity of equality, this means S = T.

By definition, we have B is not an element of T, but since S = T, we have B is not an element of S as well.

Now we have a contradiction: we've shown that B is both an element and not an element of S.


It might be possible to come up with some replacement for our axioms about equality, but I don't think its worth working that hard just to avoid Russell's paradox this way, especially when we lose things like the empty set, operations like set intersection, etc.

Another 'out' might be that it's not possible to find two unique elements A =/= B, and there's only one element A with A = {A}! Although this is probably a less interesting theory.

1
9

I think that disallowing the empty set will cause the resulting theory to lose too much. For example, the intersection of two disjoint sets, which would be the empty set, is now undefined. You will probably have trouble defining subsets too, at least if you want to define a subset of S as "the elements of S such that ...", because that could very well be an empty set, which is again not allowed. You can no longer write {x in A: P(x)} without first verifying that there exists an x in A such that P(x), which seems really inconvenient.

1
9

The axiom of foundation is not (usually) taken for a logical truth (although see about predicativism for an attempt to situate well-foundedness on the level of predication theory). Accordingly, its basic negations—"Some sets are elements of themselves," and, "All sets are elements of themselves"—seem logically possible on their own terms.

Now there is, moreover, a famous alternative set theory, or family of such theories, that heads somewhat in the direction you're going: Quine's New Foundations theory, especially one that is founded on self-singletons (AKA "Quine atoms"), which are otherwise ur-elemental terms. See the one answer to this MathOverflow question for a broad overview of how to form a cumulative hierarchy over Quine atoms (see this essay for more detailed assessment of these formations, i.e. the paradoxical-looking expression WF(At) (where WF is the class of well-founded sets and like when we talk about relative constructibility in terms of L(X), we use the class symbol to preface a parenthesis covering what the class is applied to, in this case WF(X) takes At, which is a class of Quine atoms, and outputs an externally well-founded cumulative hierarchy, despite having a internally looping base)).

Moreover, NF has a universal set, but the principle for forming it& blocks the formation of things like the deviant Russell set (I'm not sure why, actually, since on the one hand, a self-elemental term, even in negative form, apparently violates the stratification principle that Quine uses, yet there is a stratification-friendly formula that seems like it would imply a self-elemental formula, and indeed from what I have read, Quine's universal set does contain itself after all, so I'm not sure "what gives").

I noticed in one of your comments that you suspect we can reformulate Russell's paradox even if we suppose multiple types of elementhood relation. Famously, theories of proper classes modify the basic elementhood relation at least in the sense that a class C can have elements but not be an element, so this modification does reliably block a revenge Russell set. (C.f. the two flavors of extensionality in double-extension set theory.)

Finally, whether we would want a theory where every set contains itself depends on how far we press the imagery of "containers," "bags," etc. in this connection. Since it is easy to imagine empty boxes, empty cupboards, empty rooms, empty bottles, etc. then the metaphor of the terminology seems to require the possibility of empty sets. On the other hand, one might compare "being located in a space" to "being contained in a set," and then if space seems to contain itself, and indeed contain itself at every point,&& perhaps there is an alternative physical metaphor that we could avail ourselves of to describe how all sets might be elements of themselves.


&Though a somewhat "convoluted" argument is required to get from an unrestricted set to a Russell set (something based a lot on the axiom of separation, say), another argument against an unrestricted set is from the powerset axiom: if some universal set U has a powerset, and if every powerset has more/different elements than its base, then U would either be nonuniversal after all, or both universal and not universal, or a degenerate element of itself, or something else horrible-sounding. Accordingly, however, suppressing the powerset axiom is one way to get around this problem: if U doesn't have a set outside of it that contains all subsets of U, then Cantor's theorem doesn't kick in, here. Or, most acutely in this case, if U has no subsets at all, then axiom or no, there is no set of all its subsets (and indeed, to represent some elements of U as elements of a subset of U ultimately requires separation-axiom argumentation, for example). At any rate, and this floors me even more to read about, but Quine's universal set runs the risk of being larger than its own powerset, which is baffling to me (I have some intuition for incommensurable discrepancies between base sets and their powersets, as in the pre-choice dialectic of the Continuum, but a commensurably smaller relation...?).

&&Though consider also Kant's remarks in the Transcendental Dialectic about the universe as a whole being located in no specific place.


ADDENDUM

Note that there is a possibly well-founded option that is still a set and is as close to universal as can be: a couniversal set:

Xy((yX) ↔ (yX))

Accordingly ~(XX), but everything else is an element of X. If there is no fully universal set above this couniversal X, then X is not a set of all sets but it is a set of all elements. The existence of this thing is dimly possible in light of the logical contingency of the separation axiom, since we can define a concept of a set that is inconsistent with that axiom, e.g. an "impregnable set" such that its elements are necessarily confined to it and cannot be extracted from it to form external subsets. Moreover, then, we can define a set as such that it is not an element of some other set, and though it be granted that we proceed here from maybe-arbitrary to maybe-arbitrary stipulative definition, the point is that a couniversal set can be well-founded and play a sufficient counterpart role (or maybe an even better such role!) vs. universal sets proper, or (proper) classes, in other theories. Not much help for your theory, I suppose, except to form a stark contrast that can help you identify what is really at stake in the question of unrestricted quantification.

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  • 2
    Kristian. Thanks a lot for your time. Let me review and try to learn many of the things you mention in your post. I was trying to find a theory that has an universal set and no Russell's paradox, that happened to contain no contradictions. The point of the power set caught my attention. I'll be learning more about it to see if it happens to make the theory contradictory. Regards. Feb 27, 2023 at 21:37
  • 2
    Nice answer! A lot to pick from and digest.
    – Nikos M.
    Feb 27, 2023 at 21:55
  • Kristian is a beast.
    – D J Sims
    Feb 4 at 12:43
5

The question does not define a set theory, it merely investigates the domain of non-empty sets that contain at least themselves. This domain could have a name, like NESCT.

With ordinary set theory, we can describe that domain NESCT. As an example we can tell that the union, the intersection or the difference of any 2 sets in NESCT is typically not in NESCT, and that the empty set is not in NESCT. Thus NESCT is not a normal domain of sets, as opposed e.g. to the domain of sets of even numbers.

Trivially also the set of all sets that do not contain themselves is not in NESCT. But that does not make NESCT a special domain, most commonly used domains do not contain such a set.

Rewriting all that as "new definitions" does not change the outcome, it just confuses the matter.

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  • 1
    This theory says that all the sets contain themselves, and therefore, that there are no sets that do not contain themselves. Feb 28, 2023 at 1:18
  • 2
    It's not a theory, you describe a domain. It's silly to try to change facts by twisting definitions. The facts never change by changing definitions.
    – tkruse
    Feb 28, 2023 at 7:35
3

If you want to be able to do mathematics, it's useful to have some sets that don't contain themselves. For example, if I give you a quadratic equation to solve, you might be able to give me the set of all solutions to that equation. Or you might want to reason that every positive integer is either odd or even - in other words, the set of positive integers is a subset of the set of odd numbers union the set of even numbers. Except that in your version of set theory, it isn't.

So I think if mathematicians started down the road of adopting your kind of set theory, then after a while, they'd start operating in an alterative set theory - closer to the one we currently know and love. Because if they didn't, they wouldn't be able to get terribly much done.

2

I don't see why you couldn't just write down a model fitting your requirements - given a model of set theory M, I where the elements of M are proper sets and I is the proper containment relation, you could certainly construct a structure with relation M', J, U.

M' would just be the union of M with {U}, where U is the constant denoting your fake set of all sets. J is the union of relation I with the diagonal of M plus the addendum that U contains itself and everything in M.

This obeys your constraints, and has the bonus that once you take out U and ignore the diagonal part of J you can go back to useful set theory.

1
  • It feels like other answerers are adding pre-conceptions of expected behaviour of sets, but the question has defined sets for this context and doesn't need them to do useful things.
    – Cong Chen
    Feb 28, 2023 at 16:20
1

The "answer" to Russell's "set of all sets that do not contain themselves" is that such a set does not exist.

Part of being a set is that it must be unambiguous whether any given thing is a member of the set or not. For the supposed set according to Russell's form, it is not clear whether the set contains itself or not. Thus, this purported collection is not a set.

There are a number of ways that people have attempted to proceed after that. For example, some people have attempted to introduce other types of collections and relax the requirement of unambiguous membership.

But set theory is already rescued.

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  • I think that introducing other type of collections only create a new Russell's paradox for that type of collection. I have read that modern theories like ZFC don't allow for the existennce of a universal set, which I don't think make sense. Feb 27, 2023 at 19:43
  • 1
    @OtakarMolnárLópez The new type of collection is the whole point. It rescues set theory. The other type of collection is not required to have all the properties sets do, because it is not needed to be the basis of the various parts of logic and mathematics that set theory is.
    – Boba Fit
    Feb 27, 2023 at 20:34
1

I am not certain that I understand your proposal. I will explain my understanding, and critique your proposal on that basis. Please tell me if and how my understanding is incorrect, and I will revise my answer for you.

For clarity let us use the label set to denote a conventional set and newset to denote the new objects you are proposing. My understanding of the difference can be illustrated as follows...

Suppose there are n Faberge eggs. The set of Faberge eggs would have n members, and the set would not be one of those members. The newset of Faberge eggs would have n+1 members, n of which are Faberge eggs and one of which is the set of Faberge eggs. Is that correct?

If it is correct, then the newset is simply a set- a set which does not contain itself.

ADDENDUM

Thank you for clarifying that you meant that a newset is intended to mean a set in which one of the members (presumably selected at random) is nominally at the same time the set of all the other members including itself. You still are left with a problem. Consider the set of Faberge eggs, which contains n elements, each being one of the n Faberge eggs. Your newset is also a conventional set, a set containing n-1 elements that are Faberge eggs plus one element that is both a Faberge egg and a set of all Faberge eggs. That conventional set does not contain itself.

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  • No. One of the eggs is at the same time the set of all eggs. I have been thinking and I think I must define a cast test operator and a cast operator. With the cast test operator you could ask "Is this egg, at the same time, the set of all eggs?". And if the answer is yes, then, with the cast operator you could talk about the egg as the set of all eggs, e.g., "Does this egg, as the set of all eggs, contain that other egg?". I must define this in the question to stop the confusions. Thanks a lot for your time. Feb 28, 2023 at 11:59
  • Let me see if I understood the addendum. There is the set of Faberge eggs and a set with Faberge eggs that are not the set of Faberge eggs plus the egg that is the set of Faberge eggs. Aren't both the same set since both have the same elements? Feb 28, 2023 at 13:53
  • I thought not, since the egg that is both an egg and a set must be different from an egg that is just an egg. No egg is literally a set of other eggs, so the property of being the set must be assigned to one of the eggs. That egg then differs from the others by being the only egg assigned that property. I must say, it is a fascinating topic. No wonder faberge eggs are so sought after! Feb 28, 2023 at 14:06
  • Well, in this theory, the egg that is the set of all eggs is literally the set of all eggs. Feb 28, 2023 at 14:11
  • 1
    But there is a problem. Suppose I wanted to form the set of sets that have only one member. In your scheme, you would have to make one of the members of that set a special one that was also the overall set, and thus it would no longer have only one member so it could not belong in the set. There is no way out of that contradiction. Feb 28, 2023 at 14:52
1

Let me denote by # the usual "is an element of" sign, which is denoted by $\in$ in LaTeX.

Let U be a set, whose elements are called Otakar-sets. Let R be a binary relation on U. For all x,y # U, xRy is read "x is an Otakar-element of y".

Your axioms are:

  1. For all x # U, xRx.
  2. There is u # U such that for all x # U, xRu.

This is not a contradictory theory: it has a lot of models! Take any non-empty set U, any element u of U and R := =_U UNION ({(x,u) | x # U}) (recall that =_U denotes the set {(x,x) | x # U}).

0
0

These can't all be true:

  1. The set of all cats isn't a cat
  2. The set of all cats only has cats in it
  3. The set of all cats contains the set of all cats

So, which one is false? I think we can easily agree on point 1. Point 2 is true by definition: if "the set of all cats" had something in it that wasn't a cat, it wouldn't be called "the set of all cats". So point 3 must be false.

Or, there is no set of all cats.

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  • In this theory, the set of all cats is a cat at the same time. With the casting test operator the answer to questions like "Is the set of all X an X?" would always be "Yes, because all sets contain one element that is the set itself at the same time", i.e., the first axiom. Feb 28, 2023 at 13:17
  • @OtakarMolnárLópez A cat has 4 paws. How many paws does the set of all cats have? Several million?
    – user253751
    Feb 28, 2023 at 13:17
  • In this theory, the set of all cats contains, as a cat, 4 paws. Feb 28, 2023 at 13:18
  • @OtakarMolnárLópez then I think, it is obvious, that the theory makes no sense. How long is the face of the set of all cats AND horses?
    – user253751
    Feb 28, 2023 at 13:26
  • Well, it might be that a cat is, at the same time, the set of all cats AND horses. Or it might be that it is a horse. If it is a cat, then, its a small face, if it is a horse, then it is a long face. This element has both natures, and maybe more natures: its element nature (as a cat or horse that it is) and its set nature (as the set of cats AND horses). And the set of all cats AND horses is a member of itself because of this element, that is maybe a cat or maybe a horse, but not both. The criteria of which element is the set itself can be choosen freely. Feb 28, 2023 at 13:37

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