there wasn't enough clarity with regards to what it is for a set to be a member of itself, and what it is for a set to not be a member of itself. — Philosopher19

Usually, we have an intuitive notion that sets are not members of themselves. However, since 'is a member' is primitive, we will not have a formal explication of the notion.

This is handled typically in three ways (not necessarily in order of preference):

Z set theory includes the axiom of regularity. And the axiom of regularity proves that no set is a member of itself.

ZFC-R set theory drops the axiom of regularity. The existence of a set that is a member of itself is independent of those axioms.

Z-R+D set theory [where D is an appropriate anti-foundation axiom] set theory drops the axiom of regularity and adds an axiom so that the theory proves that there are sets that are members of themselves.

/

Call any set that is not a member of itself a -V. Call any set that is not the set of all sets a V'. Call any set that's simply a set, a V (the V of all Vs = the set of all sets). — Philosopher19

Your notation is mixed up (especially as you conflate 'V' as naming an object with 'V' standing for a predicate). I'll use ordinary notation:

Is the V of all -Vs a member of itself? — Philosopher19

First we must ask whether ExAy(yex <-> ~yey). The answer is no. It is a theorem of first order logic:

~ExAy(yex <-> ~yey)

However, toward a contradiction, we do start with the assumption:

ExAy(yex <-> ~yey)

Then we show xex & ~xex.

Thus ~ExAy(yex <-> ~yey).

It is impossible to have a V of all -Vs that contains all -Vs and no other sets. — Philosopher19

Correct. ~ExAy(yex <-> ~yey).

You cannot have a set of all sets that are not members of themselves that is itself not a member of itself. — Philosopher19

Correct, because there is no set of all sets that are not members of themselves anyway.

one V can contain all -Vs and something more. — Philosopher19

Incorrect. ~ExAy(~yey -> yex). The "and something more" doesn't get around that fact.

That's enough for now; I'm not at this time going on with the rest of your argument, especially to untangle your poorly chosen notation.

he set of all sets encompasses all sets. It encompasses all sets that are not members of themselves, and it is a member of itself (because it encompasses itself). No contradictions here. — Philosopher19

You didn't go far enough in the argument:

If there is a set of all sets, then it has the subset that is the set of all sets that are not members of themselves. But the claim that there is the set of all sets that are not members of themselves implies a contradiction. Therefore, the claim that there is a set of all sets implies a contradiction.

Suppose B is a set that contains itself:.{1, 2, {1, 2, {1, 2,..the task can't be completed. — TheMadFool

Set theory doesn't prove things in this kind of context by saying "the task cannot be completed".

However, the axiom of regularity disallows infinite descending membership chains and
circular memberships (so also no sets being members of themselves).

Can we construct a set of all sets that don't contain themselves? Why not? — TheMadFool

Because the axioms don't provide for such a construction while also the axioms prove that there is no such set nor such a construction.

Of course we can because ALL sets can't contain themselves. — TheMadFool

It is correct that with regularity, no set is a member of itself. But that doesn't imply that there is a set whose members are all and only the sets that are not members of themselves.

.
1. All sets don't contain themselves.
2. The set of all sets that don't contain themselves = The set of all sets.
3. The set of all sets is impossible because it can't be member of itself and so it can't be the set of all sets. (from 1)
4. The set of all sets that don't contain itself is also impossible (from 2 and 3).
5. For Russell's paradox, the set of all sets that don't contain itself must be a set.
6. The set of all sets that don't contain itself is impossible i.e. it isn't a set.
Ergo,
7. Russell's paradox is not a paradox. — TheMadFool

1. Correct. With regularity, ~Ex xex

2. You cannot refer to "the set of all sets" without first proving there is such a set. And we prove that there is not such a set.

3. Correct. If there were a set of all sets then that would violate regularity. But, we don't need regularity (used for your item 1.) to prove there is no set of all sets.

4. Correct. But again, we don't need regularity (used for your item 1.) to prove there is no set of all sets.

5. What do you mean "for Russell's paradox"? The role of Russell's paradox is this:

Suppose ExAy(yex <-> ~yey).

Then xex <-> ~xex.

Therefore, ~ExAy(yex <-> ~yey)

6. Correct.

7. What do you mean by "it's not a paradox". The paradox (the contradiction) comes from the claim that there is a set of all sets that are not members of themselves. That contradiction implies that there is not a set of all sets that are not members of themselves.

Now consider this: x, y, and z, are sets that are members of themselves. It cannot be the case that x = {x, y, z} [...] — Philosopher19

Wrong. It is possible that all these are the case: x = {x y z} (so x is a member of itself), and y is a member of y, and y is a member of x, and z a member of z, and z is a member x.

That's enough for now. At this time, I'm not going further with your argument.

But a couple of points:

Do you see the inconsistency in saying "you cannot have a set of all sets that are not members of themselves, but you can have a set of all sets that are members of themselves". — Philosopher19

It's not inconsistent. It does not imply a contradiction. A contradiction is a statement and its negation. If you claim to point out an inconsistency, then you need to show a statement and its negation that are both implied.

saying there can be no set that encompasses all sets is blatantly contradictory — Philosopher19

Nope. You have not shown that it implies a statement and its negation.

It's a contradiction for something to be a member of itself twice. — Philosopher19

"member of itself twice" has no apparent mathematical meaning.

It's a contradiction for something to be a member of itself twice. — Philosopher19

Why don't you look up a text in set theory so you would know how set theory axiomatically, clearly and unambiguously proves theorems and defines terms?

I don't like using markup. The text is plenty clear enough. — TonesInDeepFreeze

FWIW I think I've convinced myself that you're right. Under regularity, the collection of sets that are members of themselves is indeed the empty set. In fact the empty set is often defined as $\{x : x \neq x \}$. It's always bothered me that there is no containing set as required by specification. But this is a standard definition. So maybe it's ok. I don't know why it's ok. I should look into this.

I didn't go through your proof yet but perhaps I'll make a run at it.

Set theory doesn't prove things in this kind of context by saying "the task cannot be completed". — TonesInDeepFreeze

Ok. There are a couple of different ways sets can be written down as:

1. Set notation: S = {x | x is a prime number less than 10}
2. In words: S = {all prime numbers less than 10}
3. As a list of members of the set: S = {2, 3, 5, 7}


1. Set notation, C = {x | x is a set that contains itself}. Nothing seems amiss
2. In words, C = The set that contains itself. Doesn't seem to be problematic
3. As a list of members of the set: C = ??? [Problem. No such set can be]

There is no rule of set theory that we must name a set with set abstraction notation.

Moreover, for a finite set with members that are members of themselves, we can do it by extensional notation. For example:

Suppose

y = {y z}

z = {z j}

x = {x y z}

So, extensionally, we named the set whose members are x y z only.

The fact that we can continue to "expand" the notation in ways like this:

x = {x {y z} {z j}}

and

x = {{x y z} {{y z} {z j}} {{z j} j}}

ad infinitum, ad nauseum

doesn't prove without regularity that ~Ex x = {x y z} and doesn't prove ~Ex xex.

To prove something in set theory, you have to do if from the axioms. And set theory without the axiom of regularity does not prove ~Ex x = (x y z} and does not prove ~Ex xex.

C = The set of all sets that contains itself. Doesn't seem to be problematic — TheMadFool

It is problematic, because we cannot use set abstraction properly without first proving that there is a set that satisfies the defining property.

Just give me one instance of a set that contains itself. I can't. Can you?

Here's a set that doesn't contain itself: A = {7, &, troll}

Here's a set that contains itself: Your turn

Just give me one instance of a set that contains itself. — TheMadFool

I could only do that by using anti-foundation (or non-wellfounded) set theory.

These are different:

(1) The theory proves that there is not a set that is a member of itself.

(2) The theory does not prove that there is not a set that is a member of itself.

Look at those closely to see that they are different.

These hold:

Z proves that there is not a set that is a member of itself.

ZFC-R does not prove that there is not a set that is a member of itself.

/

So to make it plainly clear:

I have never claimed that ZFC-R proves that there is a set that is a member of itself.

What I have said is that ZFC-R does not prove that there is not a set that is a member of itself.

Put another way:

Ex xex is consistent with ZFC-R.

On the other hand, if you claim that ZFC-R does prove that there is not a set that is a member of itself, then the burden is on you to show such a proof. Hint: Don't bother. There is no such proof.

Please use the notation I prescribed viz. a set with its elements/members enumerated.

A set that doesn't contain itself: {1, y, $}

A set that contains itself: ???

x = {x}

Read my previous post regarding proof and lack of proof of existence.

A set that contains itself: ??? — TheMadFool

I don't think there are any common or obvious examples, but they are studied. Perhaps there are some clues here, I didn't read through this.

https://plato.stanford.edu/entries/nonwellfounded-set-theory/

No examples here either.

https://en.wikipedia.org/wiki/Non-well-founded_set_theory

It's interesting that there are articles about non well founded sets, but no specific examples. The SEP article does have some clues but nothing particularly simple.

x = {x} — TonesInDeepFreeze

So, set X = {X} = X1
That means, I can substitute X with {X}.

Here goes, X = {{X}} = X2

X1 = X2 (should be) because both are X but {X} =/= {{X}}.

Without the axiom of regularity, you can't prove

~Ex x = {x} = {{x}} = {{{x}}} ... for as finitely many iterations you want to make.

As I said, if you think there is such a proof, then try to show it.

examples — fishfry

I'm not sure, but I think a place to look might be proofs showing the relative consistency of anti-wellfoundeness. Maybe there is a construction in such proofs.

Without the axiom of regularity, you can't prove

~Ex x = {x} = {{x}} = {{{x}}} ... for as finitely many iterations you want to make. — TonesInDeepFreeze

Then, there can't be a set that contains itself.

1. All sets are sets that don't contain themselves

2. No sets that contain themselves are sets [from 1]

In other words, sets that contain themselves aren't sets which simply means the sentence, "sets that contain themselves" is meaningless.

Let's visit Russell's paradox now.

Suppose C = set of all sets that doesn't contain itself

1. If set C doesn't contain itself then, set C contains itself.

On the face of it, statement 1 looks reasonable but, as shown above, the consequent of the conditional 1 (above) is meaningless ["set" and "contains itself" in the same sentence is a contradiction]. Ergo, Russell's conditional (1 above) is gibberish.

2. If set C contains itself then, set C doesn't contain itself.
Again for the same reason as above, the antecedent of this conditional is nonsense. Ergo, Russell's conditional 2 is balderdash.

Russell's paradox can't be a paradox because the two key conditionals have no meaning at all.

Then, there can't be a set that contains itself — TheMadFool

You're not reading my posts.

Without the axiom of regularity, we cannot prove ~Ex xex.

And the rest of your post is more of your misunderstanding of how set theory works.

You started to learn the sentential calculus. That's good. Finish doing that. Then learn the predicate calculus. Then learn basic set theory. Then you'll be in a position to discuss these matters coherently.

Kindly point out a flaws in my argument if there are any.

You mentioned that without the axiom of regularity, we can't prove ~Ex xex. As far as I can tell that means,

Can prove ~Ex xex -> Axiom of regularity.
No axiom of regularity -> Can't prove ~Ex xex

Since I've proved a set can't contain itself, it follows that I've assumed the axiom of regularity then.

Challenge for you: Can you prove that a set contain itself? Feel free to use any axiom of your choice.

You mentioned that without the axiom of regularity, we can't prove ~Ex xex. As far as I can tell that means,

Can prove ~Ex xex -> Axiom of regularity. — TheMadFool

No, it is a non-sequitur to infer

ZFC-R |- ~Ex xex -> regularity

from

it is not the case that ZFC-R |- ~Ex xex

Now, clearly

ZFC-R |- regularity -> ~Ex xex

which is the same as saying

ZFC |- ~Ex xex

However, I'm not absolutely sure, but I'm really pretty sure that it is not the case that

ZFC-R |- Ex xex -> regularity.

That is, Ex xex is a clear consequence of regularity. But regularity is a more comprehensive statement than Ex xex.

Since I've proved a set can't contain itself, it follows that I've assumed the axiom of regularity then. — TheMadFool

The problem is that your argument is not a valid proof. You didn't use the axiom of regularity; instead you just used hand-waving about [paraphrasing here:] "can't show it using braces and then infinite nesting in braces".

I will say this though: It is quite reasonable to think that our common notion of a set disallows sets from being members of themselves. And it seems you have been relying on that common notion. So it is fine to say that self-membership is not in our ordinary conception of sets and that therefore we should not countenance self-membership. Very fine. But that's not proof from axioms. To make it proof from axioms you need an axiom that ensures that that common notion abides axiomatically. And that's where the axiom of regularity comes in.

Can you prove that a set contain itself? Feel free to use any axiom of your choice. — TheMadFool

(1) Again, note that I did not claim to be able to prove Ex xex. All I claimed was that ZFC-R does not prove ~Ex xex.

(2) Trivially, we can prove Ex xex, and consistently with ZFC-R by adding an axiom: Ex xex.

(3) For a non-trivial proof, we would need a non-trivial anti-foundation axiom. To see what that looks like:

https://en.wikipedia.org/wiki/Non-well-founded_set_theory

Note: I am referring to the Wikipedia article instead of the Stanford Encyclopedia of Philosophy article only because this particular Stanford article seems to start out at a pretty difficult level even from the start. I have not gone over the Wikipedia article closely, so I don't guarantee its accuracy.

Meanwhile, I hope I can take it that you now grant that you don't know how you would prove ~Ex xex in ZFC-R.

Challenge for you: Can you prove that a set contain itself? Feel free to use any axiom of your choice. — TheMadFool

It's technical to prove that non well-founded sets are consistent. See

https://math.stackexchange.com/questions/1148634/show-that-there-are-non-well-founded-models-of-zermelo-fraenkel-set-theory

I don't claim to understand the responses, but this is how people prove consistency of non well founded sets. In this case they're not showing a set that contains itself, but rather a model of set theory with an infinite membership chain $c_1 \in c_2 \in c_3 \in \dots$

The discussion thread consists of highly technical responses from professional mathematicians. There are no specific examples of sets that contain themselves; but rather, consistency proofs for ZF with the negation of the axiom of foundation/regularity.

ps here is a better answer.

https://math.stackexchange.com/questions/253818/example-of-set-which-contains-itself

The answer is that it's consistent with ZF (minus regularity) to have a set $x = \{x\}$. There's no point asking, "What is it?" It's what it is as notated. The point is simply that it's logically consistent to have such a thing.

That thread also points to this article:

https://en.wikipedia.org/wiki/Aczel%27s_anti-foundation_axiom

It says:

An accessible pointed graph is a directed graph with a distinguished vertex (the "root") such that for any node in the graph there is at least one path in the directed graph from the root to that node.

The anti-foundation axiom postulates that each such directed graph corresponds to the membership structure of a unique set. For example, the directed graph with only one node and an edge from that node to itself corresponds to a set of the form x = {x}.

I believe that last bit gives us the best visualization we're going to get for a set that contains itself; namely, a graph consisting of one node with a path that points from the node to itself.

The point being is that any set may be viewed as a graph, where the edges are the element-of relation. If you let the graph loop, you have a set that violates Regularity/Foundation.

I'm looking at 'The Joy Of Sets' by Devlin. He has a cute example (though it steps our of formal set theory by using 'explicitly referred to' and 'this book'):

"Let B be the set of all sets explicitly referred to in this book. Clearly, since B is referred to in this book (I am just now referring to it) we have

BeB."

That went over my head. Thank you for trying though.

Keep it simple for me, ok.

1. Assume whatever axiom you want to/have to assume to prove the proposition, C = sets can contain themselves.

2. I'll assume, given your knowledge, C is proven or C

3. If C then there has to be a set N ="{...} such that it contains itself

4. There has to be a set N = {...} such that it contains itself [2, 3 modus ponens]

My request to you is express N as I express the set P (prime numbers less than 4), P = {2, 3}. In other words, I want to know if it's possible to have a set N = {x, N} where x = no, one, or more members of that set.

This seems impossible for the following reason.

1. There's a set N that contains itself [assume for reductio ad absurdum]
2. Suppose, N = {x, N}
3. N is a proper subset of {x, N} [proper subset of a set]
4. If N is a proper subset of {x, N} then, n(N) < n({x, N})
5. n(N) < n({x, N}) [3, 4 modus ponens]
6. n(N) = n(N) [true for all sets]
7. n(N) < n(N) [2, 5 substitution]
8. n(N) = n(N) and n(N) < n(N) [contradiction]
Ergo,
9. There's no set N that contains itself [1 - 8 reductio ad absurdum]

consistency proofs for ZF with the negation of the axiom of foundation/regularity. — fishfry

Just to be clear, ZR-R+~R is relatively consistent with ZF-R.

And to underline your point: The conjunction of "ZF-R+~R is relatively consistent with ZF-R" with "ZF is relatively consistent with ZF-R" is equivalent to "ZF-R is consistent -> R is independent from ZF-R".

there has to be a set N ="{...} such that it contains itself — TheMadFool

N = {N}

is such a set.

where x = no, one, or more members of that set. — TheMadFool

I don't know what you mean by that.

/

3. N is a proper subset of {x, N} — TheMadFool

Wrong. N = {x N}, so N is not a proper subset of {x N}.

I'll stop there.

That went over my head. — TheMadFool

I don't think it's so over your head. Give it a bit of thought and you'll see it pretty clearly.

is no reason for supposing that you can't have a set that contains itself as a member — Amalac

This is not what I suppose. I definitely believe in a set being a member of itself. I think it is a logical necessity.

well, you've given no reason to accept that yet, except that x would “contain itself twice” — Amalac

I understand why it looks as though I haven't. Again, I'm suggesting that this has been overlooked for more than a hundred years by the likes of Russell and Frege (amongst others). It's not immediately recognisable. But I've seen what's been overlooked/misunderstood clearly. if you can see it too, I believe you will see that that there is no alternative.

A set is only a member of itself if it contains itself as a member. So when I write x (x, y, z), I mean to say set x contains items x, y, and z. Set x is a member of itself purely because it contains x (itself). WIth this in mind, consider the following:

1) x (x, y, z). Here, when we say x is a set that is not a member of itself, we get a contradiction. (because x is in x).

2) x (x, y, z). Here, when we say x is a set that is a member of itself, we get no contradiction (because x is in x).

3) x (x, y, z). Here, when we say x is a set of three sets that are members of themselves, we get a contradiction (it amounts to saying one set can be a member of itself twice, compare again with 1 and 2. It amounts to x being a member of itself twice because x is in x and the context is sets that are members of themselves. Either y and z are not members of themselves (making x the only member of itself), or x is a member of itself twice whilst y and z are members of themselves once. It cannot be the case that x is a set of three sets that are members of themselves).

Suppose I wrote: x (p, q, y). Now, x can be a set of three sets that are members of themselves (precisely because it does not include itself for it to amount to being a member of itself twice). But try having a set of all sets that are members of themselves. You cannot avoid ending up with a set that is a member of itself twice (which is absurd).

Do you see how this is like the inversion of Russell's paradox?

Everyone recognises the impossibility of a set of all sets that are not members of themselves, but nobody seemed to have recognised the impossibility of a set of all sets that are members of themselves.

If there is a set of all sets, then it has the subset that is the set of all sets that are not members of themselves. — TonesInDeepFreeze

It does not have such a subset as evidenced by such a subset being clearly contradictory. However, it is the set that contains all sets (rejecting this is also clearly contradictory). So it contains all sets that are not members of themselves (because they are all members of it, precisely because they are all sets), and it is a member of itself (precisely because it is a set and is therefore a member of itself). You cannot show me a contradiction in this. But a contradiction is clear in the following two statements:

1) There is a set that contains all sets that are not members of themselves that is itself not a member of itself.

2) No set can logically encompass all sets.

How do you possibly allow yourself to reject the set of all sets? It's the last thing one should be doing.

In any case, I don't think you read my post with enough attention to detail, otherwise I reckon you would have seen where I was coming from. But as it stands, your belief system is contradictory precisely because it sees 2 as being rational/consistent, whereas 2 is clearly contradictory.

In conclusion:

There's no such thing as a set of all sets that are not members of themselves that is itself not a member of itself. But there is a set of all sets that are not members of themselves that is itself a member of itself. That set is the set of all sets. It encompasses all sets that are members of themselves, as well as itself (hence why it is a member of itself).

There is no subset of all sets that are not members of themselves, or all sets that are members of themselves.