x, the set of all sets that are not members of themselves, is thus a member of itself. — Philosopher19

That right there is the contradiction. x is a member of itself if and only if it's not a member of itself.

Since the assumption that there is a set of all sets leads to a contradiction, we must therefore reject that assumption, and admit that no such set exists.

It's always worth pointing out in these discussions that self-containing sets are not contradictory. We generally assume an axiom, the axiom of regularity, that outlaws self-containment as well as circular chains of containment such as $a \in b \in c \in d \in a$

If we instead choose to allow self-containing sets, the resulting system is logically consistent. The study of such systems is called non well-founded set theory.

In modern set theory we avoid Russell's paradox by saying that we can not form a set merely out of the extension of a predicate, like $x \notin x$. Rather, we must start with a set we already know to exist, and then apply a predicate to it. This is the axiom schema of specification. It's a schema because it stands for an infinite collection of axioms, one for each predicate.

As an example of how this works, what is the set of all natural numbers that are not members of themselves? Well, $0 \notin 0$, so $0$ is in the set. $1 \notin 1$, so $1$ is in the set. $2 \notin 2$, so $2$ is in the set. Continuing in this manner we see that the set of all natural numbers that are not members of themselves is ... drum roll ... the set of natural numbers! No paradox. Specification saves the day.

That right there is the contradiction. x is a member of itself if and only if it's not a member of itself. — fishery

Thank you for replying and I understand where you're coming from. I will try to convey to you my understanding more specifically hoping that specification saves naive true set theory. I will ask questions to see exactly where it is that we are in disagreement.

For the sake of argument, assume we have the set of all sets. Call this x. x is a member of itself because it is a set. No paradoxes so far, agreed?

Since x contains all sets, do we agree that x contains all sets that are not members of themselves?

The set of all penguins, is a set. This set is not a member of itself precisely because it is a member of the set of all sets. By this I mean It is specifically a set, not a penguin. Are we sill in agreement? The set of all animals, is one set that contains the set of all penguins. This set is also not a member of itself, precisely because it is a member of the set of all sets. Thus, by definition, any set that truly is a set (as opposed to a penguin or animal), and is not a member of itself, is not a member of itself precisely because it truly is a set and is thus a member of the set of all sets. Agreed?

If agreed, then can you see how the set of all sets that are not members of themselves, can only be (by definition) x? I will specify this some more: All sets that are not members of themselves, truly are sets. What is the set of all these sets? Can the answer be anything other than x? The set of all sets which contains all these sets, is a member of itself (because it truly is a set).

Where do we have a paradox in what I have proposed?

I wrote a reply and then, in the process of editing out my typos, I came to understand your argument. Rather than rewrite it I just let it all stand and interspersed notes that fix it all up. Therefore this is about twice as long as it needs to be. But here it is. The bottom line is that you assume x is a set but never prove it; and Russell shows that it can't be.

Thank you for replying and I understand where you're coming from. I will try to convey to you my understanding more specifically hoping that specification saves naive true set theory. I will ask questions to see exactly where it is that we are in disagreement. — Philosopher19

Ok. But the core of the issue is the same as Russell noted in 1901. The "set of all sets that are not members of themselves" both is and isn't a member of itself, a contradiction. Therefore there is no such thing. This is not going to change. It would be helpful if you would carefully review the argument yourself and I urge you to do so.

For the sake of argument, assume we have the set of all sets. Call this x. — Philosopher19

I'm perfectly willing to agree, as this is the assumption that will soon lead to a contradiction.

x is a member of itself because it is a set. No paradoxes so far, agreed? — Philosopher19

Ah, perhaps I see the problem. Why do you think a set is a member of itself? Are you possibly confusing set membership with the subset relation? It's true that every set is a SUBSET of itself. But no set, in the presence of the axiom of regularity, is an element of itself. And even without regularity, a set can be a member of itself. But it doesn't have to be.

Is this the confusion? The set of natural numbers $\mathbb N$ is a SUBSET of itself; but not a MEMBER of itself. That is, $\mathbb N \subset \mathbb N$, but $\mathbb N \notin \mathbb N$. If this resolves your question, we're done. Because saying that x is a set does NOT in any way imply that it is a member of itself. That's an error. So if this is the problem, we're done. I'll continue, but let me know if this was the issue.

Since x contains all sets, do we agree that x contains all sets that are not members of themselves? — Philosopher19

Yes, certainly. x contains all the sets that are members of themselves and all the sets that are not members of themselves. This will lead to the conclusion that x both is and isn't a member of itself.

The set of all penguins, is a set. — Philosopher19

I have a problem. This isn't even naive set theory; it's high school set theory. Penguins are not elements of sets. In math, sets are generally "pure" sets, meaning that their only elements are other sets. There are alternate versions of set theory in which sets can contain urelements; that is, things that are elements of sets that are not themselves sets. However even in set theories containing urelements, I do not believe that penguins or any other natural objects can be urelements. I have to plead ignorance though, I don't know much about sets with urelements.

If you ask a biologist, they'll tell you that a penguin belongs to the family Spheniscidae, of the order Sphenisciformes, class Aves, phylum Chordata, kingdom Animalia. "Today I learned," as they say. Biology isn't my thing. But the classifications in biology are not mathematical sets. We could call them high school sets, but they are not even naive sets, let alone axiomatic sets. So this analogy is going in the wrong direction already. The collection -- not set, collection -- of all penguins is, as we all just learned, the family Spheniscidae. It's not a set. Whatever point you're trying to make, I'd prefer if you make it with actual mathematical sets. The set of all real numbers, or the collection of all topological spaces, which ISN'T a set (for the same reason as the set of all sets isn't a set).

This set is not a member of itself precisely because it is a member of the set of all sets. — Philosopher19

Now this I do not understand, even if I grant that the collection of all penguins is a set. Because the set of all sets contains all the sets that ARE members of themselves, AND all the sets that ARE NOT members of themselves. So even if I grant, for sake of argument, that the collection of all penguins is a set, that does not make it not a member of itself. Even granting your example, this statement makes no sense. Some sets are members of themselves (in non well-founded set theories) and some aren't. So just because something's a member of the set of all sets doesn't mean it's not a member of itself. How did you conclude that??

NOTE (after I've been through this a few times and think I understand your argument). If we assume no set is a member of itself, that still doesn't show that the collection of all non-self-containing sets is a set. You haven't shown that. And if you assume it is, you get Russell's paradox.

That is: You assume x is the set of all sets. You assume (as we all do all the time) that no set contains itself. Therefore x is also the set of all sets that don't contain themselves. HOWEVER! You still have your ASSUMPTION hanging around. And it falls to Russell's paradox. You assumed x is a set. What you have actually shown is that the collection of all sets is the same as the collection of all the sets that don't contain themselves. But you haven't shown that these collections are sets; and in fact they are not.

By this I mean It is specifically a set, not a penguin. Are we sill in agreement? — Philosopher19

I agree that the collection of all penguins is not a penguin. But I agree NOT because the collection of all penguins is a set; but rather, because the collection of all penguins doesn't happen to be a penguin. Your logic is off the rails at this point.

The set of all animals, is one set that contains the set of all penguins. — Philosopher19

As we've seen, a biologist would not agree. The concept of penguin is a subconcept or subcategory of the concept or category of animals. But they are not sets. The predicate "is a penguin" does imply the predicate "is an animal," I agree with that. But these are not sets. Still, for sake of discussion I'll grant your premise. I still fail to see your point and your claim that these classes do or don't contain themselves simply by virtue of being sets, is wrong.

This set is also not a member of itself, precisely because it is a member of the set of all sets. — Philosopher19

This is just wrong. The set of all sets (which we assume for sake of argument exists) contains some sets that DO contain themselves; and other sets which DON'T contain themselves. Just because some object (the collection of all penguins) happens to be a member of the set of all sets, doesn't let us conclude which is the case. It might contain itself or it might not. In this case the collection of all penguins does not happen to be a penguin; but that's a fact of nature and NOT just because it's a set.

Thus, by definition, any set that truly is a set (as opposed to a penguin or animal), and is not a member of itself, is not a member of itself precisely because it truly is a set and is thus a member of the set of all sets. Agreed? — Philosopher19

No, and I don't follow your reasoning. If the set of all sets contains all sets, then it contains all the sets that ARE members of themselves along with all the sets that AREN'T members of themselves. Given a set, we have to examine it carefully to determine whether or not it's a member of itself. Of course in standard set theory we have the axiom of regularity so no set is a member of itself. But if we drop regularity, then some sets ARE and some sets AREN'T member of themselves. We can't determine which is the case merely by knowing something is a set. Even granting all your premises I don't follow your reasoning. The set (if we call it that) of all penguins is not a penguin; but not because it's a set; but rather, because it doesn't happen to be a penguin!

NOTE: See Summary at the end. If you are assuming that by definition no set is an element of itself, that's perfectly fine and is the standard assumption in math. But that does not mean that the collection of all these non-self-containing sets is a set. You haven't shown that. What your argument shows is that the COLLECTION of all the sets that don't contain themselves, is identical to the COLLECTION of all the sets that there are. This is true. But you haven't shown that either of these collections are sets.

If agreed, — Philosopher19

I'm afraid not. Even granting your premises there is something terribly wrong with your reasoning. Some sets are and some sets aren't members of themselves. So even if you convince me that the collection of all penguins is a set, that doesn't tell me whether it's a penguin or not. I have to consider the specific case.

then can you see how the set of all sets that are not members of themselves, can only be (by definition) x? — Philosopher19

No. And the problem is that I have no idea how you got to this point. x includes all the sets that aren't members of themselves, AND all the sets that ARE members of themselves. x contains all possible sets, right?

NOTE (now that I think I understand your argument): You are right that if we assume no set is a member of itself (as we normally do), then the COLLECTION of all non-self-containing sets is identical to the COLLECTION of all sets. But you haven't shown that these collections are sets. You're only assuming that; and your assumption is wrong, as shown by Russell.

I will specify this some more: All sets that are not members of themselves, truly are sets. — Philosopher19

Well yes, because you said they're sets. All the fish that are left-handed are fish. Why? Because we said they're fish! A set that's painted green is a set, because we stipulated that it's a set. You've made a vacuously true statement. A set is a set, so of course a set that is painted green, or flies through the air, or is not a member of itself, is a set. We haven't said anything!

What is the set of all these sets? Can the answer be anything other than x? — Philosopher19

Yes. The set of all sets that are not members of themselves is different from the set of all sets; because x contains all the sets that are not members of themselves AND all the sets that ARE members of themselves. I don't follow why you don't see this.

A set either contains itself or it doesn't. The collection of ALL sets contains the ones that do and the ones that don't. But that collection isn't necessarily a set, and can't be. And you haven't shown that it is.

NOTE (all these notes were written after I came to understand your argument, apologies for all these interspersed notes). YOU ARE CORRECT. If no set contains itself and x is the collection of all sets, then x is also the collection of all sets that don't contain themselves. BUT the claim that x is a set was an ASSUMPTION, which you haven't justified. And Russell's paradox shows your assumption is wrong.

The set of all sets which contains all these sets, is a member of itself (because it truly is a set). — Philosopher19

No, it's not. Russell shows it's not a set. It's a collection that's not a set.

NOTE. You have CORRECTLY shown that the collection of all sets is the same as the collection of all non-self-containing sets. But you ASSUMED that this collection is a set and Russell shows that it's not. You made an assumption but never justified it.

Where do we have a paradox in what I have proposed? — Philosopher19

You didn't get there yet. You have shown that the collection of all sets is the same as the collection of all sets that don't contain themselves, under the axiom of regularity. But then you assumed that x is a set, and you never justified that. And if you then apply Russell, you'll see that x can't be a set. Just a collection that isn't a set.

Summary:

* One issue as I mentioned is that you may be confusing subsets with elements. This is a fairly common point of confusion and if that's the case, let's focus in on that.

* The business with the penguins was not helpful to me, it added confusion.

* You are assuming the axiom of regularity (perfectly normal, we all do that all the time in standard set theory) and therefore saying that everything that's a set, doesn't contain itself. This, I agree with. But that doesn't mean that the collection of all the sets that don't contain themselves is a set! That's a logic error. This is the part that I didn't realize on my first couple of readings.

That is: Let us adopt the axiom of regularity, so that no set contains itself. We can therefore form the COLLECTION (not yet proven to be a set) of all the sets that don't contain themselves; and this will indeed turn out to be the COLLECTION (not yet proven to be a set) of all possible sets. I think this is the argument you're trying to make.

But you haven't shown that either of these collections is a set! And you can't, because any such attempt runs into Russell's argument.

* So the bottom line is that we assumed the set of all sets exists, and, assuming that no set contains itself, you are correct that it must be equal to the set of all sets that don't contain themselves. But then we just apply Russell's argument to show that the set of all sets that don't contain themselves both does and doesn't contain itself, a contradiction.

* So what you need to do is, AFTER you have made your argument: that x is the set of all sets and also the set of all sets that don't contain themselves; you have to apply Russell's paradox to see that x both is and isn't a member of itself. Therefore x isn't a set. It's merely a collection, the extension of a predicate. It's not a set.

Yes. The set of all sets that are not members of themselves is different from the set of all sets; because x contains all the sets that are not members of themselves AND all the sets that ARE members of themselves. I don't follow why you don't see this. — fishery

To my knowledge, Russell's paradox concludes that you cannot have a set of all sets because he fails to non-paradoxically define a set of all sets that are not members of themselves.

No one is disputing that there can be no set of all sets that are not members of themselves that is itself, not a member of itself (call this absurd set y). But this paradox in no way logically amounts to saying that there is no set of all sets. x is the set of all sets. This set contains all sets including itself. No paradoxes.

For the sake of argument, let's say x and y are not the same. I have no problem in saying that y is absurd. But there is still a set that contains all sets that are not members of themselves. x contains them all does it not? If it does contain them, then it contains them. Why does x have to be not a member of itself??? Why are we trying to force a paradox where there is none? A set is a set. It doesn't matter if it's a member of itself or not. If it truly is a set, then it is clearly a member of the set of all sets.

Yes, x also contains other sets (actually it only contains one other set...which is itself). But it still contains all sets that are not members of themselves. Regarding sets, Russell misunderstood semantics and logic. I understand he was an important philosopher, but he made a mistake.

It is absurd to say that there is no set of all sets. Now are we in agreement?

It is absurd to say that there is no set of all sets. — Philosopher19

Rather, it is fundamental, and perfectly intuitive.

Is there an existing thing that contains all existing things? Is it not blatantly paradoxical to deny that there is an existing thing that contains all existing things? If you accept that there exists an existing thing that contains all existing things (which I will label Existence), then you must also accept that there is a set of all sets.

I don't see how you can reject without being paradoxical. To say there isn't an existing thing that contains all existing things. logically implies things can come into or go out of Existence.

Intuitively, a thing cannot be inside itself. It is itself, not inside itself.

What you say is true of all non-infinite things. It is not true of that which is actually infinite. The actually infinite has no beginning and no end.

The infinite contains the infinitesimal and the infinitesimal contains the infinite. What's the alternative, that we deny that there exists and existing thing that contains all existing things? Is this better or accepting actual infinity as being infinite through and through?

Consider the following thought experiment:

On a computer, y is the folder of all folders. Your starting/beginning position is within y and you see y amongst an infinite number of other folders. If you click y, you get the same thing (an infinite number of folders with y amongst them). This continue ad infinitum. If you go up one level of folder trying to get to the root folder, again, you get the same thing (an infinite number of folders with y (the one you just came from) amongst them). This continues ad infinitum. We cannot say y fully contains itself if you cannot go up another level ad infinitum or go down another level ad infinitum. Your starting/beginning point changes nothing. Such is the nature of true infinity, it has no beginning and no end. Mathematicians confuse potential infinity with actual infinity and you get infinity paradoxes where there really are no such paradoxes. Just poor labels chosen for semantics.

The problem with the above thought experiment, is that you can see y. You cannot see infinity or infinitesimal. In other words, you cannot see Existence, you can only see non-infinite things that are in it. The infinite is in the infinitesimal and the infinitesimal is in the infinite.

I admire and appreciate the trouble you go to here and elsewhere to put some of us back on the right track - even as some of us go right back off the rails!

I have myself learned lessons and meta-lessons from TPF, among the latter to strap in and put on my galoshes whenever an argument starts with "if," a perfectly good and useful word if it is not abused. Unfortunately, it is much abused. Hmm, maybe worth a thread....

Try again. The infinite has absolutely nothing to do with whether something can or cannot be inside itself.

If you knew more about infinity, you would not say this. Me trying again will do nothing. Try looking at my example again.

↪fishfry
I admire and appreciate the trouble you go to here and elsewhere to put some of us back on the right track - even as some of us go right back off the rails! — tim wood

Kudos to fishfry and fdrake, resident experts in set theory. Both demonstrate great patience in unraveling the queries on that subject that crop up on TPF. :up:

To my knowledge, Russell's paradox concludes that you cannot have a set of all sets because he fails to non-paradoxically define a set of all sets that are not members of themselves. — Philosopher19

You're turning the argument on its head to confuse yourself. If we can't define such a set without creating a contradiction, then there is no such set.

No one is disputing that there can be no set of all sets that are not members of themselves that is itself, not a member of itself (call this absurd set y). But this paradox in no way logically amounts to saying that there is no set of all sets. x is the set of all sets. This set contains all sets including itself. No paradoxes. — Philosopher19

If you call x the set of all sets, you quickly get a contradiction. You find that x both is and isn't a member of itself. Therefore there is no such set. You keep claiming there is but you have not provided proof.

For the sake of argument, let's say x and y are not the same. I have no problem in saying that y is absurd. But there is still a set that contains all sets that are not members of themselves. x contains them all does it not? If it does contain them, then it contains them. Why does x have to be not a member of itself??? Why are we trying to force a paradox where there is none? A set is a set. It doesn't matter if it's a member of itself or not. If it truly is a set, then it is clearly a member of the set of all sets. — Philosopher19

There is no set of all sets. You keep claiming there is, but you have not provided a proof. On the contrary, the assumption that there is a set of all sets leads to a contradiction. Therefore there is no set of all sets.

Yes, x also contains other sets (actually it only contains one other set...which is itself). But it still contains all sets that are not members of themselves. — Philosopher19

The CLASS, or COLLECTION that contains all sets may indeed be formed. It just turns out to not be a set. And you haven't shown that it can be. You keep claiming it without proof.

Regarding sets, Russell misunderstood semantics and logic. I understand he was an important philosopher, but he made a mistake. — Philosopher19

Not in this instance.

It is absurd to say that there is no set of all sets. Now are we in agreement? — Philosopher19

You trollin' me?

If you call x the set of all sets, you quickly get a contradiction. You find that x both is and isn't a member of itself. Therefore there is no such set. You keep claiming there is but you have not provided proof. — fishfry

The following is proof:

I find that if I say x isn't a member of itself, I am being paradoxical because x is a set. I find that if I say x is a member of itself, I am not being paradoxical because x is a set. Do you see?

You find that x both is and isn't a member of itself. — fishfry

How do you get to this???? x = the set of all sets. Is x a set? Yes. Thus x is a member of itself. Is x not a member of itself? Yes it is a member of itself because it is a set! Let's try the alternative. x is not a member of itself. Why not? No reason can be given. The set of all penguins is not a member of itself. Why not? Because a penguin is not a set. See?

I find that x is a member of itself. That is all I find.

You trollin' me? — fishfry

Not at all. If I am, then I'm an idiot. I just want efficiency and truth.

Patience is a virtue.

Standard set theory includes an axiom that basically says that given a set, any collection of its elements is also a set. Since sets that are not members of themselves are included in the set of all sets, then a collection of all such sets must form a set. But we know from Russell's paradox that such a collection cannot form a set on pain of contradiction.

given a set, any collection of its elements is also a set — SophistiCat

Yes I agree. But please bear in mind that all sets, are members of the set of all sets (including the set of all sets itself). So the set of all great white sharks, is a member of the set of all sharks as well as a member of the set of all sets.

Since sets that are not members of themselves are included in the set of all sets, then a collection of all such sets must form a set — SophistiCat

The set of all great white sharks is a member of at least two sets. The set of all sharks, and the set of all sets. You cannot have a set that is not a member of itself encompass/include all sets that not members of themselves precisely because (as Russell pointed out) it cannot both contain itself and not contain itself at the same time. This is clearly paradoxical. But you have the set of all sets. It includes all sets that are not members of themselves. Unlike the set of great white sharks which is a member of at least two sets, the set of all sets that are not members of themselves, is only a member of one set. That set is the set of all sets. Please look at the part in bold very carefully and recall that the set of all sets, is a member of itself, and that all sets are a member of the set of all sets (including great white sharks).

The set of all sets that are not members of themselves, is a set. Thus it is a member of the set of all sets. It is in fact the set of all sets. The set of all sets is a member of itself. No paradoxes here whatsoever.

Russell claims that there is no set of all sets that are not members of themselves whatsoever. Russell is clearly wrong. This paradox is now clearly fixed. Can we move on in a unified manner? Because academics seem to think that it's unsolvable. I'm not at a uni to get my voice heard. I don't like seeing true set theory being called naive set theory.

@fishfry already addressed a number of your misconceptions. You ignored his patient explanations and are now repeating the same mistakes here. And while you quoted my post, you did not address its content and instead repeated the same nonsensical arguments that you made earlier. I don't see a point in continuing this conversation.

Russell is clearly wrong. This paradox is now clearly fixed. Can we move on in a unified manner? — Philosopher19

Yes. Off the first page of TPF. :roll:

Yes. Off the first page of TPF. — jgill

Your bump pushes it to the front page.

The following is proof:
I find that if I say x isn't a member of itself, I am being paradoxical because x is a set. — Philosopher19

You claim x is a set but it isn't. You have no proof that x is a set.

I find that if I say x is a member of itself, I am not being paradoxical because x is a set. Do you see? — Philosopher19

Is x the collection of all sets that are not members of themselves? If so then x is not a set. You claim it is but haven't proved it is. If x is the collection of all sets that are not members of themselves, then x also isn't a set even though you claim it is. You've used x to mean both of those things at various times. In both cases they are collections, or classes, or extensions of a predicate. But they're not sets.

You find that x both is and isn't a member of itself.
— fishfry

How do you get to this???? — Philosopher19

By Russell's paradox. Say x is the set of all sets. Then let y be the set of all sets that are not members of themselves. Is y a member of itself? If it is, then it's NOT a member of y. But if it isn't, then it IS a member of y. So y is a member of itself if and only if it isn't a member of itself. That's a contradiction.

If x is the set of all sets that are not members of themselves, just run the same argument on x.

x = the set of all sets. Is x a set? Yes. — Philosopher19

No, it can't be. I just showed that if x is the set of all sets, then we can form y and derive a contradiction. If x is the set of all sets that don't contain themselves, then x itself leads to a contradiction.

Thus x is a member of itself. Is x not a member of itself? Yes it is a member of itself because it is a set! Let's try the alternative. x is not a member of itself. Why not? No reason can be given. The set of all penguins is not a member of itself. Why not? Because a penguin is not a set. See? — Philosopher19

You need to work through the proof of Russell's paradox carefully. You're just repeating incorrect ideas.

I find that x is a member of itself. That is all I find. — Philosopher19

You need to work through the argument.

Not at all. If I am, then I'm an idiot. I just want efficiency and truth. — Philosopher19

I have made no characterizations. It's a tricky argument. You need to go through it for yourself carefully.

Suppose x is the set of all sets that don't contain themselves. Then we ask if x contains itself? If it does, it's NOT a set that doesn't contain itself, so it DOESN'T contain itself.

See what just happened? If x contains itself then x doesn't contain itself.

On the other hand suppose x doesn't contain itself. Then it must be a member of x. So if it doesn't contain itself it does contain itself.

Having just shown that x contains itself if and only if it doesn't contain itself, we have a contradiction. Therefore there is no such set as x.

Now if you prefer to let x be the set of all sets, we let y be the set of all sets that don't contain themselves and we get a contradiction from y. So again, x can't be a set.

As does yours. Mine moved us eight hours ahead! Now look where we are. :worry:

As does yours — jgill

I'm not the one who has a problem with it. Russell's paradox is a deep argument. Frege was a smart guy (he invented the universal and existential quantifiers) and he missed it. It's worth discussion. I daresay @Philosopher19 is not the only person who's ever experienced confusion about the subject. SEP has 7448 words on it, they must think it has some importance. It's the paradigmatic example of all the self-referential arguments such as Cantor's diagonal argument (which influenced Russell to think of it), Gödel's incompleteness theorems, and Turing's Halting problem. It's worthy of discussion IMO regardless of the circumstances.

This website is not the proceedings of the Royal society. I often wish for a more high toned conversation around here, especially on mathematical topics; but we take what we can get. You should see the politics forums. I hope my participation in this thread doesn't inconvenience or distress you too much. It's the forum software that bumps active threads, I have no control over that. I couldn't help calling out the irony that someone who doesn't want to see this thread on the front page, bumped it to the front page themselves. How self-referential.

Deleted

You ignored his patient explanations and are now repeating the same mistakes here. And while you quoted my post, you did not address its content and instead repeated the same nonsensical arguments that you made earlier. — SophistiCat

I am not the one embracing a paradox. That would be unreasonable/nonsensical/paradoxical.

I don't see a point in continuing this conversation. — SophistiCat

Ok.

You claim x is a set but it isn't. You have no proof that x is a set. — fishfry

Ok then. A set is a set and you can have sets within sets.

By Russell's paradox. Say x is the set of all sets. Then let y be the set of all sets that are not members of themselves. Is y a member of itself? — fishfry

If y is not x, then y is absurd. How can a set that is not a member of itself, contain itself? It cannot. Say x encompasses all sets. Say z is the most encompassing set of sets after x. z is not a member of itself but it contains almost all sets that are not members of themselves. Because it does not contain itself, z is not the set of all sets that are not members of themselves because it does not contain itself despite being a set that is not a member of itself. z is contained in x because it is a set. y being absurd or z not containing itself takes nothing away from x containing all sets that are not members of themselves as well as itself. y is absurd because it claims to contain all sets that are not members of themselves whilst not being x. Why does x have to be treated as being absurd when y is absurd? The set of all sets that are not members of themselves is x and x is a member of itself. If you still think the last sentence is paradoxical, read on.

No, it can't be. I just showed that if x is the set of all sets, then we can form y and derive a contradiction. If x is the set of all sets that don't contain themselves, then x itself leads to a contradiction. — fishery

See what just happened? If x contains itself then x doesn't contain itself. — fishfry

On the other hand suppose x doesn't contain itself. Then it must be a member of x. So if it doesn't contain itself it does contain itself. — fishfry

Again, this is rooted in confusing y for x. y is absurd but x contains itself and it contains all sets that are not members of themselves because all sets that are not members of themselves, are still just sets at the end of the day. x does not have to be not a member of itself in order to contain sets that are not members of themselves. The set of all sets does not have to be a penguin to contain the set of all penguins. It just has to be the set of all sets.

Change "penguin" for "not a member of itself". Is a set that is not a member of itself, a set? If yes, then you know at least one set that it is a member of: The set of all sets. Are all sets that are not members of themselves, sets? If yes, then you know at least one set that they are all a member of: The set of all sets. There can be no other set that contains them all.

Now if you prefer to let x be the set of all sets, we let y be the set of all sets that don't contain themselves and we get a contradiction from y. So again, x can't be a set. — fishfry

Again, any set that is not a member of itself, is a member of the set of all sets. This is because it is a set. The set of all sets, is a member of itself. This is because it is itself a set. If y is absurd, that does not mean that x is also absurd. It just means y is not in x because y is not a set. Nor does it mean that x does not contain all sets that are not members of themselves when they clearly are just sets. Just for one second, consider the hypothetical possibility that all these famous philosophers after Russell, were wrong, and I'm right. I know how it sounds, yet it is still a hypothetical possibility. Just look at the proof that I am presenting without bias and without preconceived notions.

Russell's paradox applies not only to the element relation, and not just to set theory, but to any 2-place relation whatsoever and to logic in general.

Let R be any 2-place relation. It is a theorem of logic that:

There does not exist an x such that for all y, y R-relates to x if and only if y does not R-relate to itself. Symbolically:

~ExAy(Ryx <-> ~Ryy)

In particular, where R is the membership relation, there does not exist a set x such that for all y, y is a member of x if and only if y is not a member of itself. Symbolically:

~ExAy(y in x <-> ~ y in y)

Proof is simple:

Suppose, toward a contradiction, that there is an x such that for all y, y R-relates to x if and only if y does not R-relate to itself. Then x itself R-relates to x if and only if x does not R-relate to x, which is a contradiction. Symbolically:

Suppose ExAy(Ryx <-> ~Ryy). Then Rxx <-> ~Rxx, which is a contradiction.

For set theory:

Suppose ExAy(y in x <-> ~ y in x). Then x in x <-> ~ x in x, which is a contraction.

Thus trying to deny Russell's paradox by appealing to one's personal notion of the concept of 'sets' fails, since the structure of the contradiction does not rely on any concept of set. The principle that there is no set of all sets that are members of themselves is an instance of Russell's paradox, but, as I've shown, that principle does not rely on the any particular concept of 'sets'.

This is witnessed by Russell himself where he explains the paradox by reference to an arbitrary 2-place relation such as 'shaves'. It matters not whether the relation is 'is a member of', 'shaves', 'loves', or 'billwingadoobadoodles'. Seeking to dispute Russell's paradox by recourse of arguing over the concept of 'set' misses the point and is ill-conceived.

However, it is a correct that there is no 'set of all sets' is a corollary from Russell's paradox that DOES depend on a set theoretic notion that is expressed by the Axiom Schema of Separation which says that for any any formula F, and set s, there is the set y whose members are all and only those members of s such that F holds of y. Symbolically:

AxEyAz(z in y <-> (z in x & Fz))

From that axiom we derive that there is not a set of which all sets are members. Symbolically:

~EsAy y in x

Proof is simple:

Let F be the formula '~ z in z'. Suppose, toward a contradiction, that EsAy y in x. By the Axiom Schema of Separation we have EyAz(z in y <-> (z in s & ~ z in z)). So y in y <-> (y in s & ~ y in y). But, by the supposition that Ay y in s, we derive y in y <-> ~ y in y.

So one can deny that there is no set of all sets only by denying that for any property expressible by a formula and for any set, there is the subset of the set whose members are all and only those in the set and having said property.

I hope my participation in this thread doesn't inconvenience or distress you too much. — fishfry

Of course it doesn't. Your posts are uniformly excellent.

I don't see a point in continuing this conversation. — SophistiCat

That's all I meant.

Of course it doesn't. Your posts are uniformly excellent. — jgill

A number of posters around here would beg to differ, but thanks for the kind words.

Just look at the proof that I am presenting without bias and without preconceived notions. — Philosopher19

I have been doing that. I can't add anything to what I've said other than that you should carefully examine the proof of Russell's paradox. And you should carefully examine your own argument, to see that you repeatedly claim that x is a set but you never present an argument to that effect.

consider the hypothetical possibility that all these famous philosophers after Russell, were wrong, and I'm right — Philosopher19

Consider the actual reality (not just a hypothetical possibility) that the mathematicians thoroughly studied the subject matter down to its finest details and understand its rigorous axiomatization, including that set theoretic proofs are machine checkable, while on the other hand, it appears you have not read the first page in a textbook on mathematical logic or set theory.