The Absolute Russell Set Exists

I’ve already showed you how to define the class of all sets (or things, if you prefer) that are not members of themselves:

$$R = {x : x \notin x}$$

As far as your saying you don’t have to define something to show it exists, surely even you can see that’s insane.

I’m not interested in classes; I’m interested in sets. I’m skeptical of the difference between sets and classes. The difference seems rather artificial and unnecessary.

Just because a thing is not defined, doesn’t mean it doesn’t exist.

Are there any sane people on this forum?

Yes, I am a sane person on this forum.

There’s a difference between a definition and a description. All definitions are descriptions, but not all descriptions are definitions. The description I had initially given of the absolute Russell set, while on another web page, exclusive, and sufficient for my purposes, was not regarded by me as a formal definition. I believe I may have regarded so intentionally, because I may have suspected the definition of the absolute Russell set was already given in a printed textbook, the one by Barwise and Etchemendy.

I should not have to define in this thread everything that has already been defined elsewhere. That decreases the value of this thread and increases its unattractiveness to people looking for original thought.

Nonsense. The phrase “absolute Russell set” does not appear on the Internet except as a reference to the standard Russell set.

I truly doubt your text defines such a thing the way you’re using it. Feel free to post a screen shot though, maybe I’ll learn something.

It would not be possible for anyone to do that.

LOL.

I am referring to the standard Russell set. However, the Russell set, according to page 432 of the aforementioned Barwise and Etchemendy (1999, 2000, 2002, 2003, 2007, 2008), is the set of all things that are not elements of themselves, not the set of all sets that are not elements of themselves. The set of all things that are not elements of themselves is a more natural and better set to be called the Russell set than the other set. It’s more comprehensive. I concede I did coin the term the absolute Russell set myself. But my term was readily derived from page 432 of Barwise and Etchemendy (1999, 2000, 2002, 2003, 2007, 2008), where they put the word absolute in parenthesis.

Barwise and Etchemendy (1999, 2000, 2002, 2003, 2007, 2008) do not seem to formally define the absolute Russell set; they only describe and discuss it.

This is where copyright law comes into play. I’m hesitant to copy a copyrighted formal definition or symbolic expression, especially when it may not be common knowledge. I actually, in my past post at viewtopic.php?p=2699066#p2699066, which I previously cited in this thread twice already, did not use the same letter to represent the absolute Russell set as Barwise and Etchemendy (1999, 2000, 2002, 2003, 2007, 2008) do. I was afraid it might be a copyright violation or plagiarism.

I disagree. It could be worse. It could get worse.

I wish to be enlightened and to advance.

Thanks, that only took me a week to drag out of you, after you claimed several times that you found it in a math book.

So what you are calling the “absolute Russell set,” everyone else calls the proper class of all sets (or things, really doesn’t matter) that are not members of themselves. That’s really all there is to all this.

wtf:

I’m not as familiar with class theory as I am with set theory. If you are correct, then my claim is that the absolute Russell proper class exists.

I’d rather keep the discussion focused on sets rather than classes. There’s no need to speak in terms of classes when it’s evident I’m speaking in terms of sets.

Yes. It does. As I’ve noted a couple of times, (R = {x : x \notin x}) defines the Russell class. It’s the class of all things that are not members of themselves. It “exists” in the sense that its definition does not lead to contradiction. Of course here we are talking about formal logical existence, not anything in the real world. I hope we agree on that.

That’s fine. But the claim that (R) is a set leads to a contradiction.

Of course you could say it exists in an inconsistent version of set theory. Nothing wrong with that. As you’ve already pointed out, in an inconsistent system everything is true.

But even here it is not settled, since contradiction and non contradiction are in an Absolute Russell set , self inclusive sets?

The feeling I have is, that redefinition does nothing but reassert the primacy of naive logic. Can that allowance withstand other succeeding arguments? Or, is it other systems of classes , when weighed in, Change the balance ?

Meno_, I read some of your other posts on the forum and I realize that you are only communicating in your own particular style; and not at all trying to annoy me personally. Thus it was wrong for me to attack you personally. I apologize.

I still do not understand most of what you write. But some of your references are interesting as I look them up. I think there is a kernel of interestingness in your exposition, if only I could discern it.

I hope you’ll accept my apology and perhaps try to explain yourself better to a humble math guy like me.

Removed .

In the spirit of the discussion … an empty message means everything.

You’re right , and upon reassessing my obvious lack of preparation , and need to properly respond. don’t take my erasure as anything else but a preparation for a reasonably written format.

I must admit, I have a lot to do until then.
I am confident of that in general terms.

You raise two interesting points.

First, there is nothing wrong with self-inclusive sets.

Why can’t we have a set (x) such that (x \in x)? There is no reason we can’t, and in fact this is perfectly consistent with the other axioms of set theory. But for intuitive reasons – namely, (x \in x) violates our intuitions about sets – we don’t want to allow that. So we simply declare an axiom that says we can’t have (x \in x).

Of course then we might still have (x \in y) and (y \in x), or even longer chains such as (x \in y), (y \in z), and (z \in x). So there is a clever axiom that outlaws all of these circular chains of inclusion. It’s called the axiom of foundation or sometimes the axiom of regularity.

What happens if we don’t include foundation in our axioms? Then we get the study of non well-founded set theory. It’s obscure but it’s studied and even applied in some disciplines.

So first point, there is nothing inherently wrong with self-membership. It’s only outlawed in standard set theory because it doesn’t fit our intuition about what sets should be.

The second point is that browser has a misunderstanding about contradictions. Just because you have some proof that ends in a contradiction, it doesn’t mean math is broken or that everything is true. It just means that you have to throw out the assumption that led to the contradiction.

For example in Euclid’s famous proof of the infinitude of primes, we start by assuming that we have a finite list of all the primes, then we show that this leads to a contradiction. We haven’t broken math or proved everything is true. All we’ve done is shown that the assumption that there are finitely many primes is false. Nothing else.

Browser keeps saying that because we have some proof that leads to a contradiction we can use that contradiction to show that math is inconsistent. But that’s wrong. All we’re showing in the Russell proof is that the class, or collection, of things that are not members of themselves can not possibly be a set. That’s all we’ve shown. There are no implications beyond that fact.

Interesting word choice. Naive set theory is the essentially Frege’s failed set theory in which sets can be formed out of unrestricted predicates, such as the “set of all things that are not members of themselves.” Russell showed that this idea leads to a contradiction. So naive set theory fails. Sets can’t be thought of as simply collections of things satisfying some predicate. Rather, a set is something that conforms to our axioms, which are chosen carefully to avoid contradictions.

Note: “Naive Set Theory” is also the name of a standard undergrad set theory text by Paul Halmos. It’s NOT actually about naive set theory; it’s about axiomatic set theory. No idea why Halmos chose that inaccurate title but it’s a great book, highly recommended for people interested in set theory. Very readable.

en.wikipedia.org/wiki/Naive_Set_Theory_(book

Russell’s paradox shows that the collection of all things that are not members of themselves can not possibly be a set. In ZFC (the standard axiom system for math) there is no such thing as a proper class, so in ZFC we simply say the Russell set doesn’t exist. But there are other systems of set theory that formalize proper classes, and then the Russell class has official standing as a proper class: a well-defined collection that’s “too big” to be a set.

That’s incorrect; I do not have to throw out the assumption. I am permitted to keep the assumption. That permission was discussed in another thread I started, “Anything Can Be Postulated” at viewtopic.php?f=1&t=193975. An assumption and a postulate are in a sense the same; they are both a proposition that is assumed to be true.

That would be an oversimplification. The proof is not just any proof; it is a proof of a very counterintuitive claim.

That’s what’s traditionally said, but it’s not true. I’m interested in taking the other road.

I did previously encounter that book for sale online. I thought it interesting how somebody would work to such an involved level in naive set theory, since the theory’s inconsistent. Perhaps I shouldn’t judge a book by its cover.

I’ve learned fromexperience that it’s not helpful to me to follow your links to your other posts. Please say what you have to say in the thread I’m reading.

Why are you permitted to keep the assumption? If I assume there are finitely many primes and that 11 is the largest one, and then I show that 2 x 3 x 5 x 7 x 11 + 1 must be divisible by some prime larger than 11, how do you figure you can keep the assumption that 11 is the largest prime?

If an assumption leads to a contradiction we throw it out. Of course if you like working with inconsistent systems you may certainly keep assumptions that lead to contradictions, but then you get a useless system.

Math is full of counterintuitive claims. What does that have to do with keeping assumptions that lead to contradictions?

It’s not just “traditionally said.” It’s formally proved. If (R = {x : x \notin x}) and we assume (R) is a set, we get a contradiction. It’s a very well known proof. It’s disingenuous to characterize this as “traditionally said.” It’s like saying it’s traditionally said that there is no largest prime. Well yeah, it’s traditionally said because Euclid proved it and the proof is understandable to high school students.

You should definitely read that book. It’s very clear and understandable and you’ll learn a lot from it.

I am permitted to keep the assumption because there is no rule that requires me to do otherwise.

The assumption can be kept because the assumption implies no contradiction.

In a proof by contradiction, that’s what we do. But no assumption ever has to be thrown out. The first proof of ex contradictione quodlibet at en.wikipedia.org/wiki/Principle … sion#Proof actually assumes a contradiction at the very first step, and proceeds therefrom.

Inconsistent systems aren’t necessarily useless. Naive set theory, for example, has brought society much enlightenment. Inconsistent systems themselves can be used to prove trivialism, as I’ve argued in another thread I started, “Inconsistent Theories Metatheoretically Prove Trivialism” at onlinephilosophyclub.com/forums/ … =2&t=15559. That thread was somewhat inspired by the discussion I have been having with you in this thread.

If one claim is less intuitive than a second claim is, then it may be that the first claim is more difficult to keep as an assumption than the second claim is. People have limited strength. They have limited abilities. People may be uncomfortable acting on a belief that is difficult to attain or maintain. They may be unable to act on such a belief.

That’s what it is at this point in time; Russell’s paradox is disingenuous. It’s hackneyed. It’s more ingenious to talk about the existence of the absolute Russell set than it is to talk about its nonexistence.

As I said at viewtopic.php?p=2699659#p2699659, make evident that the existence of the absolute Russell set implies some contradiction. Since those proofs prove the existence of the absolute Russell set, the set exists. Since the set exists, some contradiction exists. Thus, by ex contradictione quodlibet, trivialism is true. Since trivialism is true, there is the largest prime.

There is a principle that we are seeking a consistent framework for mathematics. You are entitled to desire an inconsistent framework, but that seems pointless.

If you think 11 is the largest prime, we’re done. Actually we are done. You haven’t said anything new in quite a while.

You greatly misunderstand proof by contradiction. But if you enjoy typing Latin phrases, by all means enjoy yourself.

I would agree. But “not necessarily useless” is not the same as “We should adopt inconsistent mathematics.” The latter would be counterproductive.

You should not assume I have not looked at your other online work. I’ve seen the nudes, I’ve seen the restraining order filed against you. What of it? If you have something to say, say it here.

Whatever that word salad means.

If you reject the concept of mathematical proof, we’re done. Which we are.

All the best.