The Absolute Russell Set Exists

Contradictione quidlobet is a shift to a functional approach to verifying a second order logic , grounding a utilitarian-positivism, meaning it tries to overcome the naturalistic fallacy, utilizing a neo-Kantism.

Marbourg school-Cassirer.

It is, as if in Ayer’s behaviorist model, what signifies meaning. through the function of language where usage determines the logical structure , not some intrinsic property.

This structural hierarchy coincides with the epistemological significance of semantic usage, rather then the intractibly logical signifier, of a first order hierarchy , which does cause contradiction. The usage will will synthesize this by reasserting the logical unity between usage , function and meaning.(Typical logical progression. Is de-emphasized, in favor of logically ordered sequences).

This is the ontic, rather then the ontological take on it.

I hate to appear defensive, but a proof predicated on an assumption is only part of the story, and this is an effort to show the fallibility of that assumption.

I just learned this : in classical logic the principle of acceptance of contradictory statements is valid, but in relevance logic it is rejected. This backs up the idea that such a logical proof is not totally relevant therefore invalid.

The paradox comes in where two different types of logic are conflated.

It is an interesting mind game, however the claims upon which the assumptions are based have lost their significance.

But thanks for the opportunity !

This post includes a second proof that the absolute Russell set exists. The second proof is a modified and arguably stronger version of the first proof. All conditional statements in the modified proof are strict; that differs from the first proof, in which some conditional statements were material.

Second Proof.

Lemma 2. If a statement is false, then if the statement is true, then some contradiction exists.

Proof of Lemma 2. It is given that a statement is false. Assume the statement is true. Then some contradiction exists. Discharge the assumption. So, by conditional introduction, if the statement is true, then some contradiction exists. This concludes the proof of the lemma.

Postulate 2. If one statement implies it is not true that a second statement, then the first statement does not imply the second statement.

Argument for Postulate 2. Postulate 2 seems intuitively true, despite the fact that Postulate 2 would traditionally be false due to the case in which the first statement is necessarily false, since, as suggested by Property 4 of Theorem 2.1 on page 11 of the March 23, 2014 Edition of Want Theory #6 by me, any conditional statement with a necessarily false hypothesis is true. However, a conditional statement with a necessarily false hypothesis is only vacuously true, and treats the false hypothesis as true, yielding a contradiction that by ex contradictione quodlibet implies and does not imply everything. So, such a conditional statement is true and false simultaneously. Thus, for the sake of Postulate 2, if the first statement is necessarily false, then the conditional statement “if the first statement, then the second statement” is considered false. This concludes the argument for Postulate 2.

Assume the statement “the absolute Russell set exists” is true. By simplification, the absolute Russell set exists. As I’ve proved in my post at http://www.ilovephilosophy.com/viewtopic.php?p=2699066#p2699066, the absolute Russell set both is and is not an element of itself. wtf was critical of that citation in my first proof, so I will explain the linked proof some more here. That proof invokes the fact that the absolute Russell set is an element of itself if and only if it is not an element of itself. That fact is readily derived from a property of the absolute Russell set: a thing is an element of the absolute Russell set if and only if it is not an element of itself. As I had done in the debate I previously proposed and referred to, for more context, I cite pages 432, 433, and 434 of Language, Proof and Logic (1999, 2000, 2002, 2003, 2007, 2008) by Jon Barwise and John Etchemendy, and https://en.wikipedia.org/wiki/Russell’s_paradox#Formal_presentation.

Since the absolute Russell set is and is not an element of itself, some contradiction exists. It follows by ex contradictione quodlibet that no contradiction exists. Discharge the assumption. Thus, by conditional introduction, if the statement “the absolute Russell set exists” is true, then no contradiction exists. So, by Postulate 2, it is not true that if the statement “the absolute Russell set exists” is true, then some contradiction exists. Invoking modus tollens with Lemma 2 and the previous sentence, it is shown that the statement “the absolute Russell set exists” is not false. Thus, the statement is true. Therefore, the absolute Russell set exists. This concludes the second proof.

As to the first proof, although the truth table for material implication may be correct, it is incomplete. Material implication is not truth functional as has been believed. As suggested by my argument for Postulate 2, whenever the hypothesis of a material implication is false, the implication is not just true, but it is also false. That claim is supported by the fact that if a false statement is assumed true, then by ex contradictione quodlibet, all statements are true and false simultaneously. So, a false statement always materially implies and always does not materially imply every statement. For these reasons, the following Postulate 1, which was implicitly invoked in the first proof at the step I explicitly claimed may be questionable, is true.

Postulate 1. If one statement materially implies it is not true that a second statement, then the first statement does not materially imply the second statement.

Note that while the hypothesis and conclusion of Postulate 1 are a material implication and material nonimplication, respectively, Postulate 1 itself is a strict implication.

Bizarre that you’d write all this word salad without bothering to define the “absolute Russell set,” which has no meaning in standard math or logic.

But if you merely mean the class of all sets that are not members of themselves, I already showed you how to define it:

$$R = {x : x \notin x}$$
This is a perfectly well-defined proper class. It’s just not a set.

Proper classes are formally defined in some versions of set theory. In ZFC where they’re not formally defined, they’re used informally, meaning a class that’s “too big” to be a set. The class of all sets, the class of all Abelian groups, the class of all topological spaces, the class of all sets that are not members of themselves, etc. All those are proper classes either in the formal or informal sense.

With your interest in the subject, why don’t you study some basic set theory and logic? You’d find it interesting and fun.

I do not have to define the absolute Russell set. A property I described it to have should have sufficed for the discussion so far. I now define the absolute Russell set as the set that has a property I have ascribed to it in my previous post; I define the absolute Russell set as the set such that

I’ve already done that. And I am not convinced, despite formal proofs, that neither the absolute Russell set nor the universal set exist.

Just because a set is inconsistent, doesn’t mean it doesn’t exist.

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.