I get it that the set of all sets that are not members of themselves yields a paradox resolved by ruling that such a "set" is not a set under the rules. — tim wood

~ExAy(yex <-> ~yey) is proven from first order logic alone; we don't need any set theory axiom for that.

metaphorical — tim wood

Everything east is in the class, not set, of all sets. However, in Z set theories, that class itself is not an object; rather we refer to that class as an informal locution from "outside" the theory. But in class theories (such as NBG) the class of all sets is an object, but then there is no class of all classes that is an object in the theory.

A good mirror for us all — tim wood

I'm all for self-scrutiny, but not so overzealous that I chase demons that there is no reason to think exist. I have plenty of faults, but being a crank is not one of them.

how it could be that we can “compute that it does hold” and yet not have, at some level or another, thereby “proven” it. — Pfhorrest

Because the reckoning itself is not necessarily in a formal context, so it is not formal proof, though it could be.

Take a set {P}. If it's impossible to make this set a member of another set — TheMadFool

{P} is a member of other sets.

Meanwhile, as I already mentioned, without regularity, it is not inconsistent that

P = {P} = {{P}} = ... for finitely many iterations as you like.

A statement of set theory is inconsistent it it implies both a statement S in the language of set theory and the statement ~S.

x = {x} is inconsistent in set theory.

It is not the case that x = {x} is inconsistent in set theory without the axiom of regularity.

"something done" is not in the language of set theory, so it's not part of any statement S in the language of set theory. "always" is not in the language of set theory, so it's not part of any statement S in the language of set theory.

/

One thing ridiculous about your arguments is that they simply overlook that you get everything you want anyway just by recognizing that set theory has the axiom of regularity. That is, set theory agrees with you that no set is a member of itself. But it agrees with you via an axiom. But, as I have been making this point that you refuse to understand, if we drop that axiom, then it is not inconsistent for a set to be a member of itself. Of course, a set being a member of itself is intuitively incorrect to most people, which is fine. But inconsistency is not determined by what is intuitively incorrect, but rather by the definition: a set of formulas is inconsistent if and only if the set proves a contradiction, where a contradiction is the conjunction of some statement (in the language) and its negation.

By definition, the set of all sets encompasses all sets. — Philosopher19

By the axioms, there is no set x such that every set y is a member of x.

That's not childishness; it's axiomatic mathematics.

this means that we agree to disagree on this. — Philosopher19

Circular. I asked what is the operative import of "agree to disagree". Your answer is that we "agree to disagree".

(1) Set theory is incomplete, therefore set theory is consistent.
— TonesInDeepFreeze

If that's what you want to believe, then believe. — Philosopher19

It's not just what I believe, it's easily proven, basically from the definitions of 'incomplete' and 'consistent'.

Rejection of the set of all sets is blatantly contradictory. — Philosopher19

You keep repeating that, but without showing an actual contradiction. That is dogmatism.

You are like child in your reasoning and manner of discussion. — Philosopher19

You've shown nothing childish. Especially nothing childish in informing you of the exact mathematical formulations that rebut your ignorant dogmatism.

I shouldn't have to spoon feed you — Philosopher19

Instead, you shove nonsense down the throat.

And I shouldn't have to spoon feed you clear and correct information about the subject on which you ignorantly espouse, but I do.

am I to think that putting, say, a list of items e.g. 1, w, # inside curly braces like so, {1, w, #} amounts to doing nothing? — TheMadFool

You are to think that "doing something" is not a set theoretic predicate.

And you skipped my counterexamples to your incorrect reasoning.

You cannot have a set of all sets that are members of themselves: How are you going to logically write this? — Philosopher19

With regularity, we may prove:

E!yAx(xey <-> xex)

thus justifying abstraction notation:

{x | xex}

and the theorem:

{x | xex} = 0

1. If C doesn't contain itself then C contains itself
2. If C contains itself then C doesn't contain itself

where C = the set of all sets that don't contain themselves.

If sets can't contain themselves, the consequent in 1 above and the antecedent in 2 above become meaningless for sets can't contain themselves. At least that's what I think. — TheMadFool

What you think reflects a profound ignorance of logic. Namely, that the logic is monotonic. Adding the theorem that no set is a member of itself does not eliminate a contradiction otherwise. Once again: The logic is monotonic.

The import of Russell's paradox for set theory is this:

1 ExAy(yex <-> ~yey) [premise]
2 xex <-> ~xex [EI, UI]
3 ~ExAy(yex <-> ~yey) [2]

Then, we derive ~ExAy yex as follows:

1 ExAy yex [premise]
2 Ay yex [EI]
3 AxEzAt(tez <-> (tex & ~tet)) [instance of axiom schema of separation]
4 zez <-> (zex & ~zez) [UI, EI, UI]
5 zex [2 UI]
6 zez <-> ~zez [4 5]
7 ~ExAy yex [6]

Meanwhile, with regularity, we have ~Ex xex. But that doesn't eliminate the proof above. The logic is monotonic.

"something was done to it" is not a set theoretic predicate.

/

S = SuS

With S, "nothing was done to" S. With SuS, "something was done to S".

/

2 = 2*1

With 2, "nothing was done to 2". With 2*1, "something was done to 2".

ZF implies incompleteness in proof, theory or system. — Philosopher19

(1) Set theory is incomplete, therefore set theory is consistent.

(2) Any consistent, recursively axiomatized, arithmetically adequate theory is incomplete.

(3) There are alternative theories to set theory.

(4) You have not given even a whiff of an indication of your own alternate theory.

the post that followed it suggests that you are upset, angry, or frustrated — Philosopher19

There is plenty in this world to be angry and frustrated about. Your ignorant, arrogant, stubborn dogmatism is hardly one of them.

not a good state to be in when discussing matters of logic or pure reason. — Philosopher19

You have not shown any point on which my posts have not been clear and correct. Meanwhile, your ignorance, confusion, and stubbornness don't help you in discussing logic.

I think we should just agree to disagree. — Philosopher19

I don't know what you think the operative meaning of that is. In any case, when you post nonsense and misinformation, I will decide for myself whether to rebut it.

I'm not real knowledgeable myself, and, of course, one shouldn't expect that everyone is equally knowledgeable. But I don't get why someone would post arbitrary non sequiturs.

I might be able to offer an opinion as to whether Godel has proved that there are true propositions within mathematics that cannot be mathematically proven. — Janus

It depends on the definition of 'mathematically proven'.

Is there any proof that such a "formal system that incorporates methods going beyond ZFC." will or even could be found? — Janus

It is easy to see that there are theories that are proper extensions of ZFC . But that doesn't settle the question of whether those theories are within what we consider to be justifiable mathematics, or even the question of what it means to be justifiable mathematics.

what you [Pfhorrest] said about there always being a meta-level wherein the unprovable truths within a system can be proven seems questionable. — Janus

It is the case that there is an infinite escalation of theories, each proving arithmetical truthts not provable in the lower theories.

Would this fact render all such proofs non-exhaustive and/ or trivial — Janus

The theories are not exhaustive, indeed. But I don't see why that would make the proofs trivial.

The general idea you expressed is okay, but I suggest some clarifications and context (much of which you likely know already).

We are concerned not just with the object-language and a meta-language, but the object-theory and a meta-theory.

With a meta-theory, there a models of the object-theory. Per those models, sentences of the object-language have truth values. So the Godel-sentence is not provable in the object-language but it in a meta-theory, we prove that the Godel-sentence is true in the standard model for the language of arithmetic. Also, as you touched on, in the meta-theory, we prove the embedding of the Godel-sentence into the language of the meta-theory (which is tantamount to proving that the sentence is true in the standard model). That is a formal account of the matter. And in a more modern context than Godel's own context, if we want to be formal, then that is the account we most likely would adopt.

Godel himself did not refer to models. Godel's account is that the Godel-sentence is true per arithmetic, without having to specify a formal notion of 'truth'. And we should find this instructive. It seems to me that for sentences of arithmetic, especially ones for which a computation exists to determine whether it holds or not, we are on quite firm ground "epistemologically" to say, without quibbling about formality, that the sentence is true when we can compute that it does hold.

that lower-level system has no idea whether or not it's true (because it's unprovable) — Pfhorrest

There is an even more fundamental reason that the object-theory does not yield a determination of truth. That is that the object-language does not have a truth predicate. There's a subtle difference: A theorem of the object-theory is true in any model of the theory, so in that sense one would say that the object-theory does determine the truth of certain sentences. But the object-theory does not itself have a theorem that the sentence is true in models of the theory (or else, the object-theory would be inconsistent per Tarski's theorem).

what is the mathematical definition of 'abritrary'?

Anyway, I don't think we need 'inherent', 'actual' or 'non-arbitrary'.

It is enough to observe that sets of cardinality greater than 1 have more than 1 ordering, and that any other privileged designation is only by stipulation (such as 'the standard ordering' for specific sets is fine, since it is stipulated and defined), and there is no overall rubric of individuation for the orderings of sets in general.

It's always bothered me that there is no containing set as required by specification. — fishfry

There is. It can be any set whatsoever.

Here's a proof:

AyExAz(zex <-> (zey & ~z=z)) instance of axiom schema of separation
(zex <-> (zey & ~z=z)) ​UI, EI, UI
z=z identity theory
~zex sentential logic
ExAz ~zex UG, EG
(Az ~zex & Az ~zey) -> x=y extensionality
E!xAz ~zex from definition of '!'
x=0 <-> Az ~zex definition

It is typical that cranks confront the implications of Russell's paradox with hostility (also the other antinomies, incompleteness, the halting problem, and uncountability). Cranks just can't wrap their mind around such things. Thus, often cranks can be paraphrased as "Look at all those foolish logicians getting all twisted up about a silly little word game. They make all these crazy convoluted theories to deal with a big nothing that could be dealt with so easily like I do. Now stand back and behold my so simple way of dismissing the paradox."

In this case, we have the slight twist that it's not so much the paradox itself that has the crank's objection but rather that the paradox entails that there is no set of all sets.

What you and mainstream set theorists seem to think is that this logically entails that the set of all sets is contradictory — Philosopher19

I don't just think it. I prove it from the axioms of Z set theory. You are welcome to present your own system and axioms in which there is a set of all sets.

By the way, there are alternative systems in which there is a set of all sets, but their formation syntaxes, inference rules, and axioms are very different from Z set theory. If you were sincere in sustaining having a set of all sets, then you would look up such alternative systems instead of flaunting your utter ignorance of what proof is.

I have provided proof of this — Philosopher19

You provided muddled argument that depends on an undefined notion of "member twice". Proof, on the other hand, is from axioms with rules of inference.

You seem to have not understood why I have said "member of itself twice" — Philosopher19

You ignored my request - more than once - that you define it.

You've rightly recognised that a set of all sets that are not members of themselves that is itself not a member of itself is contradictory. — Philosopher19

It's contradictory, a fortiori, since a set of all set that are not members of themselves is contradictory.

There is nothing contradictory about the set of all sets. — Philosopher19

it contradicts the axioms of Z set theory. You don't have to accept Z set theory, of course. But then to prove that there is a set of all sets in some other system, then you would need to state your axioms and inference rules for that system.

ZF is either inconsistent, or not comprehensive enough (which ultimately means it is inconsistent). — Philosopher19

ZF has not been shown to be inconsistent. And lack of comprehensiveness does not imply inconsistency. Indeed, an inconsistent theory is ultimately comprehensive since it proves every statement.

It is typical of cranks to use the word 'inconsistent' in their own personal way. Usually for the crank it just means the crank disagrees with, or dislikes, something.

This fierce dogma needs to die. — Philosopher19

I have not asserted any dogma. Rather, I have stated the fact that the axioms of set theory prove that there is no set of all sets. That is a finitistic fact that can be objectively, mechanically checked. Meanwhile, I am quite happy to allow that one may present alternative theories that do prove that there is a set of all sets. That is the opposite of dogmatism. And it is not dogmatism to point out that your own arguments are handwaving and rely on undefined terminology. Indeed, it is your own conviction that is dogmatic.

Please stop using '=' to stand for the biconditional.

(1) N = {N} premise

(2) N e N <-> {N} e {N} from (1)

(3) ~ {N} e {N} non sequitur

Why do you waste our time?

It fallacious to argue that, because N is written as '{x N}' in '{x {x N}}' but as 'N' in '{x N}', we have that {x N} is not in {x N}. It is fallacious to argue from the mere happenstance of two different means of notating the set.
— TonesInDeepFreeze

Ok. I've run out of options. Let's get straight to the brass tacks. — TheMadFool

First, do you understand my explanation that you just quoted?

we cannot know for sure that any given proposition is true-but-unprovable, because to be sure of the first part we would have to violate that second part. — Pfhorrest

It is unprovable in the system being discussed. It is provably true in the mathematics used to discuss that system.

It might in some principled way remain the case that something or another could be true but not provable, but we could never say for sure that we had an example of that — Pfhorrest

We have a sure example. It's as sure as finitistic combinatorial arithmetic. It's a quite complicated calculation, but it is still a finite calculation.

an actual order — Metaphysician Undercover

That's just another undefined term by you as is 'inherent order'. It adds nothing to your incorrect argument.

There are many orders. You have not defined what it means to say that one of the orders is 'AN actual order' or 'THE inherent order' while none of the other orders are an actual order or the inherent order. It is a remarkable feat of stubbornness for you to persist in not recognizing that clear and simple fact.

And we still have this classic:

"Order" is defined as "the condition in which every part, unit, etc., is in its right place". — Metaphysician Undercover

Clearly, we have not been talking about that sense of 'order'. Credit to you though for your sophistical resourcefulness in looking at a dictionary to find a different sense of a word to distract from the sense that has been used (even by you) throughout the discussion.

x is either no elements or x is some elements — TheMadFool

That's not phrasing I've ever seen in set theory. I already told you I don't know what that means. I don't know why you present it again without saying what it means.

{x, {x, N}} - {x, N} = {{x, N} — TheMadFool

Wrong.

{x, {x, N}} - {x, N} = {y | y e {x {x N}} & ~ y e {x N}}.

But {x N} = N, so {x N} e {x N}, so ~ {x, {x, N}} - {x, N} = {{x, N}}.

It fallacious to argue that, because N is written as '{x N}' in '{x {x N}}' but as 'N' in '{x N}', we have that {x N} is not in {x N}. It is fallacious to argue from the mere happenstance of two different means of notating the set in the way you have.

You're talking about Russell's paradox in context of sets, and about set membership, and using extensional braces.

But your posting shows that you have no understanding of set theory. So, contrary to your claim, of course you don't understand my points.

Do you the know primitive of the language? Do you know the axioms? Do you know the proof system? Do you know the method of definitions? Do you know any of the basics? Do you know anything at all about it?

I've understood your points — Philosopher19

No, you skipped my points.

x is a member of itself twice — Philosopher19

You see, right there, you skipped my point, posted at least three times now, that "member of itself twice" has no apparent set theoretic meaning. Or if you think it does have meaning, then define it in set theoretic terms.

Moreover, even if you did define it, you would still need to show that "member of itself twice" implies a contradiction.

or x is a member of itself, with y and z not being members of themselves — Philosopher19

So? There's no contradiction with x = {x y z} and ~yey and ~zez.

[Of course, we always bear in mind, per context, whether we are working with or without the axiom of regularity.]

It does not have such a subset as evidenced by such a subset being clearly contradictory. — Philosopher19

Look at the word "IF" [emphasis added] I wrote.

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

So, indeed, since there is no set of all sets that are not members of themselves, there is no set of all sets.

How do you possibly allow yourself to reject the set of all sets? — Philosopher19

By deriving a contradiction from the assumption that there does exist a set of all sets. I made that clear.

a contradiction is clear in the following:

1) 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. — Philosopher19

Yes {1) 2)} is an inconsistent set of statements, since {1)} itself is an inconsistent set. It is inconsistent since "There is a set that has as members all sets that are not members of themselves" implies a contradiction, so, a fortiori, "There is a set that has as members all sets that are not members of themselves and that set is not a member of itself" implies a contradiction.

I don't think you read my post with enough attention to detail — Philosopher19

I read it in such detail that I pointed out exactly its fatal errors.

your belief system is contradictory precisely because it sees 2 as being rational/consistent, whereas 2 is clearly contradictory. — Philosopher19

We prove from the axioms that there is no set that has all sets as members. If you want to have a set that has all sets as members, then state your axioms and prove from those axioms that there is a set that has all sets as members..

I definitely believe in a set being a member of itself. I think it is a logical necessity. — Philosopher19

How is it logically necessary? (With an ordinary understanding of 'logically necessary'.)

x (x, y, z).

Do you mean x = {x y z}?
— Philosopher19

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 — Philosopher19

I have said:

(1) "member of itself twice" has no apparent set theoretic meaning. Of if you think it does, then you should define it set theoretically.

(2) A contradiction is a statement and its negation. You have not shown that x = {x y z} implies a contradiction.

nobody seemed to have recognised the impossibility of a set of all sets that are members of themselves. — Philosopher19

I have said, it's not impossible. It is correct that there is the set of all sets that are members of themselves. It is the empty set.

You just keep repeating yourself without coming to grips with the key points that refute you.

If L is true then L is false = L is false or L is false = L is false — TheMadFool

The equal sign there and the ones following it are not syntactical. Perhaps you mean the biconditional. If so, rewrite to see whether your argument still holds up,

"[...] a self-referential sentence which “says of itself” [...] Such figures of speech may be heuristically useful, but they are also easily misleading and suggest too much." - https://plato.stanford.edu/entries/goedel-incompleteness/#DiaSelRef

Read more there for more explanation.

On the other hand, see page 44 of Franzen's'Godel's Theorem: An Incomplete Guide To Its Use And Abuse' for a different take on the matter.

In any case, in whatever sense we may reasonably say the proof uses "self-reference", the proof is not vitiated by it, as the proof can be carried out in finitistic combinatorial arithmetic.

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.

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.

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".

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."

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.

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.

"Order" is defined as "the condition in which every part, unit, etc., is in its right place" — Metaphysician Undercover

That's not the sense of the word 'order' we're talking about! You have yourself even being using 'order' in a sense not expressed as "in its right place". Wow, you really do have a cognitive problem.

It's a spatial order, each dot has its own specific position on the plane. — Metaphysician Undercover

That is not specification of an order, let alone of "THE inherent order".

Of course, we could move left to right and also up and down, zigzagging to specify an ordering of the dots. But we could also do it right to left, or from spiraling from the center outward, or from one particular dot anywhere. So there is not one single order that is "THE inherent order".

It's amazing that you don't get this. Or that you simply evade, as you are Metaphysician Undercover from Evasionsville.

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.

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

Gödel used a modified version of the liar paradox, replacing "this sentence is false" with "this sentence is not provable" — Wikipedia

Yes, there is an analogy and similarity, but also the modification used makes a great difference too.