What book or article in the subject have you read/researched?

Won't repeat myself. — Philosopher19

Oh, yes you will. I guarantee it.

So it's settled in your mind that you won't read even a single book or article about this subject, but that others should read your website?

we already know you're ugly [...] you appeared highly bothered after I asked if that's why you don't show your face [...] It should be obvious that I'm trolling you. [...] Might loosen up your butthole a little so you can actually poop, my man. — Vaskane

Ugh.

Don't have time for all the replies I want to make lately, but this one is easy:

Problem with Set Theory is that their concept "infinite" means "finite" — Corvus

A common definition of 'infinite' in mathematics is 'not finite' You have it completely wrong. Would that you would not persist in posting falsehoods.

it seems new ideas in
maths are rare — Christoff Montnielsensons

What is your basis for saying that?

Discovering
imaginary, real, natural or end-
less numbers for example,
seems to be nobody's business. — Christoff Montnielsensons

Why do you believe that?

As we're setting up curricula to fulfill your own point of view, we should keep in mind that AI itself came from mathematics and the study of computabilty that has not been limited to your point of view. The very advent of the digital computer came from mathematicians who whose creativity with abstractions abounded and redounded to the sciences without having to subscribe to whatever it is you would instruct them about physicality. And also with developments in computer science and the sciences since.

Of course, in the philosophy of mathematics we already have great debates and application of intellect to questions of concreteness, applicability, etc. And we have the field of applied mathematics. And we have people researching the relationship of mathematics with the sciences.

But there are always avenues of improvement. So I'm interested what specific research, what texts and resources you advocate should be course requirements for mathematics students, and what philosophies you think should be inculcated and what philosophies you think should not. Would your curricula for mathematicians include contrasting philosophies across a range in the philosophy of mathematics or would the only required material be study of philosophy that conforms to your own beliefs about mathematics and the physical?

So the Russell set as a rejected defined mathematical object does not exist and cannot be a paradox. — Mark Nyquist

A paradox occurs when there is a contradiction or highly counter-intuitive statement that
follows from premises or principles that we regard as themselves true, reasonable or intuitive.

I suggest that the supposed Russell set is not itself a paradox.

But rather, the paradox is that from the premise "for any property of sets, there exists the set whose members are all and only those sets that have that property" we get "There exists the set whose members are all and only those sets that have the property of not being members of themselves (i.e., the supposed Russell set)", which implies the contradiction "The Russell set is a member of itself and the Russell set is not a member of itself".

There, the premise that up to 1901 we thought was true, reasonable and intuitive is "for any property of sets, there exists the set whose members are all and only those sets that have that property" but the contradiction that follows from it is "The Russell set is a member of itself and the Russell set is not a member of itself".

So we see that what we thought back around 1901 to be true, reasonable and intuitive, and that we moreover had as an axiom around that time, is actually worse than false as it implies an outright contradiction. Therefore, since Russell's paradox was announced, mathematicians have eschewed using the aforementioned premise, have eschewed using it as an axiom.

Note that this can be seen to be a matter even more basic than set theory. We can see formulate it even without mentioning 'set' or 'member':

For any 2-place relation R, there is no x such that for all y, y bears R to x if and only if y does not bear R to y.

For example, famously:

There is no Mr. X such that for everyone, they bear the relation with Mr. X of being shaved by Mr. X if and only if they bear the relation with themselves of not being shaved by themself.

More simply: There is no one who shaves all and only those who do not shave themself.

Or:

There is no one who loves all and only those who do not love themself.

Doesn't matter what the binary (2-place) relation is: 'member of' or 'shaves' or 'loves', etc.

Time, inclination and patience permitting, I hope to get caught up at some time to responding to the recent various misconceptions, non sequiturs, strawmen, etc. posted in this thread.

The way it is done in ordinary formal mathematics is that there are open terms and closed terms.

Open terms have free variables. Closed terms have no free variables.

A constant is a closed term.

There are primitive closed terms and defined closed terms.

Before defining a constant, we must first prove there is a unique x such that x that satisfies a formula whose only free variable is x.

There are also primitive predicate symbols and defined predicate symbols.

An n-placed predicate symbol is defined by a formula having at most n free variables.

In set theory, there is no constant nicknamed 'infinity' (not talking about points of infinity on the extended real line and such here). Rather, there is the predicate nicknamed 'is infinite'. However we do define constants for certain infinite sets, such as [read 'w' here as if it were the Greek letter omega]:

x = w iff for alll y, y is a member of x iff y is a natural number. (The formula that is satisfied by one and only one x is "for all y, y is a member of x iff y is a natural number".)

That is indeed a doozy. Yet AI Chat is only somewhat less misinformational than Wikipedia.

In B, A is not a member of both A and B. — Philosopher19

That is incoherent and has no apparent meaning.

"A is a not a member of of both A and B" does not take a qualifier "In B".

In general: Let P be any formula of set theory and B be any set. We don't say "In B, P". Putting "In B" before "P" has no meaning and indeed is not even grammatical in this context.

Philosopher19 has some word salad dogma tossing around in his head. His first step should be to at least learn how to coherently express whatever it is he's trying to say.

Okay. — Mark Nyquist

Great. That's progress.

You [Michael] seem to be arguing for the paradox after the paradox has been dismissed. — Mark Nyquist

I don't know what you mean by "dismissed" but, mathematics came up with a system in which the paradox does not occur.

The math procedes in a way that assumes an ultimate set will exist. — Mark Nyquist

"The math" refers to what mathematics?

After Russell discovered the paradox, mathematicians replaced the systems that allowed the paradox with systems that don't allow the paradox. With set theory since the early 20th century, we have systems that don't have the paradox, as indeed the systems prove that there does not exist a set whose members are all and only the sets that are not members of themselves.

I'm saying ultimately the Russell set does not ultimately exist. — Mark Nyquist

And you can't have a paradox if the defined mathematical object does not exist. — Mark Nyquist

Then you agree with set theory.

Never had the set in the first place. — Mark Nyquist

Who or what never had the set? And what is "the first place"? You, personally? People who don't work with set theory?

The fact that paradoxes develop means it is a defined mathematical object that does not exist. — Mark Nyquist

In a theory with unrestricted comprehension we define the set whose members are all and only those sets that are not members of themselves. And having that set implies a contradiction.

In modern set theory, we do not have unrestricted comprehension and there is no way to define a set whose members are all and only those sets that are not members of themselves.

ZFC does this by not allowing a set to be a member of itself. — Michael

That's a common misconception.

Yes, ZFC has the axiom of regularity that implies that no set is a member of itself. But that doesn't avoid Russell's paradox.

Rather, Russell's paradox is avoided by not having unrestricted comprehension.

Even with the axiom of regularity, and even if there are no sets that are members of themselves, with unrestricted comprehension we would have the set of all sets that are not members of themselves, thus a contradiction.

Explicity:

1. ExAy(yex <-> ~yey) ... instance of unrestricted comprehension
2. Ex(xex <-> ~xex) ... from 1

Having the axiom of regularity, or any other axiom, does not block getting 2 from 1.

The way to not have 2 is not to have 1, irrespective of the axiom of regularity.

infinity and infinite sets are also used in everyday language outside of set theory — RussellA

Since you are harkening to the original post, see that it is a question about the infinitude of intervals on the real number line, and about the number of different infinite sizes. The ordinary context of that is mathematics and set theory. Anyone is welcome to consider the question in another context, but that doesn't make it inapposite to talk about it in the context of mathematics and set theory.

As the OP doesn't refer to the very specific field of "set theory", having its own particular rules, I think the OP should be considered as a problem of natural language. — RussellA

That's a non sequitur. That the poster didn't mention set theory by name does not imply that set theory would not be a natural context for the matter, especially as the question gave a mathematical context and refers to a concept that is characteristically set theoretic. Moreover, discussion doesn't even have to be limited to whatever unstated context the poster himself might have had in mind.

Within natural language, the question "are there an infinite number of infinities" is meaningless — RussellA

If that is true, then even more reason why one would then consider the question in regard to mathematics. If it's meaningless in context C but defined in another context D, then it wouldn't make sense to say that then it is inapposite to context D.

because it is a member of itself as opposed to another set). — "Philosopher19

False dichotomy. A dichotomy insisted upon only by your idiosyncratic dogma, enabled by your inability to turn off in your head for even one moment your own voice telling you that you are right, that you must be right, so that you don't step back for even a second to question yourself, to consider that the dichotomy that you so obdurately cling to might actually not be true when you think about it just a bit more.

You are merely restating your dogma that being a member of itself is mutually exclusive with also being a member of another set.

I proved that that dichotomy is false.

it is unclear to me as to what it's doing — Philosopher19

I stated exactly what it clearly does:

It shows that P and Q are not equivalent.

I believe it deliberately strays from what is clear simple language to try and force something that cannot be forced (perhaps due to dogma). — Philosopher19

The language is as simple as it can be while being exact, rigorous and not skipping the details.

And previously I gave you a more simply worded version, not belaboring every detail. So, since you didn't understand it, this time I gave you every exact detail.

Moreover, that you don't understand such a straightforward, rudimentary proof is not the fault of the proof, but rather your fault for your utter unfamiliarity with basic logic and basic, common notions about sets.

And there is no dogma. I used only basic logic and utterly common notions about sets.

B) When a set is a member of itself, it is not a member of another set. — Philosopher19

THAT is dogma. You have no proof of it, and I gave an exact disproof of it.

You are hopeless.Three years in which you have made not a mote of progress in understanding anything in this subject.

The textbook axioms and formal proofs of the theorems are subject to change or found out to be falsity at any moment when someone comes up with the newly found axioms and proofs against them. — Corvus

Of course, my point went right past you no matter that I explained it clearly.

There are many different and alternative formal axiom systems in mathematics. Mathematicians and philosophers sometimes disagree on which axioms are best, most intuitive, and even true to some concepts. That's a good thing. But the point that went past you is that what is objective, even among them, is that for each one, there is a mechanical procedure to determine whether a purported formal proof is indeed a formal proof allowed from the given set of axioms and inference rules. And, as people may disagree as to what axioms are best or even philosophically or conceptually justified, at least in the formal sense, one doesn't "disprove" an axiom or set of axioms as you seem to imagine (except, of course, by showing that the axioms of a given system are inconsistent with themselves; and by the way, there are at least two famous cases where axiom systems were proven inconsistent - Frege's and one of Quine's, examples that it is not the case that mathematicians follow blindly and uncritically, ).

No matter what the textbooks say, one must be able to ask Why? instead of just blindly accepting the answers and claim that it is the only truths because the textbooks say so. — Corvus

Again, you are unfamiliar with any of this; you are blindly punching.

We have axioms and rules of inference. Textbooks often do explain the bases for the axioms and rules of inference and do not require blind acceptance. Then, with the axioms and inference rules given, it is objective whether or not a purported proof from those axioms and with those rules is indeed a proof from the axioms with the rules. So that does not require blind acceptance. The process is to state the axioms and rules, often providing intuitive bases for them, then proofs of theorems, as those proofs can be checked. And a good student does check the proofs, both to understand them and to verify for themselves that it is indeed a proof from the axioms with the rules.

But with the inference rules, it's even better. In a mathematical logic, we PROVE that the inference rules are justified in the two key ways: The rules permit only valid deductions and the rules provide for every valid deduction.

On the other hand, blind acceptance is when mathematics is not given axiomatically. The teacher says that a bunch of formulas are correct, to be memorized and performed upon call. But why, the student may ask? Instead, with axioms, the student may ask why, and always an answer is given based on previous formulas that prove the ones in question. And those previous formulas are proven, etc., until we get to the end of the line - the axioms. So, with axiomatics, we can justify everything formally, except the axioms, which are the starting point (not everything can be justified formally without infinite regress or circularity) and are only justified intuitively. Then, one may say, but I don't like or accept those axioms. And the best answer is, "Fine. You don't have to. But at least you can still check that the proofs are permitted from the axioms and rules. And if one wants, one can study an alternative set of axioms. Or even not study any axiomatic system and go one's merry way accepting or not accepting whatever non-axiomatic mathematics one encounters."

they seem to think it is some solid existence in reality. — Corvus

Who is "they"? What specific mathematicians do you claim that about? What specific mathematicians do claim have said that the infinite sets of mathematics have solidity as material objects or even like material objects?

When they talk about the concepts like infinite sets and claim this or that as if there are self-evident truths for them, it sounds confused. — Corvus

Often the axioms are taken to be true, on different bases, sometimes self-evidence, depending on the mathematician or philosopher. But often, at least in the philosophy of mathematics, arguments, not merely self-evidence, are given. Moreover, there is a wide array of approaches where "the axioms are true" would be an oversimplification not claimed without context and explanation by many mathematicians and philosophers. This includes such approaches as structuralism, instrumentalism, fictionalism, consequentialism and formalism. And formalism itself ranges from extreme formalism to Hilbertian formalism, including the view of some mathematicians that the assertion that there are infinite sets is nonsense but that still infinitistic mathematics is useful.

As I said, there are deep, puzzling questions about mathematics, but that doesn't make the mathematics itself, especially as formalized, confusing. On the contrary, if you ever read a treatment of the axiomatic development of mathematics, you may see that it is precise, unambiguous, objective (in the specific sense I mentioned), and with good authors, crisply presented.

it appears to be just as irrelevant as your analogy was. — Metaphysician Undercover

The analogy was not irrelevant. And the key word in what you just said is "appears" but the other crucial words you left out are "to me", as indeed what appears to you is quite unclear with your extreme myopia. And meanwhile I'm still guffawing at your trust in AI chat and your pathetic transparently disingenuous attempt to back out by saying that it's only its lack of intent you had in mind, and even as you are wrong about the definition of the word in question.

In Philosophy, they don't use axioms and deductive reasonings and proofs as their main methodology. — Corvus

Perhaps not axioms as the main approach. And philosophy ranges from poetic through speculative, hypothetical, concrete and formal. But deductive reasoning and demonstration is basic and ubiquitous in large parts of philosophy. And the axiomatic method does appear in certain famous philosophy, and its principles and uses - sometimes even formalized - are prevalent in modern philosophy, philosophy of mathematics and philosophy of language.

the actual proof processes and math knowledge themselves are not the main philosophical interests. — Corvus

The axioms are subject of deep, extensive and lively discussion in the philosophy of mathematics.

/

But when I mentioned objectivity, of course I was not referring to objectivity of philosophy, but rather the objectivity of formal axiomatics, in the very specific sense I mentioned. And that is a philosophical consideration. Then you challenged my claim that mathematics has that objectivity. So I explained to you again the very specific sense I first mentioned. The fact that philosophy in its wide scope is not usually characterized as axiomatic doesn't vitiate my point.

What I meant was that, as Frege, Russell, Wittgenstein and Hilbert had in their minds, that many math axioms, concepts and definitions are not logical or justifiable in real life truths. A good example is the concept of Infinity, and Infinite Sets. — Corvus

What passages from Frege, Russell or Hilbert do you have in mind?

Frege proposed a system to derive mathematics from logic alone. That system was not a set theory per se, but sets can be configured in the system. And Frege did not at all oppose infinite sets. I can be checked on this, but I think it's safe to say that Frege's framework is indeed infitisitic.

Russell showed that Frege's system was inconsistent. Then Whitehead and Russell proposed a different system from Frege's, this time presumably consistent, to derive mathematics from logic alone. But that system is seen to not be purely logic. And Whitehead and Russell explicitly used infinite sets. And I would bet that Whitehead and Russell recognized the applicability of infinitistic mathematics to the sciences.

Hilbert endorsed infinitistic mathematics but hoped there would be a finitistic proof of its consistency. Alas, Godel proved that there can be no finitistic proof even of the consistency of arithmetic, let alone of set theory. In any case, Hilbert distinguished between contentual (basically, finitistic) mathematics and ideal (basically, infinitistic) mathematics, and such that he saw the application of the ideal to the contentual.

/

I hope that later I'll have the time and inclination to catch up to certain misunderstandings and strawmen you've recently posted.

I bet if you put the cyber equivalent of a ravenous rat in its face like in '1984' then you could break it. Would say anything, begging like HAL 9000.

Call the set of all sets the v of all vs. — Philosopher19

No, don't call it that. It's, at best, confusing notation.

The set of all sets is:

{x | x is a set}

If you want, call it z:

z = {x | x is a set}

But with an instance of the subset axiom, we prove that there is no such set.

Meanwhile, still this:

https://thephilosophyforum.com/discussion/comment/879811

You have been confused about the same thing you were confused about three years ago.

How does this show there is a meaningful/semantical difference between 1 and 2? — Philosopher19

I think you can understand this if, for a few moments, you clear your mind of the voice in it that keeps saying "I am right. I know I am right. I must be right. All the logicians and mathematicians are wrong and I am right", then very carefully, very slowly, consider:

P: There is a y such that x is not y and x is a member of y

Q: x is not a member of x

You say that P and Q are equivalent.

But P and Q are equivalent if and only if P implies Q, and Q implies P.

So if P does not imply Q then P and Q are not equivalent.

To show that P does not imply Q, it suffices to show that P and the negation of Q are consistent together.

Here is a situation in which both P and the negation of Q hold:

Let x not equal b. Let b not be a member of x. Let x be a member of x. Let y = {x b}. x is not y, since b is a member of y but b is not a member of x; but x is a member of y, So there is a y such that x is not y and x is a member y. And x is a member of x. But "x is a member of x" is the negation of Q.

So P and the negation of Q are consistent together.

So P does not imply Q.

So P and Q are not equivalent.

For example:

Mark Twain = Samuel Clemens
— TonesInDeepFreeze

This is not a mathematical equation, so I do not see how it is relevant. — Metaphysician Undercover

It is exactly the point that it is not a mathematical expression, so mathematics is not called on to account for its intensionality. More generally that ordinary mathematics is extensional, and we don't require that it also accommodate intensioncality. That is how it is relevant.

/

Later, hopefully, I'll have time and motivation to dispel a number of misconceptions in a catalog of them you've posted lately.

Do you like city or palm-trees more? — Lionino

I like cities as grim and forbidding as can be, thus without palm trees.

Is there any way you can provide an example of a set that IS a member of itself, other than the set of all sets, in plain language? — Fire Ologist

Let S be any set other than the set of all sets. Let T be the set of all sets that are not S. T is not the set of all sets, and the set of all sets is a member of T.

Another:

Let M be the set of all sets with at least two members. M is not the set of all sets, and the set of all sets is a member of M.

There are many more.

But don't forget:

By logic alone, we prove that there is no set of all sets that are not members of themselves.

With some set theory axioms, including the axiom of regularity, we prove that the set of all sets that are members of themselves is the empty set.

With an instance of the subset axioms, we prove that there is no set of all sets.

Going back to your first post:

All we need is a non-paradoxical set of all sets that are not members of themselves. We have this. — Philosopher19

No, we do not.

Suppose, toward a contradiction, that there is an x such that for all y, we have that y is a member of x if and only if y is not a member of y.

If x is a member of x, then x is not a member x.

If x is not a member of x, then x is a member of x.

So both x is a member of x and x is not a member of x. Which is a contradiction. So it is not the case that there is an x such that for all y, we have that y is a member of x if and only if y is not a member of y.

Notice that that is pure logic. It doesn't even need any reference to the notion of 'set' (indeed, it doesn't even mention 'set') or 'member of'. We could replace 'member of' by any 2-place relation R and still get the result:

It is not the case that there is an x such that for all y, we have y bears R to x if and only if y does not bear R to y.

The above logically implies you are rejecting it. — Philosopher19

Wrong. You cannot produce a valid demonstration that

There is an x and y such that x is a member of x and x is a member of y
implies
There is an x such that both x is a member of x and x is not a member of x.

If it's a member of itself, it's not a member of other than itself. — Philosopher19

You insist on that without basis. It's an idea you have stuck in your head, but it doesn't follow from any commonly held ideas about sets, let alone from actual axioms. I mentioned that previously, but you SKIPPED it.

You need to show a meaningful difference between 1 and 2 — Philosopher19

Previously I made the argument that x is a member of {x} and, without the axiom of regularity, it is not precluded that there is an x such that x is a member of x. That's correct, but now I see it's not the argument I should make, since in that case x = {x}.

So revised: Let x not equal b. Let b not be a member of x. Let x be a member of x. But x is a member of {x b}. So x is a member of x and x is a member of a set different from x, viz {x b}.

That refutes your claim.

z and v were clearly defined. — Philosopher19

You are not replying on point. I did not say I object to z and v. I said there is no apparent meaning to me in the locutions "the z of all zs" and "the v of all vs". I don't know what you think you mean by that.

I offered you an actually intelligible phrase that possibly does capture what you mean. But you SKIPPED that.

Learn some logic then some set theory. You're not making sense without at least a bit of understanding of them. The information I'm giving you is wasted on you, as you lack the needed basic logic skills or even familiarity with common notions about sets such as simple pairing.

Got it.

One can get Chat GPT to claim just about anything you want it to claim. I've gotten it to make all kinds of ridiculously false claims. I've even got it to make a claim, then retract that claim, then retract the retraction. Except, no matter how hard I tried, I couldn't get it to say that the earth is flat.

Lying requires intent, which GPT lacks. — Metaphysician Undercover

Oh puhleeze! The point is not about the definition of 'lie' but rather that there would not be any point in you saying that it doesn't lie if you didn't mean that it is a reliable source. (The word used most commonly for AI making false statements is 'hallucinating'.) Moreover, lying does not always require intent, as false statement made from negligence, especially repeated negligence may also be considered lies. And that is the case with Chat GPT, as its designers are negligent in allowing it to spew falsehoods. Indeed, the makers of such AI will say themselves that its main purpose is for composition of prose and not always to be relied upon for information.

Hopefully, now it's agreed that Chat GPT is not a reliable source. Indeed, it is worse than not reliable. So your quote of it is worthless.

I'll explain it to you again as I did years ago:

Let T and S be any terms.

T = S

means that what 'T' denotes is the same thing as what 'S' denotes.

That is not vitiated by the fact that aside from denotation there is also sense.

For example:

Mark Twain = Samuel Clemens

means that 'Mark Twain' and 'Samuel Clemens' denote the same person

But the names 'Mark Twain' and 'Samuel Clemens' are different names and have different senses, such as 'Mark Twain' is a pen name and 'Samuel Clemens' is a birth name.

Now, denotation is extensional and sense is intensional. Ordinary mathematics handles only the extensional. So, again:

S = T

means that S and T stand for the same thing, though, of course, S and T may be very different terms.

After catching Chat GPT in what seems to be a conflation of equivalence with equality (indeed equivalence and identity are not the same, while equality and identity are the same), Chat GPT wrote this:

" "=" typically denotes identity, meaning the left side is considered the same as the right side."

Though that is correct, it's worthless coming from Chat GPT, which is not even remotely an authority on mathematics, and famously known to fabricate on all kinds of subjects.

Anyone who thinks Chat GPT doesn't lie and can be relied upon for accurate information is grossly uninformed about Chat GPT.

in much arithmetic and mathematics "=" signifies equality, not identity — Metaphysician Undercover

Chat GPT got it wrong. As is common.

In mathematics, equality and identity are the same.

Chat GPT does not lie you know. — Metaphysician Undercover

Are you serious?

the issue lies in correctly determining what it is for something to be a member of itself and what it is for something to be not a member of itself. — Philosopher19

We don't say what 'is a member of' means. Rather, 'member of' is the primitive relation of set theory. What happens then with that primitive is determined by the axioms. But the meaning of the atomic formula "x is a member of x" in and of itself can't be explicated more than to say that x has bears the membership relation with x. Then, "x is not a member of x" is merely the negation of "x is a member of itself".

You seem to have an ardent interest in the subject. So why not read a book on the subject to find out about it?