Interesting. It had not occurred to me to consider rules of construction as part of the set, even though in our case at the forefront ("sets that do not contain themselves"). On that consideration, the language is a claim without a possible referent. Meaning the demon would not even have encountered it, meaning it is not even a thing. Suggesting the paradox is an artifact of language and no real part of set theory.

The constraint of things (in the west) would seem to rule out non-things, and in any case certainly keep them from the east (in our model if that still holds).

But set theory can't do that and be recursively axiomatized, because there is no algorithm to determine whether a given sentence contradicts a previous set of statements. — TonesInDeepFreeze

With this the discussion becomes one of language and sentences. Should I cry foul? Or is it just a matter of the game itself and knowing the rules? Taking sets as groups of objects, whether bricks or numbers, I don't see them in themselves leading to paradoxes. Language, on the other hand, a different animal. As the Youtuber mathematician Michael Penn says - which I paraphrase here as a question - Is this a good place to stop?

Suggesting the paradox is an artifact of language and no real part of set theory. — tim wood

If you're referring to Russell's paradox, it is a matter of logic and is not peculiar to set theory, but rather applies to any 2-place relation R:

For any 2-place relation R, there is no x such that, for all y, y bears relation R to x if and only if y does not bear relation R to itself. (With set theory, R happens to be 'is a member of'.)

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. — TonesInDeepFreeze

You have not proven this. You have just stated it. Here's my response:

Call the set of all sets X. Call any set that is not X, a Y. X contains all Ys plus itself. Every set Y is a member of X. Show me how this is contradictory.

Again, you are defending a contradiction. That contradiction being "there is no set of all sets". Consider that it is you who is being dogmatic and not me. Who is possibly dogmatic here? The one that is defending a contradiction, or the one that is against it?

Let me try. Subject to correction. Start with a blank slate. Define
"set" in the usual way. Consider three sets A, B, C that do not contain themselves. There are a lot of such sets, but let's suppose these three are all of them. No problem so far.

Consider set D defined as the set of all sets that do not contain themselves. That is, D contains A, B, C. D is certainly a set. And D does not contain itself. By the definition of D, it should contain itself. But if it does, then it doesn't. And so forth.

Now consider S set of all sets. Does it contain D? What is D? S chokes on D. Meaning that sets cannot be "defined in the usual way." Sense?

Consider three sets A, B, C that do not contain themselves. There are a lot of such sets, but let's suppose these three are all of them. — tim wood

Ok.

Consider set D defined as the set of all sets that do not contain themselves. — tim wood

Ok, so that means that D necessarily contains A, B, and C.

That is, D contains A, B, C. D is certainly a set. — tim wood

Ok.

And D does not contain itself. — tim wood

D not containing itself or not being a member of itself is a contradiction because you said:

Consider three sets A, B, C that do not contain themselves. There are a lot of such sets, but let's suppose these three are all of them.

If A, B, and C are ALL of them, then by definition, D is a member of itself because D is the set that contains ALL sets that do not contain themselves. D can only be such a set if it contains itself (and it does).

You cannot have a set of all sets that are not members of themselves that is itself not a member of itself. This does not mean that the set of all sets is contradictory. Rejecting the set of all sets is blatantly contradictory. It is the last thing western philosophers should have done with regards to being sincere to the semantic of "set". I reckon because this has gone on for a 100 years, it has become a fierce dogma.

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

You have not proven this. You have just stated it. — Philosopher19

That is false.

https://thephilosophyforum.com/discussion/comment/546798

That is not a proof. There are a few problems with it, but most glaring:

There is no set D such that for all y, y is a member of D if and only if y is not a member of itself. So, it makes no sense to say that S "chokes" on D (whatever a clear definition of 'chokes' might be) since there is no such D for S to choke on.

Again, you are defending a contradiction. — Philosopher19

AGAIN, you have not shown that "there does not exist a set of which all sets are a member" is contradictory. Au contraire, I have shown a proof that "there does exist a set of which all sets are a member" is inconsistent with taking subsets.

Consider that it is you who is being dogmatic and not me. — Philosopher19

I considered it:

(1) I refer to axioms and inference rules such that it is objectively, publicly, mechanically verifiable whether a given sequence is or is not a proof from the axioms with the rules. I show certain proofs with those axioms and inference rules.

(2) I use standard definitions in mathematical logic. I point out when criticisms of mathematical logic are based in misunderstandings of the definitions. But I allow that other people may have offer different definitions, though they should be clear, non-circular, and properly formulated, and as long as the context is made clear and there is not confusion by mixing contexts with conflicting definitions.

(3) I happily embrace that other people may propose a wide range of different axioms and rules and show proofs relative to those axioms and rules. These include, among others, constructivist and intuitionist, finitist, strict finitist, predicativist, multi-valued, relevance logic, free logic, and even para-consistency. Even ersatz proposals and even non-axiomatic sketches, though I may criticize if they are not coherent or are presented with ignorant, confused, and incorrect criticism of abiding mathematics.

(4) I don't opine whether mathematics is or is not to be regarded merely as application of axioms and inference rules. And without commitment to a philosophical stance, I do countenance considerations in a wide range of alternatives to mere extremist formalism (which itself I don't opine to be necessarily incorrect), including truth regarded in different ways such as reasonable formalism, intuitiveness, correspondence, coherence, realist, structuralist, fictionalist, consequentialist, contextualist, operationalist, pragmatist. constructivist and intuitionist, common sense everyday notions, and I even allow the legitimacy of interest in para-consistency, and even (brainstorming) contrarianism. So not only am I not dogmatic, but to the contrary, I am liable to criticism for being too agnostic and. lacking philosophical commitment, too philosophically timid.

(5) I happily admit that set theory and mathematical logic themselves have certain difficulties (call them even 'mysteries') that I can't completely explain.

(6) I happily admit that my knowledge of set theory, mathematical logic, mathematics, philosophy of mathematics, and philosophy only extends to some basics and that I am not an expert. And I welcome being corrected on anything I've posted that is indeed incorrect, and am happy to post recognition of the correction and to retract as needed. And, when I have myself noticed that I made mistaken remarks or claims that I realize are not on solid footing, then I post a correction.

You:

(1) Keep demanding that your position is correct and that mathematics is incorrect, by repeating your argument while ignoring the specific points demonstrating the errors in your argument.

(2) Tendentiously make incorrect denunciations of mathematics and mathematicians that you don't know anything about.

(3) Accuse non-dogmatists of dogmatism while not facing your own dogmatism.

AGAIN, you have not shown that "there does not exist a set of which all sets are a member" is contradictory. — TonesInDeepFreeze

It is in the definition of the semantic of "set" that you can have a set of ALL things of which you can have more than one of. Examples of such things include: Numbers, people, shapes, trees, sets. If you can have more than one set, you can have a set of all sets. It is clearly contradictory to say: You can have more than one X, but there's no such thing as a set of all Xs.

I refer to axioms and inference rules — TonesInDeepFreeze

That which you described to me as an axiom I showed as being false. You have not addressed this. Again:

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

To which I replied: Call the set of all sets X. Call any set that is not X, a Y. X contains all Ys plus itself. Every set Y is a member of X. Show me how this is contradictory.

To which you replied: AGAIN, you have not shown that "there does not exist a set of which all sets are a member" is contradictory.

To which I will repeat the last part of the beginning of this post again: It is contradictory to say you can have more than one X, but there's no such thing as a set of all Xs.

You have yet to address that which I have underlined for you. I have addressed the "axiom" which you present as an objection to the set of all sets.

Bear in mind that it was me who suggested that we agree to disagree, to which you decided to hurl insults at me, but still decided to provide your "axiom" as a refutation of the set of all sets.

That is false.

https://thephilosophyforum.com/discussion/comment/546798 — TonesInDeepFreeze

You're posting a reply to another person. Do you not see that you have hurled insults at me accusing me of not giving you proof and suggesting that you gave me proof and then provide a link to something that you said to someone else? Do you not see the problem with this?

In any case, I checked the link (in an attempt to be charitable). I am neither rejecting a set that is a member of itself, nor a set that is not a member of itself. I am rejecting the rejection of the set of all sets. I suggest you read the OP carefully.

It is in the definition of the semantic of "set" — Philosopher19

It is not unreasonable to desire a set theory that upholds our everyday notion of 'set'. However, certain difficulties arise. For example, our everyday notions might include that for every property there is the set of all and only those sets having that property. But that doesn't work, as Russell's paradox reveals.

And people also like to have a theory grounded axiomatically. Trying to have both - our everyday notions and axiomatization - presents more difficulties.

So, for those inclined to axiomatics, we have to admit that not all everyday notions will be preserved. But that is not contradiction; rather it is modesty.

an axiom I showed as being false — Philosopher19

Which axiom do you claim is false?

Call the set of all sets X. Call any set that is not X, a Y. X contains all Ys plus itself. Every set Y is a member of X. Show me how this is contradictory. — Philosopher19

As long as we have taking of subsets, the inconsistency comes with the assumption that there is a set of all sets. I proved that.

I will repeat the last part of the beginning of this post again: It is contradictory to say you can have more than one X, but there's no such thing as a set of all Xs. — Philosopher19

In context of the mathematics you are denouncing, a contradiction is a sentence and its negation. If you cannot show that set theory implies both a sentence and its negation , then you have not shown that set theory is inconsistent.

You may have your own notion of 'contradiction'. Your own notion of 'contradiction' may be that something is to you incompatible or counterintuitive or not in accord with everyday notions. And if that is the definition of 'contradiction' that we use, then, of course, I cannot deny that set theory has results that you find incompatible or counterintuitive or not in accord with everyday notions, so under that notion of 'contradiction', set theory is contradictory. But, again, that is not what mathematicians mean by 'contradiction'. And as 'contradiction' is regarded in mathematics as a sentence and its negation, you have not shown set theory to be contradictory.

"you can have more than one X"

Exy ~x=y

"there's no such thing as a set of all Xs"

~EyAx xey

Observe that Exy ~x=y and ~EyAx xey is not a statement and its negation, nor have you shown how they imply both a statement and its negation.

I have addressed the "axiom" which you present as an objection to the set of all sets. — Philosopher19

The axioms I used is an instance of the axiom schema of separation. You have not addressed that.

Also, you putting 'axiom' in scare quotes is silly and jejune.

Do you not see that you have hurled insults at me accusing me of not giving you proof — Philosopher19

(1) Pointing out that you have not proved something is not an insult.

(2) You began the volley regarding 'dogmatism' as you claimed that mathematicians are dogmatic (arguably a general insult).

(3) Whatever insults I might post do not make me dogmatic.

suggesting that you gave me proof and then provide a link to something that you said to someone? — Philosopher19

I didn't say that I addressed a proof to you personally. You said that I had not proved my claim. So I correctly said that is false and I gave you (and whomever else is reading) a link to a post in which I did prove the claim.

I checked the link (in an attempt to be charitable) — Philosopher19

It's no charity to me. It's your own improvement that would be gained by understanding the proof.

I am rejecting the rejection of a set of all sets. — Philosopher19

And I proved that there is no set of all sets, using only an instance of the axiom schema of separation.

Essentially, I was looking for a reply to:

Call the set of all sets X. Call any set that is not X, a Y. X contains all Ys plus itself. Every set Y is a member of X. Show me how this is contradictory.

Which axiom do you claim is false? — TonesInDeepFreeze

The above which I have underlined. The best that I can see from your last reply as addressing the underlined is:

As long as we have taking of subsets, the inconsistency comes with the assumption that there is a set of all sets. — TonesInDeepFreeze

Then there is an issue in the manner in which you take subsets. Call any set that is not the set of all sets a V', call any set that is not a member of itself a -V, call any set that is simply a set a V.

Can you have a -V as the V of all -Vs? No. Can you have any V as the V of all -Vs? If by a V of all -Vs you mean a V that encompasses ALL -Vs and no other Vs, then you are asking for a -V as the V of all -Vs (in which case what you are asking for is contradictory). But where you are not asking for a -V as the V of all -Vs, then the V of all Vs is such that it encompasses all -Vs, and it is not a -V (because it is a member of itself). -V is only meaningful in the context of the V of all Vs. -V' (any non-set-of-all-sets set that is not a member of itself) is only meaningful in the context of the V' of all V's.

Either we say:

A) There is the V of all Vs. It encompasses all Vs. All other Vs are -Vs in this context.

Or

B) There is the V of all -Vs. It encompasses all -Vs. It is not a -V because it is a V.

If you think a V of all Vs is contradictory, then fine, but then you cannot say a V of all -Vs is contradictory. Do you see? You cannot reject both A and B and be consistent at the same time. It's either A or B. But to my understanding, you reject both A and B.

I've put effort into understanding you and trying to accommodate you in this discussion. I've also put effort into giving you explanations that are easily accessible. If I do not feel you reciprocate this with regards to the last reply I sent you, then I will stop trying.

Essentially, I was looking for a reply to:

Call the set of all sets X. Call any set that is not X, a Y. X contains all Ys plus itself. Every set Y is a member of X. Show me how this is contradictory. — Philosopher19

I showed that the very first sentence is inconsistent with subsets.

Which axiom do you claim is false?
— TonesInDeepFreeze

The above which I have underlined. — Philosopher19

What? The underlined passage is not an axiom of set theory. And it's your claim, not mine, so there's no reason for me to defend it. You are extremely confused.

Then there is an issue in the manner in which you take subsets. — Philosopher19

Finally we're making progress. Yes, if you require that the theory uphold "There is a set of all sets" then you must reject the axiom schema of separation. And you are welcome to propose an alternative (otherwise, good luck doing any set theory without subsets).

And I've read your 'V' stuff before, and I commented on it with exactness. But if you wish to engage me with the additions you've made now, then, to start, you need to clean up the incoherent notation. In a previous post, I suggested how you could do that, but you ignored.

I suggest you parlay the progress you made when you identified that it is the subset axiom (axiom schema of separation) that is the root of your disagreement with set theory. But you should also check out the rest of the axioms to see which you might also reject.

And I've read your 'V' stuff before, and I commented on it with exactness. But if you wish to engage me with the additions you've made now, then, to start, you need to clean up the incoherent notation. In a previous post, I suggested how you could do that, but you ignored. — TonesInDeepFreeze

I have no interest in trying to accommodate you any further. What you have said to me and what I have replied to you is clear. I think you have failed to prove your position, whilst I have proved my position left, right, and centre. Evidently you disagree with this.

It's like I said before, we'll have to agree to disagree.

I have no interest in trying to accommodate you any further. — Philosopher19

I don't seek accommodation from you. Cleaning up your notation would be a favor to yourself.

I think you have failed to prove your position — Philosopher19

My main point is that ExAy yex is inconsistent in set theory. I proved it.

You made your only progress for yourself thus far when you replied that, given such a proof, you think the axiom used is false.

we'll have to agree to disagree — Philosopher19

As I asked before, what is the operative meaning of that? Your reply was the circlarity that 'agree to disagree' means to agree to disagree.

I am curious how you got so mixed up here:

Which axiom do you claim is false?
— TonesInDeepFreeze

The above which I have underlined.
— Philosopher19

What? The underlined passage is not an axiom of set theory. And it's your claim, not mine, so there's no reason for me to defend it. You are extremely confused. — TonesInDeepFreeze

Somewhere you asked about having a set of all sets except the Russell set. I didn't send a proof that that doesn't work, but I want to now, because it's cute (basically reapplying the Russell argument.

1, EUAx(xeU <-> ~Az(zex <-> ~zez)) [premise]

i.e, assume, toward a contradiction, that there is a set of all sets except the Russell set.

2. Ax(xeU <-> ~Az(zex <-> ~zez)) [EG]

3. ERAy(yeR <-> (yeU & ~yey)) [separation]

i.e. take the subset of U that is the set of all sets in U that are not members of themselves.

4. Ay(yeR <-> (yeU & ~yey))

5. ReR <-> (ReU & ~ReR) [UI]

6. ReR [premise]

7. ~ReR [5, 6]

i.e. we have ~ReR since the premise ReR implies ~ReR.

8. ~ReU or ReR [5, 7]

9. ~ReU [7, 8]

10. ~ReU <-> Az(zeR <-> ~zez) [2 UI]

11. Az(zeR <-> ~zez) [9, 10]

12 ReR <-> ~ReR [UI]

13. ~EUAx(xeU <-> ~Az(xex <-> ~zez)) [12]

Read 'em and weep.

Ty. Looks good, and chewy. But it will take me a while to chew through it.

A set can't contain itself. Period!

Suppose K can't be contained in another set like so {K}, any attempt to do so will result in K.

The set that allegedly contains itself is K = {K}. Take K and make it an element of a set thus, {K} and what happens? It's, according to how it's defined, K again. That didn't quite go as planned, did it? Let's work with {K} then; making {K} a member of another set like this {{K}} and what happens? {{K}} = {K} = K which in plain English means K can't be made a member of another set.

The whole exercise involving K = {K} is akin to claiming that b × 1 =/= b. Multiplying by 1 doesn't change b into some other number.

Basically, it's just saying:

Suppose we have a set U whose members are all only those that are not the Russell set.

Let R be the subset of U such that R has every member of U except sets that are not members of themselves.

Some reasoning shows that R is not a member of itself, and that leads to some more reasoning that R is not a member of U. And some more reasoning leads to R is a member of itself if and only if R is not a member of itself.

So the initial supposition that there is a set whose members are all and only those that are not the Russel set is contradictory.

/

Basically, you wondered about avoiding Russell by a cutdown that deletes the Russell set from a universal set. So, instead of applying the Russell argument itself, I applied it to the cutdown, and still got the contradiction.

A set can't contain itself. Period! — TheMadFool

I already gave you multiple pointers to articles about non well-founded sets, and I showed you how a graph with a single node and an edge from that node to itself models a set that contains itself.

Here is that thread. You might be interested in rereading it and looking at the references I gave.

https://thephilosophyforum.com/discussion/comment/546388

The essential point is that there is no actual definition of a set. A set is defined by its behavior under a given collection of axioms. If you include the axiom of regularity, no set can contain itself, and also there are no circular membership chains like $a \in b \in c \in a$.

On the other hand, it turns out that the negation of Regularity is consistent with the other axioms. That's the key point. There's no a priori reason a set can't contain itself. After all the collection of all ideas is an idea, and the collection of all collections is a collection. Self-containing entities are natural.

The ONLY reason there isn't a set of all sets is that we typically adopt an entirely arbitrary axiom saying so. Drop the axiom, and sets can contain themselves.

https://en.wikipedia.org/wiki/Axiom_of_regularity

A set can't contain itself. Period! — TheMadFool

With Z set theory, there is no set that is a member of itself.

With ZFC-R, it is not inconsistent that there is a set that is a member of itself.

I gave you copious explanation about that. But you just go on your merry way ignoring the information you've been given.

Take K and make it an element of a set thus, {K} and what happens? It's, according to how it's defined, K again. That didn't quite go as planned, did it? — TheMadFool

No, it was as "planned", and consistent (without reguarity).

{{K}} = {K} = K which in plain English means K can't be made a member of another set. — TheMadFool

Wrong. That kind of nesting is consistent (without regularity). And it does not mean that K can't be a member of another set. I informed you about that multiple times. Proof:

Suppose K = {K}. Let ~x=K and ~xeK. Then ~ {x K} = K but K e {x K}.

What is wrong with you that you can't see that a set being a member of itself doesn't prohibit the set from being a member of other sets?

The whole exercise involving K = {K} is akin to claiming that b × 1 =/= b. — TheMadFool

That is argument by analogy, which is not valid for deduction such as mathematics. And the analogy even works against your claim.

Multiplication by 1 is idempotent.

With Z, the singleton operation is not idempotent. (So your analogy is backwards.)

With ZFC-R, it is consistent that there are sets for which the singleton operation is idempotent. (So your analogy DOES work in that context, and it works AGAINST your claim.)

So often, you are plainly illogical.

No, it was as "planned", and consistent (without reguarity). — TonesInDeepFreeze

Sorry, I beg to differ. K =/= {K}.

I gave you a very good reason. Please address the proof. I'll restate it here for your viewing pleasure.

1. There's a set N that can't be a member of another set. Either the attempt to make N an element of another set won't make sense or it will.

If it's the former, case closed. Like trying to multiply 2 with $. It's nonsense.


If it's the latter, all attempts to make N a member of another set will result in N, the set itself. Like multiplying 2 with 1: 2 × 1 = 2; 2 × 1× 1 = 2. Multiplying by 1 does nothing to 2.

2. You claim there's a set K = {K}. Let's try and make K a member of another set like so {K}. However, {K} = K. Let's try something different like so {{K}}. However {{K}} = {K} = K. Basically, this is analogous to multiplying 2 with 1 (see vide infra): just as "× 1" does nothing to the number 2, if K = {K}, making K a member of another set literally does nothing too.

...{...{...{...{K}...}...}...}... = K. The curly braces, an infinity of them return the same value K. Just like 2 × 1 × 1 × 1 x... = 2 in which case, multiplying by 1 changes nothing about the number 2.

In essence, there's no difference between set N, a set that can't be a member of any set (thus can't contain itself) and set K [K = {K}] defined as a set that contains itself. It's a paradox! The set K = {K} cannot exist. Period!

That is argument by analogy, which is not valid for deduction such as mathematics. And the analogy even works against your claim. — TonesInDeepFreeze

The analogy was meant for you and is definitely not something I would submit for publication although I just might if given the opportunity - it makes so much sense.

Please address the proof. I'll restate it here. — TheMadFool

Since it's a restatement, I don't need to address it again, since I've replied to your "proofs" already, in quite detail. And you have not gotten back to me on my replies. Typically, all you do in reply is to restate your incorrect argument that had just been refuted.

Moreover, I just gave you a proof in my last post, and you have not addressed it.

Suppose K = {K}. Let ~x=K and ~xeK. Then ~ {x K} = K but K e {x K}. — TonesInDeepFreeze

[the analogy] makes so much sense. — TheMadFool

I gave you specific detail why the analogy doesn't work for you. Instead of responding to that, you merely reiterate your claim that you are right.

Since it's a restatement, I don't need to address it again, — TonesInDeepFreeze

You haven't addressed it and that's why I'm restating it.

I gave you specific detail why the analogy doesn't work for you. — TonesInDeepFreeze

The analogy is perfect. There's a precise 1-to-1 correspondence between 2 and K, creating the set that contains itself and multiplying by 1, and last but not the least, making K a member of itself matches perfectly with 2 × 1 = 2.

You haven't addressed it — TheMadFool

If you link to where you first posted it, then I'll link to where I answered it.

The analogy is perfect. — TheMadFool

You still have not addressed my rebuttal.

There's a precise 1-to-1 correspondence between 2 and K — TheMadFool

Incorrect.

2 = {0 1} and has cardinality 2.

K = {K} and has cardinality 1.