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.

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

I believe I've understood your point, but I don't think you have understood mine. So from my point of view, this is what you are doing.

You have not shown that x = {x y z} implies a contradiction. — TonesInDeepFreeze

I have shown that x = {x y z} results in a contradiction when we say:

1) x is not a member of itself (because x is in x along with y and z. Thus, x is a member of itself whilst y and z are not members of themselves).

2) x is a member of itself as well as y and z (because x is in x along with y and z. Thus x = {x y z} has to be interpreted as either meaning x is a member of itself twice with y and z being members of themselves once, or x is a member of itself, with y and z not being members of themselves).

Do you see that if I wrote p = {x y z}, then it is not contradictory for me to say x is a member of itself as well as y and z?

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'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.]

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?

@TonesInDeepFreeze

1. Sets can contain themselves. [assume for reductio ad absurdum]
2. Suppose N is a set that contains itself
3. Let N = {x, N} where x is either no elements or x is some elements
4. N = {x, {x, N}} [substituting N with {x, N}]
4. N - N = {x, N} - {x, N} = { } [N = {x, N} from 3]
5. N - N = {x, {x, N}} - {x, N} = {{x, N}} = {N} [N = {x, {x, N}} and N = {x, N} from 4 abd 3]
6. N - N = { } and N - N = {N} [from 4 and 5] [contradiction]
Ergo,
7. Sets can't contain themselves [1 - 6 reductio ad absurdum]

What do you think?

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.

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.

A set that doesn't contain itself = {1, &, :sad: }

Now, consider the set {...} = N1 where "..." stand for element(s). Suppose {...} = {N}

Let's try and make N1 contain itself, {...{...}} = {N, {N}}

But N2 =/= N1 as {...} =/= {...{...}} or {N} =/= {N, {N}}

In other words, N1 couldn't be made to contain itself.

Let's try to make N2 contain itself, {...{...{...}}} = N3

But N2 =/= N3 as {...{...}} =/= {...{...{...}}} or {N, {N}} =/= {N, {N, {N}}}

So and so forth.

In other words, no set can be made to contain itself as trying to do that results in the new set being different from the old set i.e. the (say) set A that contains the set A is different from the set A.

The only option then is to try out the empty set N1 = { }.

Let's try and put N1 inside N1, {{ }} = N2

Unfortunately, { } =/= {{ }}

So and so forth...

I've proved here that N = {N} isn't possible

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?

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

Sorry, if it seemed as though I hadn't paid attention to it but my argument in the previous post seems to make your well-meaning explanation moot.

1. Take N = {N}, the set that contains itself.

2. N contains itself = {N} contains itself

3. {N} doesn't contain itself.

QED

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?

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. — TonesInDeepFreeze

Everyone recognises such a set is contradictory. What you and mainstream set theorists seem to think is that this logically entails that the set of all sets is contradictory, whereas it does not (and I have provided proof of this).

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. — TonesInDeepFreeze

It seems that you are not reading what I'm writing with enough attention to detail. You seem to have not understood why I have said "member of itself twice" (which is my proof of why a set of all sets that are members of themselves is as contradictory as a set of all sets that are not members of themselves).

By deriving a contradiction from the assumption that there does exist a set of all sets. — TonesInDeepFreeze

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. What you've wrongly concluded is that this means the set of all sets is contradictory.

Only one set contains all sets that are not members of themselves (the set of all sets). Again, there is nothing contradictory about the set that encompasses all sets that are members of it (because they are sets), and it is a member of itself (because it is a set). There is nothing contradictory about the set of all sets. It's rejection is blatantly contradictory and all set theorists know this (including you). ZF is either inconsistent, or not comprehensive enough (which ultimately means it is inconsistent).

This fierce dogma needs to die.

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.

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.

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

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? — TonesInDeepFreeze

Thanks for the clarification and I'm sorry if I've wasted my time but I suppose for people likey yourself who have to deal with those less knowledgeable than themselves, it's part of the territory.

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.

But I don't get why someone would post arbitrary non sequiturs. — TonesInDeepFreeze

Work in progress.

ZF has not been shown to be inconsistent. And lack of comprehensiveness does not imply inconsistency. — TonesInDeepFreeze

ZF implies incompleteness in proof, theory or system. Perhaps some are happy with such standards, I am not.

In any case, your last reply to me suggests that it's a waste of time to continue this discussion with you. Also, the post that followed it suggests that you are upset, angry, or frustrated, which is not a good state to be in when discussing matters of logic or pure reason. I presented what I say is clear proof, you say that I have not. This has happened twice now, there's no point in there being a third time. I think we should just agree to disagree.

Suppose a set P

1. P = P [reflexivity. Nothing was done to set P]

2. {P}. P was made an element of the set {P}. Something was done to P]

3. P = {P} [P is the set that contains itself] [assume]

4. Nothing was done to P = Something was done to P [contradiction]

5. P =/= {P} [3 - 4 reductio ad absurdum]

Thanks for starting this thread. I'd like to share with you my finding regarding Russell's paradox.

Russell's paradox boils down to whether the set of all sets that doesn't includes themselves in includes itself? or, for easier comprehension of my attempt at a resolution of the paradox, does the set of all sets that doesn't contain themselves contain itself?

As you can see, one of the central issues is whether a set can contain itself or not because Russell's paradox can be rephrased as,

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.

If sets can contain themselves, there should be a set N ={N}

1. N = {N}
2. N contains itself, {N} contains itself
3. If {N} contains itself, {N} = {{N}}

Did you see what happened there?

The outermost curly braces "{...}", in a sense, collapsed or behaves as if it didn't exist at all: {N} = { {N} }.

Likewise,

4. N = {N} = {{N}} = {{{N}}} = {{{{N}}}} =...

Same things's happening or rather not happenning as any attempt to make the set N an element of another set like {N} returns, to use a computer terminogy, the set N itself.

In short, the set N, though defined as {N} can't be contained in another set for the reasons provided above.

The entire series N = {N} = {{N}} =... is an illusion so to speak.

Thus, a set N such that N = {N} can't exist. In other words, no set can contain itself and so Russell's paradox is a none issue.

Thanks for starting this thread. — TheMadFool

You're welcome.

Thus, a set N such that N = {N} can't exist. In other words, no set can contain itself and so Russell's paradox is a none issue. — TheMadFool

I think I get what you're saying, and I recognise that what you say holds true when N is finite. But when N is not finite, what you say does not hold true:

Imagine having four folders on your computer. These are folders '1', '2', '3', and 'all folders on this computer'. The last folder must contain itself in order to meaningfully qualify as a folder of all folders on this computer.

If it is the case that when you open 'all folders on this computer', you get the following folders: '1', '2', '3', and 'all folders on this computer', and then you click the last folder from these four folders and you again get the following folders: '1', '2', '3', and 'all folders on this computer', and you do this again and again forever and this holds true, then arguably the folder 'all folders on this computer' contains itself.

But you're looking at this from a metaphysical perspective (in which case the computer at hand would have to be non-finite because a finite computer does not have the capacity to really contain a folder that contains itself, because it does not have endless energy or potential, and endlessness is needed to sustain a self-containing folder as discussed in the previous paragraph).

I'm approaching this from a purely logical angel:

A) Assume that the letters a-z were representative of all sets that are not members of themselves. You cannot have a set of all sets that are not members of themselves: How are you going to logically write this? a = {a b c...} Here, a is a member of itself. Whatever letter you choose from a-z, it will be a member of itself, so it can't be a set that is not a member of itself.

B) Now assume that the letters a-z were representative of all sets that are members of themselves. You cannot have a set of all sets that are members of themselves: How are you going to logically write this? a = {a b c...} Here, a is a member of itself twice (and whatever letter you choose from a-z, this problem will occur), and such a thing is as contradictory as a set that is not a member of itself, that is in fact a member of itself (as highlighted in A).

Do you see how B is the overlooked part of Russell's paradox?

1. There's a set N
2. {N} does something to N
3. N = {N} [a set can contain itself][assume for reductio ad absurdum]
4. If N = {N} then {N} does nothing to N
5. {N} does nothing to N [3, 4 modus ponens]
6. {N} does something to N and {N} does nothing to N [2, 5 conj]
7. False that N = {N} or no set can contain itself [3 - 6 reductio ad absurdum]

I see your argument. What would you say to the following:

The list of all lists, lists all lists (including itself). So this list contains itself as an element. So this list is a member of itself. Do you agree?

Call the list of all lists L. L = L. Given my interpretation of your argument, it's not the case that L = {L}.

So I don't think it's a case of {N} doing something to N. I think it's a case of N being N and N being such that it contains itself as an element (like the list of all lists).

I strongly recommend you have a read of this:

http://philosophyneedsgods.com/2021/05/22/the-solution-to-russells-paradox-and-the-absurdity-of-more-than-one-infinity/

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.

"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".

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

So, 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?

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.

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

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. — TonesInDeepFreeze

If I say I did x, and you say I did not do x, and neither of us changes his position, this means that we agree to disagree on this. I hope that's clear to you now.

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

If that's what you want to believe, then believe. Rejection of the set of all sets is blatantly contradictory. I say I have provided proof, you say I have not, except you accuse me of arrogance but seem to not apply it to yourself. This is despite the fact that you are defending a contradiction.

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

You are like child in your reasoning and manner of discussion. I shouldn't have to spoon-feed you, but they say feeding the needy is good, and per the dictates of pure reason, Karma is real (I've provided proof of this in another thread).

The list of all lists lists itself. In this list, one item is a member of itself whilst all other items are not members of themselves (precisely because they are members of it, and the reference is it).

By definition, the set of all sets encompasses all sets. This includes itself. Thus in the context of sets, the set of all sets encompasses all sets that are not members of themselves, as well as itself. Because it encompasses itself, it is a member of itself. Because all other sets are encompassed by it, they are members of it, and not themselves. This is when the reference is sets (as opposed to lists).

If you directly show a contradiction in the above, I might reply to you. If not, I'm done trying to spoon-feed you. You are in need because your belief system is contradictory, yet you act like you are not, and you complain about the world like some spoilt child (you set these standards for discussion. I am reciprocating). You are contradictory/unreasonable/inconsistent. I suggest you reconcile.