The set of all sets is really the set of all members, which is also how we use the word "all" in the first place. — Fire Ologist

Isn't the set of all sets equivalent to the set of all members? — Fire Ologist

And by definition all those members are sets.

There aren't actually any sets within the set of all sets. There are only members. — Fire Ologist

I get where you're coming from. I believe 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. This is something I address in my last reply to TonesInDeepFreeze (which is literally the post that is above/before this post) in case you're interested.

I am not skipping. You say:

So whether a set x is a member of itself or not a member of itself, if there is a set of all sets then x is a member of that set of all sets.
— TonesInDeepFreeze — Philosopher19

You're confused. What you skipped is my refutation (posted twice) of your claim that a set can't be both a member of itself and a member of another set.

So whether a set x is a member of itself or not a member of itself, if there is a set of all sets then x is a member of that set of all sets.
— TonesInDeepFreeze

This point implies that a set v can have more than one set that is a member of itself, as a member of itself — Philosopher19

"is a member of itself, as a member of itself" has no apparent meaning to me.

you perhaps fairly said "the z of all zs has no apparent meaning to me" — Philosopher19

Yes, so fix it.

L = the list of all lists
LL = the list of all lists that list themselves

Is L a member of itself in L? — Philosopher19

I gave you guidance before in how the notion of 'list' is couched in set theory. But you skip it. To avoid incoherence with a clash between 'lists itself' and 'member of itself', you could go back to what I wrote.

The v of all vs [...] The z of all zs — Philosopher19

You said that it is perhaps fair to say that such locutions have no apparent meaning, but then you proceed to post them again, as, for the second time, you've ignored my suggestion of the exact way you could reformulate so that you make sense.

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?

What you skipped is my refutation (posted twice) of your claim that a set can't be both a member of itself and a member of another set. — TonesInDeepFreeze

You agreed with me that a set cannot be both a member of itself and not a member of itself. You said it was an important point in your refutation to me. The above logically implies you are rejecting it. First look at the part I underlined in the quote above. Then look at how there is no difference between:

1) A member of other than itself (which is the same as saying a member of another set that is not itself)
2) Not a member of itself

If it's not a member of itself, it's a member of other than itself. If it's a member of itself, it's not a member of other than itself. You need to show a meaningful difference between 1 and 2 since you're the one claiming that on the one hand a set can't be both a member of itself and not a member of itself, and on the other hand, a set can be both a member of itself and a member of another set (to be a member of another set is to be a member of other than itself)

"is a member of itself, as a member of itself" has no apparent meaning to me. — TonesInDeepFreeze

This is perhaps why you didn't get my point/refutation. Perhaps if you try to reply to my above point, you will start to get my point.

You said that it is perhaps fair to say that such locutions have no apparent meaning, but then you proceed to post them again — TonesInDeepFreeze

No, z and v were clearly defined. What I did not say (which is why I said "perhaps fair to say") was that the v of all vs = the set of all sets and that the z of all zs = the not-the-set-of-all-sets-set of all not-the-set-of-all-sets-sets. I thought they could be inferred from v and z, but still, in my next post to you I clearly defined them, yet you did not address that main refutation of mine.

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.

I get where you're coming from. I believe the issue lies in correctly determining what it is for something to be a member of itself... — Philosopher19

I'm not sure I get what I'm saying.

Help me out. Besides the set of all sets, what is an example of a set that is a member of itself?

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.

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. — TonesInDeepFreeze

How does this show there is a meaningful/semantical difference between 1 and 2? Therefore, how has this refuted my claim? It is clear that "not itself" and "other than itself" mean the same thing.

Wrong. You cannot produce a valid demonstration that — TonesInDeepFreeze

I did. I think it was very clear. Here it is again in a numbered format. Tell me which number doesn't follow from (or is irrelevant to) which number or tell me which number is wrong if you are sincere in this discussion.

1) There is no meaningful/logical difference between "not a member of itself" and "a member of other than itself".

2) A set cannot be a member of itself and not a member of itself

3) If a set is a member of itself, it is not a member of other than itself (precisely because it is a member of itself)

4) If a set is a member of itself, it is not a member of another set (precisely because it is a member of itself. If it is a member of another set, it is not a member of itself precisely because it is a member of another set).

The 1-4 point I'm making is clear. If I get a reasonable/meaningful response to it, I believe I will respond to that response.

I'm not sure I get what I'm saying. — Fire Ologist

I think your instincts/intuition is in the right place (or at least trying to get to the right place)

Help me out. Besides the set of all sets, what is an example of a set that is a member of itself? — Fire Ologist

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

I recommend the following: http://godisallthatmatters.com/2021/05/22/the-solution-to-russells-paradox-and-the-absurdity-of-more-than-one-infinity/ — Philosopher19

I read the article. I certainly get how this discussion needs to be analytical to be precise, but I am not fluent enough in the symbolic language to keep up.

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? I can't think of one.

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? I can't think of one. — Fire Ologist

That's because speaking in absolute terms, only the set of all sets is a member of itself.

In short, if we were to focus on absolutely all sets, then only the set of all sets is a member of itself. However, if we were to only focus on all sets other than the set of all sets, then another set is a member of itself (but then we are not speaking in absolute/complete terms). I will try and show this to you:

Call any set a v.
Call the set of all sets the v of all vs.
If we were to focus on all sets, we would be focused on all vs.

Now call any set that is not the set of all sets a z.
If we were to focus only on all sets other than the set of all sets, we would be focused on all zs as opposed to all vs.

Since z = any set that is not-the-set-of-all-sets, the z of all zs means "the not-the-set-of-all-sets-set of all not-the-set-of-all-sets-sets. The z of all zs, is a member of itself as a z, but it is not a member of itself as a v. As a v, it is a member of the v of all vs.

So if we were to talk about all vs, then only the set of all sets is a member of itself.
If we were to only talk about all zs, then only the z of all zs is a member of itself.

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.

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.

Meanwhile, still this:

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

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

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.

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: — TonesInDeepFreeze

I tried to look at your reply, but it is unclear to me as to what it's doing. I believe it deliberately strays from what is clear simple language to try and force something that cannot be forced (perhaps due to dogma).

A) When a set is not a member of itself, it is a member of a set other than itself.
B) When a set is a member of itself, it is not a member of another set.

Rejection of either A or B is blatantly contradictory, yet, you seem to be arguing that rejection of B is not contradictory. Until you acknowledge that the rejection of B is contradictory, I don't see how we can progress.

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.

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.

@Philosopher19

These are two different claims:

1. A is not a member of itself
2. A is a member of some other set

Given this:

$A=\{A\}\\B=\{A,0\}$

(1) is false and (2) is true.

@Philosopher19

Regarding Russell's paradox, it is simply this:

1. $x$ is a member of $R$ if and only if $x$ is not a member of $x$.

Is $R$ a member of $R$?

Either answer entails a contradiction, and so (1) is a contradiction. Given that naive set theory entails (1), naive set theory is shown to be inconsistent.

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

My proof was here:

z = any set that is not the set of all sets
v = any set
The v of all vs = the set of all sets
The z of all zs = The not-the-set-of-all-sets-set of all not-the-set-of-all-sets-sets — Philosopher19

Take z to be any set that is not the set of all sets, and take v to be any set. The z of all zs is a member of itself as a z (as in in the z of all zs it is a member of itself). But it is not a member of itself in the v of all vs, precisely because in the v of all vs it is a member of the v of all vs as opposed to a member of itself. If we view the z of all zs as a z, it is a member of itself. If we view the z of all zs as a v, it is a member of the set of all sets. You can't view it as both a member of the z of all zs and a member of the v of all vs at the same time. That will lead to contradictions. In other words, we can't treat two different standards/contrexts as one (as in are we focused on the context/items of vs or the context/items of zs to determine what is a member of itself or not a member of itself?) — Philosopher19

As if "the rejection of the set of all sets is by definition contradictory" is not proof enough.
Or as if "a set is not a member of itself when it is a member of another set" is not proof enough.

These are two different claims:

1. A is not a member of itself
2. A is a member of some other set

Given this:

A={A}B={A,0}

(1) is false and (2) is true. — Michael

Evidently, in A, A is a member of itself.
Evidently, in B, A is not a member of itself because A is a member of B.

So you have not shown that 1 and 2 are two different claims.

I repeat:

A) When a set is not a member of itself, it is a member of another set
B) When a set is a member of itself, it is not a member of another set

Regarding Russell's paradox, it is simply this:

1. x is a member of R if and only if x is not a member of x.

Is R a member of R?

Either answer entails a contradiction, and so (1) is a contradiction. Given that naive set theory entails (1), naive set theory is shown to be inconsistent. — Michael

I understand Russell's paradox. Here is what I say in response:

z = any set that is not the set of all sets
v = any set
The v of all vs = the set of all sets
The z of all zs = The not-the-set-of-all-sets-set of all not-the-set-of-all-sets-sets — Philosopher19

Take z to be any set that is not the set of all sets, and take v to be any set. The z of all zs is a member of itself as a z (as in in the z of all zs it is a member of itself). But it is not a member of itself in the v of all vs, precisely because in the v of all vs it is a member of the v of all vs as opposed to a member of itself. If we view the z of all zs as a z, it is a member of itself. If we view the z of all zs as a v, it is a member of the set of all sets. You can't view it as both a member of the z of all zs and a member of the v of all vs at the same time. That will lead to contradictions. In other words, we can't treat two different standards/contrexts as one (as in are we focused on the context/items of vs or the context/items of zs to determine what is a member of itself or not a member of itself?) — Philosopher19

A is a member of both A and B.

I'll explain it to you in non-math terms:

I am a member of the football team and a member of the tennis team.

These are two different claims:

1. I am not a member of the football team
2. I am a member of a non-football team

(1) is false and (2) is true.

What you say in response doesn't prove that Russell's paradox isn't a contradiction.

1. x is a member of R if and only if x is not a member of x
2. Let x = R
3. R is a member of R if and only if R is not a member of R

(3) is a very obvious contradiction. You don't even need to know maths to see that.

A is a member of both A and B.

I'll explain it to you in non-math terms:

I am a member of the football team and a member of the tennis team.

These are two different claims:

1. I am not a member of the football team
2. I am a member of a non-football team

(1) is false and (2) is true. — Michael

So at least you're engaging me in clear meaningful language. I will respond in kind.

You are not a set. You can never be a member of yourself. But a set can either be a member of itself or a member of other than itself. And it is logically the case that when a set is a member of itself, it is a member of itself. And when it is a member of other than itself (as was the case with A in B), it is not a member of itself.

So your non-math example doesn't apply because you are talking about something that by definition can't be a member of itself, whereas in your A and B example, you were talking about something that by definition was a member of itself in its own set whilst a member of other than itself in another set.

I repeat:

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

But a set can either be a member of itself or a member of other than itself. — Philosopher19

A set can be a member of more than one set. You just don't understand the basics of set theory.

You should really take a few math lessons before you start telling mathematicians that they're wrong about maths.

I believe you're not paying attention to what I'm saying. I'm not saying a set can't be a member of more than one set. I am saying:

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

See again my replies to you:

Evidently, in A, A is a member of itself.
Evidently, in B, A is not a member of itself because A is a member of B. — Philosopher19

You are not a set. You can never be a member of yourself. But a set can either be a member of itself or a member of other than itself. And it is logically the case that when a set is a member of itself, it is a member of itself. And when it is a member of other than itself (as was the case with A in B), it is not a member of itself. — Philosopher19

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

And this is a fundamental misunderstanding of set theory.

If A = {A} and if B = {A, 0} then A is a member of A and a member of B.

If A = {A} and if B = {A, 0} then A is a member of A and a member of B. — Michael

Yes. But what you're not responding to is the following:

In the case of A = {A}. A is a member of itself.
In the case of B = {A, 0}, is A a member of A/itself, or is A a member of B/non-itself?

Also, don't forget that your non-math example doesn't apply because you are an x that by definition cannot be a member of itself, whereas a set is an x that can be a member of itself as well as other than itself.