Why argue with others? — jgill

It could be that arguing with someone is what it takes for them to be sincere to a truly perfect existence/being (or God/Goodness/Truth). It could be that that is what it takes for them to favour good/truth over evil/falsehood in a particular instance/moment. It could also be the opposite. I'd like to think I got into this discussion as a part of my efforts to favour/prioritise Goodness/Truth (or to strive in the cause of a truly perfect existence/being).

Your p and q make no sense in set theory — Michael

I believe it makes perfect sense to say set x is only a member of itself in its own set, and I tired to prove it to you, and we even narrowed it down to p and q, but you have ceased to engage on the grounds that "p and q make no sense in set theory". This truth (the truth of a set only being a member of itself in its own set) has logical implications that would have shown (contrary to those who reject the universal set because they believe a set can be a member of itself outside of its own set) that the universal set is not contradictory in any way.

Russell’s paradox:

Assumption: S is the set of all sets that are not members of themselves.

Option 1:

S = {}

S is not a member of itself. But, as per the assumption above, it ought be a member of itself.

Option 2:

S = {S}

S is a member of itself. But, as per the assumption above, it ought not be a member itself.

Neither option 1 nor option 2 work. Therefore, the assumption is a contradiction. — Michael

I believe if I respond to this, we'll just go around in circles, and then perhaps we'll come back to the p and q point again. But instead of doing that, I will just say that my question remains unanswered:

Which is correct: p or q or both or neither? — Philosopher19

When a set is a member of another set it is still a set with members of its own. — Michael

I understand/agree.

3. In B, A is a set with 1 member, and that member is itself — Michael

Let's focus on 3. There is a difference between:

3) In B, A is a set with 1 member, and that member is A/itself. (Correct)
3a) A is a set in B that is a member of itself in A/itself (Correct)
3b) A is a set that is a member of itself in B

Here is why 3b is incorrect. Compare:

p) In B, A is a set that is a member of itself in A
q) In B, A is a set that is a member of itself in B

p and q are not the same and they both highlight a meaningful difference. p is correct (because in B, A is a member of A/itself in A). q is false (because in B, A is not a member of A/itself in B)

To my understanding, you either wrongly view p and q as the same thing, or, you wrongly view q as correct. But how can you view q as correct given that p is truth and p and q are opposing things?

So my question to you is:
Which is correct: p or q or both or neither?

Forgot to say thank you for the detailed reply. Thanks.

v = any set
z = any set other than the set of all sets
V = the set of all v
Z = the set of all z

Am I right in saying that the RANGE of V is greater than Z because there are more v than z? Is this what you mean by RANGE? — Philosopher19

but all those lists have the same RANGE, which is {John Paul George Ringo}, whose members are John, Paul, George and Ringo. — TonesInDeepFreeze

So I think I understand what you mean by RANGE, but I feel any further conversation will not be fruitful. I feel like I've already said all that I should have said. And anything that could have added to what I should have said I think would be in the post to which I provided the link to before. In the event that I've not done justice to what you've been saying to me, I apologise.

not the members of the RANGE of the set. — TonesInDeepFreeze

Let me see if I've understood what you mean by RANGE.

v = any set
z = any set other than the set of all sets
V = the set of all v
Z = the set of all z

Am I right in saying that the RANGE of V is greater than Z because there are more v than z? Is this what you mean by RANGE?

You said that L is a member of itself "in L" but not a member of itself "in LL". So you're saying that L "in L" has one more member (itself) than L "in LL". — Michael

That is not what I'm saying. Nor do I see what I'm saying as amounting to that.

If L has n members then L has n members "in L" and L has n members "in LL". — Michael

L does not have n members in LL. L is one item in LL. And of all the items in LL (of which L is one), only one item is LL.

Whether an item in a set is a member of itself or not is dependent on what it is and in what set it is in.

LL is an item.
It is in both L and LL.
In L it is not a member of L/itself.
In LL it is a member of LL/itself.

Your position entails that B = C, which is false. — Michael

I do not see how it does. Again, L does not have n members in LL. L has n members in L, and LL has n members in LL.

What exactly would need to be the case for "L is a member of itself" to be true while "L is a member of itself in L" is false? And what exactly would need to be the case for "L is a member of itself" to be false while "L is a member of itself in L" is true? — TonesInDeepFreeze

L is a member of itself in L is true.
L is a member of itself is true (but it is logically implied that L is a member of itself in L).

Set theory does not have a "where". — TonesInDeepFreeze

It makes sense to say x is a member of itself because x is in x. It makes sense to say x is a member of a set other than itself in a set other than itself. If you reject this because "set theory does not have a "where"", then our conversation cannot progress.

"L lists itself in L". What does that mean other than "L lists itself"? — TonesInDeepFreeze

"L is a member of itself in L". What does that mean other than "L is a member of itself"? — TonesInDeepFreeze

If I ask the question where does L list itself, the answer is in L. If I ask where else does L list itself, the answer is nowhere else. This shows that L is only a member of itself in L.

What does that mean? — TonesInDeepFreeze

What does 'not-L' stand for? Does it stand for the set of all things not in L? That is, {x | x not in L}. Or does it mean the set of all things that are not L? That is, {x | x not= L}. — TonesInDeepFreeze

It stood for LL. I meant to highlight that LL is not-L to try and draw attention to L is not a member of L/itself in LL/not-L

What is the difference between asking if L lists itself and asking if L is a member of itself? — Michael

There is no difference. I asked the question to highlight a point about the property of "being a member of oneself".

If L is a member of itself "in L" but not a member of itself "in LL" then L has n members "in L" and n−1 members "in LL".

But this makes no sense. A set is defined by its members. — Michael

Every member of LL (including L) is a member of LL because there is a list wherein which it lists itself.
So how does it follow that L has n-1 members in LL? Now note how of all the members of LL there is only one item that is actually LL. This is what instantiate LL lists itself in LL. Again, L is not a member of itself in LL (even though it is in LL because it is a member of itself in L). L is only a member of itself in L.

Q2, Q4, Q6, and Q7 are redundant/confused questions. We only have to consider Q1, Q3, and Q5. — Michael

I'm not asking how many members does x or y have. So I don't see how your example is relevant to what I asked.

So returning to your questions, they should simply be:

1. Does L list itself?
2. Is L a member of itself? — Michael

My questions were reasonable/relevant/meaningful questions. I believe they were what they should have been. They highlight that a list only qualifies for the property of 'listing itself' when it is listed in its own list. The property of "being a member of self" is instantiated depending on what the item is and what set it is in.

So the questions to be asked are:

Does L list itself in LL?
Is L a member of itself in LL/not-L?

I will leave you to it. — Lionino

Ok

Here is what you said:

t does not mean anything in mathematics.
In B, A is not a member of anything, A simply exists. Because it exists in B, it is a member of B. But that has no bearing on Russell's paradox. It is a semantic point — Lionino

See the parts I underlined? Here they are for more clarity:

In B, A is not a member of anything
in B, it is a member of B

I have not because those two are different sentences. — Lionino

You literally said two things that contradict each other:
In B, A is not a member of anything.
In B, A is a member of B.

If in B A is not a member of anything, then how is it a member of B in B?

And I said that the exact answer to "Does L list itself?" is yes. — TonesInDeepFreeze

I believe you used your notation to avoid/miss the semantical point I was trying to make. Here are the answers to the questions I asked:

Does L list itself in L? Yes
Does L list itself in LL? No

Additionally

Is L a member of itself in L? Yes
Is L a member of itself in LL? No

It has everything to do with what you said. — TonesInDeepFreeze

I don't believe it does. I am talking about the semantical implications of being a member of self and not being a member of self, I believe you are talking about something else. I believe you are in essence removing/ignoring the semantical implications of "member of self" with the way you engage.

No, I explained the difference.

I'll say it again, a list is a sequence. A sequence is a function whose domain is an ordinal. So the members of a list are ordered pairs. The members of the range of a list are the items listed by the list. — TonesInDeepFreeze

I believe this has nothing to do with what I said.

I answered those questions exactly. — TonesInDeepFreeze

I believe the exact answer to "Does L list itself in L?" is yes.

I believe the conversation won't progress beyond this point.

Peace

So on the one hand you say:

In B, A is not a member of anything, A simply exists. — Lionino

On the other hand you say:

Because it exists in B, it is a member of B — Lionino

Do you see how you have contradicted yourself?

And then you say

It is a semantic point — Lionino

Yes, proper attention has not been paid to the semantics of "member of self" and "not member of self". Whether a set is a member of itself or not, is determined by what item it is and in what set it is in.

L lists itself in L. L does not list itself in LL. L is a member of itself in L/itself. It is not a member of itself in LL/not-itself.

I don't know what you mean by that. I don't know what you mean by a property being instantiated in this context — TonesInDeepFreeze

L is listed in both L and LL. In L, it has the property P of being a member of itself/L. As in the fact that L is in L is what instantiates the property P (P = self-membership) in L's case.

In terms of function and logic, there is no difference between "lists itself" and "is a member of itself". Listing oneself and being a member of oneself are both matters of self-reference. So:
Does L list itself in L?
Does L list itself in LL?

In mathematics, it makes no sense to ask, "In which set does x have the property P?" Rather, we ask, "Does x have property P?" If x has property P, then that is not qualified by "x has property P in some sets but not others". So one can't give sensical answers to nonsensical questions such as 1) and 2) in the previous post. — TonesInDeepFreeze

The property P is instantiated based on what set the item x is in. 1 and 2 were solid meaningful questions that highlight precisely this point. L is listed in both L and LL. In L it has the property of being a member of itself/L, in LL it has the property of being a member of other than itself/L. L is not LL. L is L.

Hi Michael

In case you are interested in discussing this further:

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

1) In which list does L list itself?
2) In which list is L a member of itself?

Can you answer both questions consistently and non-contradictorily?

I don't think the process of continuing forever amounts to anything infinite. I see infinity as the reason for the process of continuing forever as being possible/meaningful.

To me, Infinity and Existence denote the same.

I see Existence as the set of all trees, humans, numbers, existents/cardinalities. I see the set of all existents/cardinalities as Infinite. I'm not sure if I should describe Infinity as a quantity here or not. But I think something like 'the cardinality of absolutely all existents (so that's all numbers, letters, trees, semantics, hypothetical possibilities and so on), amounts to Infinity'. I don't see Existence as incomplete because such a view runs into contradictions, hence the need for Existence to equal Infinite (and possibly the need for Infinity to equal a quantity representative of the cardinality of absolutely all existents).

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

I watched a YouTube video amongst other things. Didn't read a book on it. Had a look at wiki and Stanford but maybe or maybe not with massive amounts of focus. Just enough to understand the problem. Also had an exchange with a math lecturer who specialises in set theory (but we disagreed with each other fundamentally, but perhaps the exchange helped me better understand the other's position)

In any case, given the responses I have seen, I see no point in continuing this discussion.

Peace TonesInDeepFreeze

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? — TonesInDeepFreeze

I believe I understand the matter well enough. I don't feel it sincere to Truth/Goodness to read/research any more on it than I already have. The link to what I've written on Russell's paradox and Infinity is there should anyone choose to read it. Whether they should read it or not, is not for me to say. It may be sincere to Truth/Goodness for them to read it, it may be sincere to Truth/Goodness for them not to read it. It may even be that Truth/Goodness is not their priority. I don't know, I don't have their self-awareness.

No, it's nonsense. That's not how set theory works.

1 is a member of N and R.
A is a member of A and B.

That's it. — Michael

Already responded to it at least twice. Won't repeat myself.

I'm saying whether something is a member of itself or not is determined by whether it is in its own set or not. This matter of pure reason/definition seems to have been disregarded and faulty notation/language has then been used to falsely conclude there is no universal set. It seems to me foundationally corrupt theories have been built on a misunderstanding and so much has been invested in them that there is either unwillingness to let them go, or there is so much dogma that the obviously true is not paid attention to.

Your argument seems to be that A is not a member of B in A because B is not defined in A — Lionino

My argument is that A is not a member of itself in B because A is a member of B in B. To preserve this truth and take into account what set theorists seem to be focusing on, I wrote the following:

here is a demonstration of how a set can only be viewed as a member of itself in its own respective set:

v = any set
The v of all vs = the set of all sets
z = any set that is not 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

The z of all zs is a member of itself in the the z of all zs, 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 v of all vs. We can't treat two different things/references/contexts/standards/items 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

For the whole of what I'm saying, read:

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

Fine, I will more or less repeat myself but this is probably the last time.

If A is the set {A} then A is a member of both A and U. — Michael

p) In A, A is a member of itself/A.
q) In U, A is not a member of itself/A.

p and q are true by definition.

If definitional truths are not enough (and they really should be), here is a demonstration of how a set can only be viewed as a member of itself in its own respective set:

v = any set
The v of all vs = the set of all sets
z = any set that is not 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

In the z of all zs, the z of all zs 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 v of all vs. We can't treat two different things/references/contexts/standards 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?)

You've argued that there is a set of all sets, U.

If A is the set {A} then A is a member of both A and U. — Michael

It seems evident to me that you've not been paying attention to what I've been saying. I've already given more than one relevant reply to this. I'm not going to repeat myself.

1. x is a member of A if and only if x is a member of x — Michael

I have been saying to you that x is not a member of A if x is a member of x. I have not been saying 1. 1 is blatantly contradictory. So it seems to me that you have misrepresented/misunderstood what I have been saying to you.

Yes, you’re confused. A is a member of A and B. 1 is a member of N and R. That’s all there is to it. — Michael

I believe I addressed this point both in my reply to your non-math example and in your N and R example and in my other replies. No point doing in doing it again. Plus there is my z and v example.

And you’re confused. It’s not the case that “in A” it’s a member of one thing and “in B” it’s a member only of something else. — Michael

Am I the one that's confused? So it's not the case that in A it's a member of one thing and in B it's a member of another thing? So it's not the case that in A it's a member of itself and in B it's not a member of itself?

These are blatantly obvious things. As far as I can see, only dogma and bias and insincerity to Truth would cause one to fail to see/recognise them.

Maybe the following will help. I don't know.

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

Scenario 1
— Michael

Scenario 2
B = {0, A}, where A = {A} — Michael

I have not disagreed with scenario 2. I have said that in B, A is not a member of itself precisely because it is a member B (as opposed to itself), and in A, A is a member of itself.

I believe I have already said more than enough. Check my replies to you because I believe I am now repeating myself where I shouldn't have to.

Yet you seem to think that in B, A is both a member of A and B (which is contradictory)

So why is it that A can be both a member of B and C but not a member of both A and B? — Michael

I never said A can't be both a member of A and B. I said, in A, A is a member of A/itself, and in B, A is a member of B/other-than-itself.

N is the set of natural numbers.
R is the set of real numbers.

Every natural number is a member of both N and R. We don't say "in R, the natural numbers are not members of N". — Michael

I addressed a similar point in my reply to your non-math example. Every natural number can be both a member of N and R. I am not denying this. But what a natural number cannot be (and what N and R cannot be) is members of themselves. When you talked about A, you talked about something that was by definition a member of itself precisely because A = {A}. But in the case of B = {A, 0}, A is not a member of itself because now, by definition, A is a member of B.

↪Philosopher19 A is a member of both A and B. This is basic set theory. Take a math lesson. — Michael

And in B (as opposed to in both A and B), A is not a member of itself because it is a member of B (and not A). This is basic.

It is just the case that the symbol "A" is defined recursively as "{A}" and that the symbol "B" is defined as "{A, 0}", which is the same as "{{A}, 0}" given the recursive definition of "A". — Michael

And now you have strayed from clear language. What I have given you is clear should you choose to pay attention to it:

I asked you:

In the case of B = {A, 0}, is A a member of A/itself, or is A a member of B/non-itself? — Philosopher19

to which you said:

Both a member of itself and a member of B. — Michael

to which I said:

Look at what you're saying:

"In B" does not equal to "in both A and B".
Do you see your contradiction? You have treated "in B" as the same as "in both A and B". — Philosopher19

Both a member of itself and a member of B. — Michael

Look at what you're saying:

In B, A is both a member of A and B.

"In B" does not equal to "in both A and B".
Do you see your contradiction? You have treated "in B" as the same as "in both A and B".

Both — Michael

But in B, A is a member of B. In B, A is not a member of both A and B. So once again, in B, is A a member of itself or not a member of itself?

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.