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.

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?

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

That is not what I said, which is why you typed that out instead of quoting me.

how is it a member of B in B? — Philosopher19

It is not. That means nothing in mathematics.

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

See the parts I underlined? — Philosopher19

That is cute that you highlighted half of a sentence to make it seem it says something that it does not. Let me fix it:

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

You are having trouble figuring out the difference between a locative sentence and an explanatory one. I will leave you to it.

I will leave you to it. — Lionino

Ok

"L is a member of itself in L" has no apparent meaning if it does not simply mean "L is a member of L". Nothing is added by saying "in L". If L is a member of L then L is in L. Nothing is qualified by saying "in L" again.

There are two matters:

Whether L is a member of L. It is not, since the members of L are ordered pairs and L is not an ordered pair.

Whether L is a member of the range of L. It is.
I.e. whether L is one of the items listed by L. It is.

I believe you are in essence removing/ignoring the semantical implications of "member of self" with the way you engage. — Philosopher19

Saying 'semantical' adds nothing substantive in this context.

The notion of "member of self" is not more than "x is a member of itself if and only if x is a member of x".

And I addressed exactly whether L is a member of L. It is not. Rather, L is a member of the range of L.

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? — Philosopher19

Let's consider these four lists:

Months

• January
• February
• March
• April
• May
• June
• July
• August
• September
• October
• November
• December

Planets

• Mercury
• Venus
• Earth
• Mars
• Jupiter
• Saturn
• Uranus
• Neptune

Lists

• Months (January, February, ...)
• Planets (Mercury, Venus, ...)
• Lists (Months, Planets, Lists, Lists that list themselves)
• Lists that list themselves (Lists, (?) Lists that list themselves)

Lists that list themselves

• Lists (Months, Planets, Lists, Lists that list themselves)
• (?) Lists that list themselves (Lists, Lists that list themselves)

Q1. How many members does Months have?
A1. 12

Q2. How many members does Months have when it is a member of Lists?
A2. 12

Q3. How many members does Planets have?
A3. 8

Q4. How many members does Planets have when it is a member of Lists?
A4. 8

Q5. How many members does Lists have?
A5. 4

Q6. How many members does Lists have when it is a member of Lists?
A6. 4

Q7. How many members does Lists have when it is a member of Lists that list themselves?
A7. 4

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

So returning to your questions, they should simply be:

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

Assuming that these mean the same thing, the answer to both is yes. L lists itself/is a member of itself.

And so it is also listed by/a member of LL.

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?

Does L list itself in LL?
Is L a member of itself in LL/not-L? — Philosopher19

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

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. — Philosopher19

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.

If L has $n$ members then it has $n$ members "in L" and it has (the same) $n$ members "in LL".

So if L is a member of itself then it is a member of itself "in L" and it is a member of itself "in LL".

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

A list is a sequence. S lists x if and only if x is in the range of S. In that case, x is not a member of S, but rather it is a member of the range of S.

For example:

B = {John Paul George Ringo} is just a set.

T = {<1 John> <2 Paul> <3 George> <4 Ringo>} is a list of the members of B. Sometimes represented as the equivalent ordered 4-tuple:

<John Paul George Ringo>

or

1 John
2 Paul
3 George
4 Ringo

or

John
Paul
George
Ringo

B = range(T)

So John is not a member of T, but John is a member of the range of T = B.

Here's some JavaScript code to demonstrate:


Can test here. Click "Run" in top left. Bottom right will show "true".

The terminology in this context needs to be exact.

(I use 'K' instead of 'LL' because it is not good notation to use a letter 'L' as a standalone constant and also concatenated with itself to form another constant.)

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

I hope it is recognized that these are equivalent:

x is in S
x in S
x is an element of S
x is a member of S
xeS {read 'e' as the letter epsilon)

So, these are equivalent:

L is in L
L in L
L is an element of L
L is a member of L
LeL

These are equivalent:

T lists y
T is a list & y in range(T)

These are equivalent:

T lists T
T is a list & T in range(T)
T lists itself

So, these are equivalent:

L lists L
L is a list & L in range(L)
L lists itself

And these are equivalent:

K lists K
K is a list & K in range(K)
K lists itself

/

Does L list itself in L? — Philosopher19

"L lists itself" is sensical.

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

In other words, what exactly would need to be the case for "L lists itself" to be true while "L lists itself in L" is false? And what exactly would need to be the case for "L lists itself" to be false while "L lists itself in L" is true?

More generally, what exactly would need to be the case for "S lists y" to be true while "S lists y in x" is false? And what exactly would need to be the case for "S lists y" to be false while "S lists y in x" is true?

Even more generally, what exactly would need to be the case for "S has property P" to be true while "S has property P in x" is false? And what exactly would need to be the case for "S has property P" to be false while "S has property P in x" is true?

Is L a member of itself in L? — Philosopher19

"L is a member of itself" is sensical.

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

In other words, 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?

More generally, what exactly would need to be the case for "S is a member of y" to be true while "S is a member of y in x" is false? And what exactly would need to be the case for "S is a member of y" to be false while "S is a member of y in x" is true?

Again, even more generally, what exactly would need to be the case for "S has property P" to be true while "S has property P in x" is false? And what exactly would need to be the case for "S has property P" to be false while "S has property P in x" is true?

LL/not-L — Philosopher19

What does that mean?

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}.

What does '/' stand for?

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.

"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

Set theory does not have a "where".

My question remains:

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

You are using your own personal terminology for an unclear notion that no one other than you can make sense of.

You need to define your terminology in already understood mathematical phrasing. Otherwise, whatever you have in mind won't be understood by others.

So how does it follow that L has n-1 members in LL? — Philosopher19

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".

Which is nonsense.

If L has $n$ members then L has $n$ members "in L" and L has $n$ members "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. — Philosopher19

This is wrong.

1. L is a member of L.
2. L is a member of LL.
3. L is a member of L "in LL"

4. January is a member of Months
5. Months is a member of L
6. January is a member of Months "in L"

Let's make this simple:

A = {A}
B = {A, 1}
C = {{}, 1}

When A is a member of B, is A the empty set?

The answer is no. "In B" the set A contains itself as the only member.

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

When in fact B = {A, 1} = {{A}, 1}

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.

You didn't answer the question. I won't bother to post it yet again.

If you can find even one mathematician who can say what you mean by "L is a member of itself in L" then I'd welcome hearing about it.

And if you can define 'where', in whatever sense you mean by it, from the terminology of set theory, then I'd welcome hearing about it.

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.

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. — Philosopher19

If x in x, then x in x no matter what other sets x is or is not in.

If you don't agree or don't understand, then either you need to state your alternative system and terminology or you need to learn basic set theory. I recommend starting with the latter.

As I said, you have a highly idiosyncratic notion of sets and a highly idiosyncratic terminology to go with that. If you wish to be understood by other people, then you need to either state your system of notions and terminology, or show how they can be defined from notions and terminology that other people do already understand.

/

You can start at the very beginning:

'is an element of'
'is a member of'
'is in'
'in'
'e'

are just variants of the same primitive relation of set theory.

The symbol 'e' itself is a primitive relation symbol. It is not defined. Rather, there are axioms that determine what theorems are derived with 'e' in them. However, from 'e' we define a number of other symbols that intuitively stand for various concepts, including 'list'. With those definitions we formulate yet more theorems (though they all are in principle reducible to just the primitive 'e').

Your notions though, as so far stated, don't resolve to that process. Therefore, if you want to be understood, then you need to either show how your notions do resolve to those of set theory, or you need to state your own primitives, axioms and definitions that do explicate your own notions, or at least give some coherent outline about that.

/

Meanwhile, at least you should pay attention to the fact that there is a difference between 'set' and 'list' as a list is a certain kind of set. For example:

{John Paul George Ringo} = {George Paul Ringo John}

but

{<1 John> <2 Paul> <3 George> <4 Ringo>} not= {<1 George> <2 Paul> <3 Ringo> <4 John>}

and

<John Paul George Ringo> not= <George Paul Ringo John>

{John Paul George Ringo} is the set whose members are John, Paul, George and Ringo, in whatever order you want to mention them.

{<1 John> <2 Paul> <3 George> <4 Ringo>} is a set theoretic list, which is itself a set whose members are <1 John>, <2 Paul>, <3 George> and <4 Ringo>, in whatever order you want to mention them.

The range of {<1 John> <2 Paul> <3 George> <4 Ringo>} is {John Paul George Ringo}. Or, we can say that the entries of the list are John, Paul, George and Ringo.

<John Paul George Ringo> is the ordered 4-tuple that corresponds to the above list. (It actually unpacks to nested ordered pairs that unpack to certain unordered pairs, but we don't need to spell the details of that now.) Or we can say that the coordinates of <John Paul George Ringo> are John, Paul, George and Ringo.

Crucial takeway: A set is not in any particular order, except a list is a special kind of set that does convey a particular order. And the members of a list are ordered pairs, not the members of the RANGE of the set.

So, I've given you information that is at least a start for you to use common mathematical terminology, so that you may be understood by people other than yourself.

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?

V = the set of all v
Z = the set of all z — Philosopher19

That doesn't mean anything.

We don't say "the set of all z". We say "the set of all z such that [fill in some property here]"

Examples:

the set of all z such that z is an even number
which is
{z | z is an even number}

the set of all z such that z is a subset of the set of natural numbers
which is
{z | z is a subset of the set of natural numbers}

/

'domain' and 'range' are defined:

domain(S) = {x | Ey <x y> in S}

range(S) = {y | Ex <x y> in S}

That is most applicable when S is a relation (a relation is a set of ordered pairs).

And every function is a relation.

The notion of 'list' is captured by a certain kind of function.

{<1 John> <2 Paul> <3 George> <4 Ringo>} is a list. It maps

1 to John
2 to Paul
3 to George
4 to Ringo

It tells you not just the members of the set {John Paul George Ringo} but also a particular ordering of that set.

The set of objects that the function maps to is the range of the function. In this case

the domain is {1 2 3 4} and the range is {John Paul George Ringo}.

Think of it this way:

Suppose I tell you that B is the set whose members are the Beatles. I haven't told you an order of the musicians in the Beatles, just that B is the set of them no matter in what order.

There are 24 ways to list the Beatles.

Set theory conveys the notion of a list with functions:

{<1 John> <2 Paul> <3 George> <4 Ringo>} is one list.

{<1 George> <2 John> <3 Paul> <4 Ringo>} is another list.

and there are 22 more.

In other words, we exhibit the order of the list by putting the number of the item in the list before the item.

Again:

the members of {John Paul George Ringo} are John, Paul, George and Ringo, in no particular order.

the members of one LIST of the Beatles are <1 John>, <2 Paul>, <3 George> and <4 Ringo>.

the members of another LIST of the Beatles are <1 George>, <2 John>, <3 Paul> and <4 Ringo>.

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

/

So, if L is the list of all lists, then L is a function. And the items listed by L are the range of L. In this case, the range of L is the set of all lists. So, since L is itself a list, L is in the range of L. But L is not in L, since L is not an ordered pair.

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.

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

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

Given this:

A = {A}
B = {A, 1}

One of these must be true:

1. In B, A isn't a set
2. In B, A is a set with 0 members
3. In B, A is a set with 1 member, and that member is itself
4. In B, A is a set with 1 member, and that member isn't itself
5. In B, A is a set with more than 1 member, one of which is itself
6. In B, A is a set with more than 1 member, none of which is itself

So which of these claims are you making? It must be one of them.

The correct answer is (3):

A = {A}
B = {A, 1} = {{A}, 1}

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?

Your p and q make no sense in set theory. Only 3a is meaningful in set theory (although as already mentioned, the “in B” part of the sentence is vacuous). And using that meaning, the set of all sets that don’t contain themselves is a contradiction. Russell proved this.

Whatever you’re trying to argue has nothing to do with Russell’s paradox and nothing to do with set theory.

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.