And I said that the exact answer to "Does L list itself?" is yes. — TonesInDeepFreeze
It has everything to do with what you said. — TonesInDeepFreeze
I have not because those two are different sentences. — Lionino
In B, A is not a member of anything.
In B, A is a member of B. — Philosopher19
how is it a member of B in B? — Philosopher19
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? — Philosopher19
Because it exists in B, it is a member of B. — Lionino
I believe you are in essence removing/ignoring the semantical implications of "member of self" with the way you engage. — Philosopher19
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
Q2, Q4, Q6, and Q7 are redundant/confused questions. We only have to consider Q1, Q3, and Q5. — Michael
So returning to your questions, they should simply be:
1. Does L list itself?
2. Is L a member of itself? — Michael
Does L list itself in LL?
Is L a member of itself in LL/not-L? — Philosopher19
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
What is the difference between asking if L lists itself and asking if L is a member of itself? — Michael
// Lists const l = {} // Add Lists to itself l.l = l // Lists that list themselves const ll = {} // Add Lists to Lists that list themselves ll.l = l // Get all the members of L-in-LL const members = Object.values(ll.l) // Is L a member of L-in-LL? console.log(members.includes(l))
Does L list itself in L? — Philosopher19
Is L a member of itself in L? — Philosopher19
LL/not-L — Philosopher19
What is the difference between asking if L lists itself and asking if L is a member of itself? — Michael
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
"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
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
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
So how does it follow that L has n-1 members in LL? — Philosopher19
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
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
Set theory does not have a "where". — TonesInDeepFreeze
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
If L has n members then L has n members "in L" and L has n members "in LL". — Michael
Your position entails that B = C, which is false. — Michael
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
not the members of the RANGE of the set. — TonesInDeepFreeze
V = the set of all v
Z = the set of all z — Philosopher19
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
When a set is a member of another set it is still a set with members of its own. — Michael
3. In B, A is a set with 1 member, and that member is itself — Michael
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