Why argue with others? — jgill
Your p and q make no sense in set theory — Michael
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
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
3. In B, A is a set with 1 member, and that member is itself — Michael
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
not the members of the RANGE of the set. — 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
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
"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 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
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
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
I have not because those two are different sentences. — Lionino
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
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 answered those questions exactly. — TonesInDeepFreeze
In B, A is not a member of anything, A simply exists. — Lionino
Because it exists in B, it is a member of B — Lionino
It is a semantic point — Lionino
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
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
What book or article in the subject have you read/researched? — 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
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
Your argument seems to be that A is not a member of B in A because B is not defined in A — Lionino
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
If A is the set {A} then A is a member of both A and U. — Michael
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
1. x is a member of A if and only if x is a member of x — Michael
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
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
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
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
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
↪Philosopher19 A is a member of both A and B. This is basic set theory. Take a math lesson. — Michael
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
In the case of B = {A, 0}, is A a member of A/itself, or is A a member of B/non-itself? — Philosopher19
Both a member of itself and a member of B. — Michael
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
Both — Michael
If A = {A} and if B = {A, 0} then A is a member of A and a member of B. — Michael