x, the set of all sets that are not members of themselves, is thus a member of itself. — Philosopher19
That right there is the contradiction. x is a member of itself if and only if it's not a member of itself. — fishery
Thank you for replying and I understand where you're coming from. I will try to convey to you my understanding more specifically hoping that specification saves naive true set theory. I will ask questions to see exactly where it is that we are in disagreement. — Philosopher19
For the sake of argument, assume we have the set of all sets. Call this x. — Philosopher19
x is a member of itself because it is a set. No paradoxes so far, agreed? — Philosopher19
Since x contains all sets, do we agree that x contains all sets that are not members of themselves? — Philosopher19
The set of all penguins, is a set. — Philosopher19
This set is not a member of itself precisely because it is a member of the set of all sets. — Philosopher19
By this I mean It is specifically a set, not a penguin. Are we sill in agreement? — Philosopher19
The set of all animals, is one set that contains the set of all penguins. — Philosopher19
This set is also not a member of itself, precisely because it is a member of the set of all sets. — Philosopher19
Thus, by definition, any set that truly is a set (as opposed to a penguin or animal), and is not a member of itself, is not a member of itself precisely because it truly is a set and is thus a member of the set of all sets. Agreed? — Philosopher19
If agreed, — Philosopher19
then can you see how the set of all sets that are not members of themselves, can only be (by definition) x? — Philosopher19
I will specify this some more: All sets that are not members of themselves, truly are sets. — Philosopher19
What is the set of all these sets? Can the answer be anything other than x? — Philosopher19
The set of all sets which contains all these sets, is a member of itself (because it truly is a set). — Philosopher19
Where do we have a paradox in what I have proposed? — Philosopher19
Yes. The set of all sets that are not members of themselves is different from the set of all sets; because x contains all the sets that are not members of themselves AND all the sets that ARE members of themselves. I don't follow why you don't see this. — fishery
It is absurd to say that there is no set of all sets. — Philosopher19
↪fishfry
I admire and appreciate the trouble you go to here and elsewhere to put some of us back on the right track - even as some of us go right back off the rails! — tim wood
To my knowledge, Russell's paradox concludes that you cannot have a set of all sets because he fails to non-paradoxically define a set of all sets that are not members of themselves. — Philosopher19
No one is disputing that there can be no set of all sets that are not members of themselves that is itself, not a member of itself (call this absurd set y). But this paradox in no way logically amounts to saying that there is no set of all sets. x is the set of all sets. This set contains all sets including itself. No paradoxes. — Philosopher19
For the sake of argument, let's say x and y are not the same. I have no problem in saying that y is absurd. But there is still a set that contains all sets that are not members of themselves. x contains them all does it not? If it does contain them, then it contains them. Why does x have to be not a member of itself??? Why are we trying to force a paradox where there is none? A set is a set. It doesn't matter if it's a member of itself or not. If it truly is a set, then it is clearly a member of the set of all sets. — Philosopher19
Yes, x also contains other sets (actually it only contains one other set...which is itself). But it still contains all sets that are not members of themselves. — Philosopher19
Regarding sets, Russell misunderstood semantics and logic. I understand he was an important philosopher, but he made a mistake. — Philosopher19
It is absurd to say that there is no set of all sets. Now are we in agreement? — Philosopher19
If you call x the set of all sets, you quickly get a contradiction. You find that x both is and isn't a member of itself. Therefore there is no such set. You keep claiming there is but you have not provided proof. — fishfry
You find that x both is and isn't a member of itself. — fishfry
You trollin' me? — fishfry
given a set, any collection of its elements is also a set — SophistiCat
Since sets that are not members of themselves are included in the set of all sets, then a collection of all such sets must form a set — SophistiCat
Russell is clearly wrong. This paradox is now clearly fixed. Can we move on in a unified manner? — Philosopher19
Yes. Off the first page of TPF. — jgill
The following is proof:
I find that if I say x isn't a member of itself, I am being paradoxical because x is a set. — Philosopher19
I find that if I say x is a member of itself, I am not being paradoxical because x is a set. Do you see? — Philosopher19
You find that x both is and isn't a member of itself.
— fishfry
How do you get to this???? — Philosopher19
x = the set of all sets. Is x a set? Yes. — Philosopher19
Thus x is a member of itself. Is x not a member of itself? Yes it is a member of itself because it is a set! Let's try the alternative. x is not a member of itself. Why not? No reason can be given. The set of all penguins is not a member of itself. Why not? Because a penguin is not a set. See? — Philosopher19
I find that x is a member of itself. That is all I find. — Philosopher19
Not at all. If I am, then I'm an idiot. I just want efficiency and truth. — Philosopher19
As does yours — jgill
You ignored his patient explanations and are now repeating the same mistakes here. And while you quoted my post, you did not address its content and instead repeated the same nonsensical arguments that you made earlier. — SophistiCat
I don't see a point in continuing this conversation. — SophistiCat
You claim x is a set but it isn't. You have no proof that x is a set. — fishfry
By Russell's paradox. Say x is the set of all sets. Then let y be the set of all sets that are not members of themselves. Is y a member of itself? — fishfry
No, it can't be. I just showed that if x is the set of all sets, then we can form y and derive a contradiction. If x is the set of all sets that don't contain themselves, then x itself leads to a contradiction. — fishery
See what just happened? If x contains itself then x doesn't contain itself. — fishfry
On the other hand suppose x doesn't contain itself. Then it must be a member of x. So if it doesn't contain itself it does contain itself. — fishfry
Now if you prefer to let x be the set of all sets, we let y be the set of all sets that don't contain themselves and we get a contradiction from y. So again, x can't be a set. — fishfry
Just look at the proof that I am presenting without bias and without preconceived notions. — Philosopher19
consider the hypothetical possibility that all these famous philosophers after Russell, were wrong, and I'm right — Philosopher19
Get involved in philosophical discussions about knowledge, truth, language, consciousness, science, politics, religion, logic and mathematics, art, history, and lots more. No ads, no clutter, and very little agreement — just fascinating conversations.