[Guille]

Originally Posted by dustin_archibald

I don't know if someone already said this, in this thread; there is a lot to read and understand.

One way to avoid Russell's Paradox is to prove that it isn't a paradox at all.

One way we can do this (prove it's not a paradox) is by proving that it's based on a false assumption; that sets can contain themselves, more specifically, sets have the ability to contain themselves. How do we do this? We need to prove that a set cannot contain itself; that it cannot exist if it contains itself.

A second way to do this (prove it's not a paradox) is by proving that it's based on a false assumption; that a set cannot contain itself; a set does not have the ability to contain itself. How do we do this? We need to prove that a set must contain itself; that a set cannot exist without containing itself.

If you can prove that either (one or the other) assumption is incorrect, the paradox will fall apart since it is based on a false assumption

Answering the question "Does the set of all sets contain itself?" is important since, in my opinion, it is the only set with the possibility of containing itself.

I had thought of disprovng a starting assumption. In some threads I've stated clearly that the logicians have foughtthe paradox by saying that sets dont' contain sets, and that if there is a collection of any kind of sets, it is to be called a group. A collection of groups is a class...etz, there's a point where it all stops, but that's nto important: the important thing is that the claim to be false the assuption that sets can contain any set at all.

Now, both of the assumptions you propose that could be proved false are good points for they are not as strong as the other assuptions (there are a few other: that a set can contain sets, that a set can contain sets that don't contain themselves, that a set can contain sets that do contain themselves....). But any attack to them is simply impossible. They are right by definition. If you read my starting post, I give two examples one of a set that contains itself and one that doesn't.

I give you these proof: the set of all Dustins is not a Dustin itself, therefore it doesn't contain itself. The set of of all sets that have infinite to 0 members contains itself, as it has a number of members (even if it's infinite). By the first I proof that sets have the ability to not contain themselves, and by the second I proof that sets have the ability to contain themselves.