I am a little thrown off with this.
Can we express Russell’s paradox as "A dog chases all and only those dogs that don’t chase themselves"? What is a formal proof showing that it is impossible for such a dog to exist?
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1What have you tried so far?– Joseph Weissman ♦Commented Dec 2, 2020 at 16:43
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1Already "discovered"as Barber paradox– Mauro ALLEGRANZACommented Dec 2, 2020 at 17:14
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It will be nice if you provide your attempt so that people who want to answer will get a clearer idea of what exactly you are looking for. If you already know the formal proof that the set R cannot exist, then I guess you can also proceed with the example you gave by setting R as the set of all dogs that do not chase themselves. The proof should be a reductio proof. Note that here I am assuming a naive set theory and R := the set of all sets that do not contain themselves.– Abdul Muhaymin -Free PalestineCommented Dec 2, 2020 at 19:41
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1@user4894 I disagree, the proof to show this is the same, as is the problem. It's just a renaming of the stuff the proof is about. Rename "Set" to "Barber/Dog/Object" and "contains" to "shaves/chases/has relation R with" and the result is "Barber paradox" or this "Dog paradox" or some other arbitrary paradoxical situation of the same form.– kutschkemCommented Dec 4, 2020 at 8:17
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2@user4894 it depends how strict you're being. Russell's paradox is a particular instance of a more general logical argument, that we can never have a relation on a domain such that there is some element related to exactly those elements not related to themselves. Precisely what the domain and relation are doesn't affect this. So picking out the specific set-theoretic context is somewhat extraneous.– Noah SchweberCommented Dec 5, 2020 at 1:51
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2 Answers
So the reason this isn’t paradoxical is that there’s nothing inherently wrong with that definite description; it just fails to refer. There is no such dog!
The Russell class results in a paradoxical outcome for any theory of predicative set definitions in that if the description itself is perfectly legitimate then the method of specifying sets by description is unsound. The Russell set is not itself the Russell paradox!
Does the dog chase itself?
Assume it does:
Then it doesn't chase itself because it is not not chasing itself. Contradiction.
Assume it doesn't:
Then it does chase itself because it is not chasing itself. Contradiction.
Both cases lead to a contradiction, so such a dog can't exist.
And yes, this is the same situation as Russel's paradox.
Replace dog with Set, replace chase with contains.
Many paradoxes boil down to this or a similar proof. The impact or point of Russels paradox however is that you can't use arbitrary properties to define sets. This is an insight that means that naive set theory leads to a contradiction (because it assumes just that), and lead to set theory being reformulated a few years later.
ZFC does not assume that, for every property, there is a set of all things satisfying that property. Rather, it asserts that given any set X, any subset of X definable using first-order logic exists. The object R discussed above cannot be constructed in this fashion, and is therefore not a ZFC set. In some extensions of ZFC, objects like R are called proper classes.
