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So I understand Russels paradox (barber) but I do not understand the proof, I've looked everywhere online and youtube videos but it doesn't seem to make sense.

Please note, I have compensated dyslexic so I find it hard to interpret and understand text. I think this is the reason why I am finding it difficult to understand the proof.

This is as much as I understand: P(x) <----> x is not a member of itself. What I don't understand is if it means the element x is not a member of itself? And if so, I do not understand how a element can be a member of itself.

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  • $\begingroup$ The issue with trying to find a set that is an element of itself isn't really possible, with the traditional definition of a set (in fact, in ZFC there is an axiom which implies that a set cannot be an element of itself). However, the axioms for Naive Set Theory (which is the setting of Russell's Paradox) don't forbid it, so there could be really weird sets out there that are elements of themselves, and in fact in Naive Set Theory the universe is a set (and hence contains itself); Russell's Paradox shows that this is an issue. $\endgroup$
    – Hayden
    Mar 24 '15 at 18:03
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HINT:

Don't get hung up on the set-theoretic $\in$ relation. The proof works for every binary relation.

Consider the statement: $$\exists r: \forall x:[E(x,r) \iff \neg E(x,x)]$$ where $E$ is just some logical predicate having nothing to do with set membership.

Show how, from the above statement, you can obtain the contradiction: $$E(r,r) \iff \neg E(r,r)$$ What would this tell you about the original statement?

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Don't use set-theoretic notation for the barber paradox. The proof is just:

Take any barber who cuts the hair of exactly those who don't cut their own hair.

Either the barber cuts his own hair or he does not.

If he does cut his own hair, then by his own rule he is not supposed to cut his hair.

If he does not cut his own hair, then by his own rule he is supposed to cut his hair.

In all cases we get a contradiction, and so such a barber does not exist.

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  • $\begingroup$ Using a set-theoretic analysis, you can assume the existence of a barber who is man in the village ($barber\in village$) and show that there cannot exist a relation $shaves$ among the men in the village that satisfies the requirement, $\forall x\in village:[(barber,x)\in shaves \iff (x,x)\notin shaves]$. See the video demo at my website dcproof.com $\endgroup$ Apr 23 '15 at 2:21
  • $\begingroup$ @DanChristensen: My point is that it is unnecessary to use set theory. One could also use type theory, or just first-order logic, or ... But the actual content of the barber paradox has nothing to do with any of them. $\endgroup$
    – user21820
    Apr 23 '15 at 2:29
  • $\begingroup$ My point is that you can obtain more "solutions" using set theory. Using only FOL, you can only conclude that the barber does not exist -- not a very satisfying conclusion IMO. Using set theory, you could also conclude that no shaving relation can possibly satisfy the given requirements. That makes more sense. It points to an actual solution to the dilemma: change the requirements for the shaving relation. $\endgroup$ Apr 23 '15 at 5:16
  • $\begingroup$ @DanChristensen: "Such a barber does not exist" is a perfect satisfying conclusion. If it isn't, then "Such a shaving relation does not exist" is equally unsatisfying. You can't say that using set theory points to an actual solution without also saying that first-order logic does the same: change the requirements for the barber!! $\endgroup$
    – user21820
    Apr 23 '15 at 5:19
  • $\begingroup$ @DanChristensen: I've also seen your attempts at formal proofs and your other questions on Math SE, and I recommend that you work through at least an equivalent of an undergraduate mathematics course, so that you get a better overall picture. It is well-known that most mathematicians do not have the slightest idea what ZFC is, nor do they need it. There are reasons for that, but to appreciate those reasons you would need to learn enough of various branches of mathematics first. $\endgroup$
    – user21820
    Apr 23 '15 at 5:25

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