
Re: Russell's Paradox
Posted:
Jun 19, 1996 8:01 PM


In article <4q59nn$8kg@tornn1hb0.netcom.ca>, dchris@netcom.ca(Dan Christensen) writes: > In <zunger.834989322@rintintin.Colorado.EDU> > zunger@rintintin.Colorado.EDU (B. L. Z. Bub) writes: > >> ... in ZF it is possible to prove that Russell's construction is not >> a set, and so no contradiction arises therefrom.
I have never seen this paradox as much of a paradox myself. And you don't need monsters like ZF to knock it on the head.
It suffices to note that
S = `the set of all sets that are not members of themselves'
is not properly defined. We can consider a set to be a rule which sorts objects into two piles  member and nonmembers. The definition above is not a set, not because of some fancy ZF exclusion, but simply because the purported rule fails to adequately specify in which pile the object S belongs. The rule is incomplete! One can turn S into a valid set simply by specifying what it does to S. But if you do this, the paradox vanishes. The Russell paradox is no more a paradox than the so called Zeno paradox, and you don't need to retreat into formalism as ZF does to get rid of the problem.
Ian H

