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Replies: 15   Last Post: Jul 3, 1996 8:49 PM

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 hawthorn@waikato.ac.nz Posts: 14 Registered: 12/12/04
Posted: Jun 19, 1996 8:01 PM

In article <4q59nn\$8kg@tor-nn1-hb0.netcom.ca>, dchris@netcom.ca(Dan Christensen) writes:
> 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 non-members.
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

Date Subject Author
6/16/96 Dan Christensen
6/17/96 B. L. Z. Bub
6/18/96 Dan Christensen
6/19/96 hawthorn@waikato.ac.nz
6/19/96 hawthorn@waikato.ac.nz
6/22/96 Toby Bartels
7/3/96 Brot675
6/21/96 Toby Bartels
6/22/96 Toby Bartels
6/20/96 Michael Abbott
6/21/96 Mark Hopkins
6/22/96 Michael Hardy
6/22/96 Ilias Kastanas
6/27/96 hawthorn@waikato.ac.nz
6/28/96 ilias kastanas 08-14-90
6/30/96 Toby Bartels