Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Replies: 3   Last Post: Jun 21, 1996 2:39 PM

 Messages: [ Previous | Next ]
 hawthorn@waikato.ac.nz Posts: 14 Registered: 12/12/04
Posted: Jun 19, 1996 9:14 PM

A discussion on sci.math prompted the following. On reflection
I don't think this way of looking at the paradox is common. I
therefore post it here for comment.

> zunger@rintintin.Colorado.EDU (B. L. Z. Bub) wrote:
>> ... in ZF it is possible to prove that Russell's construction
>> is not a set, and so no contradiction arises therefrom.

I then replied:
> 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 - members 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.

Is this reasoning not sufficient by itself to rescue naive
set theory from the immediate perils of Russels paradox without
any need for formal machinery?

Note that if you adopt this view, then in order to properly define
a set, you really should specify whether or not it is to contain
itself at the time of definition. Hence you obtain a naive set theory
where there are two kinds of sets - those that contain themselves, and
those that do not. Yet there is no Russell paradox! To see this note
the following two sets

S = the set which contains itself and every other set
which fails to contain itself.

T = the set which does not contain itself and which contains
every other set which fails to contain itself.

Note that T is notin T, T is in S, S is in S, S is notin T .
No problems!

In order to try to construct a Russell paradox you must try to
sell an invalid definition which deliberately confuses the two
simply asking someone whether the set they are defining is S or T.
If the question is answerable, there is no paradox. If the question
the inadequacy of the definition, not in naive set theory itself.

Comments please! Note that I am not claiming that naive set theory is
perfect. I simply believe that Russells paradox is not in and of itself
a big problem for naive set theory.

Ian H

[Russell's "paradox" is definitely an error in at least one formal
axiom system for set theory that was proposed in Russell's time.
On the other hand, one can certainly envision valid axioms that
allow self-containment of things called sets, but are they are really