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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 27, 1996 8:02 AM

In article <4qie9f\$qtj@gap.cco.caltech.edu>,
ikastan@alumnae.caltech.edu (Ilias Kastanas) writes:

> [ ... ] S is the set of all sets... and that
> in itself is contradictory; e.g. Powerset(S) has greater cardinality
> than S... but is a subset (as well as an element) of S...

The proof of this that I am familiar with is an adaptation
of Russell's paradox itself, namely ...

proof: Assume there exists a bijection f:S ---> P(S) . Form the set
T = { x : x \notin f(x) } . As f is a bijection, there exists y \in S
so that f(y) = T . Now if y is in T then it is not in T, and vice versa.
This is a contradiction, so such a bijection f cannot exist.
QED

I am unhappy with this proof. Russells paradox itself is
resolved by declaring (for whatever reason you prefer) that the
purported definition does not give a valid set. The definition
of T above is extremely similar to Russells definition'. Doesn't
this raise another possibility in the proof above which may enable
us to escape the conclusion. Namely, could we not conclude that
perhaps our definition of T was, in classic Russell fashion,
not a set!

To see better what I mean, let us test the proof above against
a specific example.

Assume for a moment the existence of a universal set U. Then P(U)
would be a subset of U with the obvious embedding. We would also
have an injection of U into P(U) mapping element x \in U to
set {x} \in P(U) . The Bernstein-Shroder trick then would give us
a bijection f between U and P(U). In fact we can explictly write out
what f would be.

Definition: A singleton of index n is the set {{{...{x}...}}} where
the element x of U is not a set, and there are n pairs of nested
brackets. Note that U\P(U) consists of all singletons of index zero.

Define f : U ----> P(U) as follows.

f(A) = | A if A is not a singleton of any index
|
| {A} if A is a singleton of some index.

Clearly f looks like a bijection. The proof given above should thus
show in some fashion that f does not exist.

Application of proof: We form the set T = { x \in U : x \notin f(x) }
Note that T contains no singletons, and in fact consists precisely
of those sets which are not members of themselves - i.e. EXACTLY the
Russell definition. Clearly T is not a singleton, and so f(T) = T.
The proof then asks us to note that if T \in T then T \notin T and

So what can we conclude? When considering Russell's paradox in
isolation, the usual conclusion from the apparently contradictory
properties of T is that the definition of T does not give rise to
a valid set. In the context of this proof, we are asked to accept
that there is nothing wrong in principle with the definition of T,
and that the only conclusion we can draw from its apparently
contradictory properties is that f does not exist. It looks to me
like my fears are justified, and there is a hole in the proof.

The obvious way to patch the hole is to declare or prove somehow
that U itself is not a set. But you certainly cannot use the result
that |P(S)| > |S| for any set in order to do this, that would be most
circular.

> Circles, vicious and virtuous

I agree. I just wish there was another way off the merrygoround. I find
it very unsatisfying to have to speak of the class' of all groups, or
the class' of all sets. What is a class'? - we think of it as a set,
we treat it like a set, but we are not allowed to call it a set. And only
a specialist in set theory can explain the difference, and then only by
resort to a formal axiomatic model - yuk!

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