Making Light:  Open Thread 88 ::: July 27, 2007, 11:53 AM

Xopher (#578, 585, 599):  In case sleeping on it didn't get you
anywhere, here's the problem:

The set of all sets of which the empty set is not a subset IS, in
fact, the empty set (by 2), but since it's a subset of itself (by 1),
it contains the empty set as a subset, which means that it isn't what
it was defined to be at the beginning of this sentence.

Let S be the set of all sets of which the empty set is not a subset.
That is, for all T, T is an element of S exactly when the empty set is
not a subset of T.  You've noticed that the empty set is a subset of
S.  This is not a contradiction.  The definition of S requires that
the empty set not be a subset of any element of S, and that is true
(because there aren't any elements of S).  The definition of S does
not require that the empty set not be a subset of S itself.  So
everything is right except the last clause asserting a contradiction
that does not exist.

It's easy to make the mistake of mistaking properties of a set with
properties of its elements.  I think Plato did the same thing when he
defined things called forms that captured the essence of some kind
of object.  The form of all dogs would be a trait that contained all
aspects of dogginess.  But if I recall correctly, he decided that the
form of all dogs would itself be some kind of dog.  This runs into all
kinds of paradoxes.

For instance, we could consider what kind of objects are ordinary
specific objects, as opposed to forms.  Then there would have to be a
form of all ordinary specific objects.  But if forms are instances of
the thing they describe then that form would have to be an ordinary
specific non-form, which is inconsistent with it being a form.

A final caution--defining "the set of all sets with property P" is not
guaranteed to work, even though it does in this case.  The way it's
done rigorously is to define the class of all sets with property P,
then prove (if you can) that the class is actually a set.  In this
case, most mathematicians wouldn't bother, because it's clear that S
is a set.