Re: Axiomatic ontology
Rob,
There is no contradiction there:
RF> I thought Russell's paradox might provide a hint where any
> complete conception could tie itself in knots. A set that
> both is and is not a member of itself. Maybe "the universe"
> is that way.
There is no contradiction whatever in the assumption that a set
could be a member of itself. The contradiction arises between
two other assumptions:
1. The axiom that for every monadic predicate P(x), there
exists a set of all the x's for which P is true.
2. The construction of a special set:
{x | ~memberOf(x,x) }
In English, the set of all x's such that x is not a member of x.
There are two straightforward methods for getting rid of the
contradiction:
1. The simplest is to drop axiom #1. That means that there may be
predicates such as P(x) for which there is no corresponding set.
2. The other is to define a method of constructing sets that makes
it impossible to form a set that contradicts axiom #1.
Both of these methods are used in various theories.
Finally, I have no idea what you mean by "that way". One thing
that is certain is that anything that exists can be described by
a list (possibly a very long list) of simple statements that do
not contain any negations. Since it is impossible to have a
contradiction without having at least one negation, everything
that exists must be describable by a list of consistent statements.
Therefore, nothing that exists can be self contradictory.
John