Re: Proper class.Proper class ?
- From: "Ross A. Finlayson" <raf@xxxxxxxxxxxxxxx>
- Date: 10 Nov 2006 21:12:22 -0800
zuhair wrote:
Ross A. Finlayson wrote:
Jonathan Hoyle wrote:
still my question about what is A.B? when A is the set of all sets that
subsets their power sets, and B is the set of all sets that has
bijection to S is till now not solved.
First of all, the class of all sets is a proper class, not a set.
Secondly, what do you mean by the predicate "that subsets their
powersets"? And what is this S that you speak of in the second
sentence? You have undefined (or at least badly defined) terms here.
In any case, I believe the question you asked was answered definitely:
The intersection of two non-disjoint proper classes, where one is not a
subclass of the other, can either result in a set or another proper
class.
Jonathan Hoyle
Eastman Kodak
Classes are non-sets. Non-sets in a set theory are non-sense.
There can be only one proper class or none, in a set theory it would be
a set.
There is no class of classes in set theory with classes.
There is no universe in ZF, ZF is inconsistent.
Ross
ZFC is inconsistent because it doesn't have a universe, that's why I am
trying to develop a set theorum having a universe.
In reality the whole idea of axiomatization is silly really. Since e
is not defined and a set is not defined then the whole set theory
mounts to nothing. And those people seems to think that they can
produce things from nothing, which is philosophically absurd.
Zuhair
Actually, to produce things from nothing is the requirement.
That's because there is obviously not nothing. There is existence, E,
as I recently was illustrating, in a technical way. The point of
deaxiomatization and deriving truths from that process, reverse
mathematics of a sort, is to attempt to gain insight from the most
primitive truths and how they could be at all.
It resolves basically to what looks like a paradox, on reexamination
always true because its opposite is the same.
That can be stated in the terms of mathematical logic, conveniently.
Ross
.
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