Re: Question about universal set

On 26 Feb 2006 19:33:29 -0800, agapito6314@xxxxxxx wrote:
I've had some trouble understanding exactly how the ZF Axiom of
Separation overcomes the problem of contradiction first pointed out by
Russell. Is the following reasoning correct?
The axiom states:

Ap Az Ey Ax [x e y <=> (x e z & p(x)]

Inserting Russell's predicate ~ (x e x) for p(x) one obtains (after
simplification)

Az Ey [~ (y e y) & ~ (y e z)]

So the axiom is contradicted if either

1.- Ay (y e y) or
2.- Ez Ay (y e z) or, in other words z is the universal set.

Now 1.- apparently can be shown to be false using other ZF axioms. Is
the non-existence of the universal "set", then, proved indirectly by
its negation of the Axiom of Separation? No text I have come across
makes this clear, they all simply state that it overcomes Russell's
famous objection.

Others have answered this, but I would like to point out that there is
another way to prove the nonexistence of the universal set that doesn't
use the axiom of separation. The idea is based on Cantor's Paradox.

Suppose a universal set U exists. Let P(U) be its power set. By
Cantor's theorem, there must exist an x in P(U) that is not in U,
contrary to the assumption that U is a universal set.

When people say that the axiom of separation overcomes Cantor's paradox,
I suspect what they really mean is that we can't use just a predicate
alone to define a set. That's the assumption that leads to Russell's
Paradox. In place of that flawed concept, sometimes called "unbounded
comprehension", we have instead "bounded comprehension", which is another
name for the axiom of separation. That means, you need both a predicate
AND a bounding set in order to establish the existence of another set
using that axiom. Thus, it is not the addition of an axiom (bounded
comprehension) that resolves the paradox, but the subtraction of an axiom
(unbounded comprehension).


--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>
.

Relevant Pages

  • Re: GCH vs. Axiom of Choice.
    ... |The axiom of separation says that for every existing sentence and set, ... the axiom scheme of separation doesn't talk about ... quantifier in front. ...
    (sci.math)
  • Re: Axiom of Pairing, Scheme of Replacement from others
    ... separation follows from the axiom of replacement together with the axiom ... empty set is required. ...
    (sci.math)
  • Re: GCH vs. Axiom of Choice.
    ... |separation is vacuous or it has content. ... seem related to the axiom of separation? ... |insisting that the power set contains all the subsets and that all ... In the language of ZFC, ...
    (sci.math)
  • Re: GCH vs. Axiom of Choice.
    ... > |separation is vacuous or it has content. ... > seem related to the axiom of separation? ... So since in ZFC there are ... that the power set "contains them all", ...
    (sci.math)
  • Re: Why? [was Re: Cantor`s powerset theorem is false?]
    ... prove, without the axiom of regularity, Ax~x=, as we've discussed ... we use the axiom of separation. ... But that is not a definition of the successor ... What is PRECISE is the mathematics. ...
    (sci.logic)