Re: Russell's paradox and ZF(C)

On Sun, 30 Dec 2007 06:15:27 -0800 (PST), Zaljohar@xxxxxxxxx wrote:


It is often told that ZF(C) avoids Russell's paradox, because of its
limited comprehension schema (Separation). This is a false!

Look man:

unlimited comprehension ---> Russell's paradox derivable

limited comprehension of ZFC (Separation) ---> Russell's
paradox not derivable

Got it?


The correct statement is that Separation prevents the existence of the
set of all sets that are not in themselfs USING the negation Russell's
paradox [...], and that is not prevention of Russell's paradox due to
separation per se.

Nonsense.


What prevents the existence of Russell's paradox without using the
negation of the paradox itself is actually Regularity and Pairing.

(*sigh*)


ZF(C) without regularity doesn't really avoid Russell's paradox due
to its axiomatization per se,

It does.


[...] So separation is not really a solution to Russell's paradox.

It is. (Oh boy!)

So WHY does Separation block the derivation of "Russell's paradox"?
Let's see.

We will use the predicate "x !e x" (which allows to derive the paradox
if unrestricted comprehension is available).

Then Separation allows to deduce that for any set A there is a set B
such that for all x: x e B <-> x e A & x !e x. Now let's specialize that
to B:

B e B <-> B e A & B !e B.

From that we immediately get that B is not in A: B !e A. (Here's a short
proof: Assume B e A, then a contradiction follows. Hence B !e A by RAA.)

Since B is not in A it is also not in B. That's all. No Russell paradox.

We might derive the theorem:

Ax e A: x e B <-> x !e x.

So we see that Separation "restricts" the "domain" of the "considered"
sets to some set (A). IF we had unrestricted comprehension available we
would get here

Ax: x e B <-> x !e B.

(Immediately leading to Russell's paradox.)


F.

--

E-mail: info<at>simple-line<dot>de
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