Re: Question about universal set

In article <1141011209.811639.284730@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<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?

[.snip.]

As others have pointed out, you basically have it.

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.

When I first learned it, my teacher made a point of explaining exactly
how. Namely, one can use the Axiom of Separation to prove that:

For any set X, there exists a set A such that A is not an element of
X.

From this, it is clear that there is no universal set.

The proof is: given X, let

A = { x in X : x is not an element of x}.

By separation this is a set. If A is in X, then we have two cases:

A is an element of A; then A is not in A. Since [p->~p]->~p, this
means that A is not an element of A.

But if A is not an element of A, then A is an element of A. Again,
this means that A is an element of A.

The contradiction arises from assuming that A is in X; hence A is not
an element of X. Thus, there exists at least one set A which is not
an element of X.


You will see that we are in fact using "Russell's Paradox" together
with separation to deduce that there is no universal set.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

.

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