Re: 'Navie set theory': why when S(x) is (x = x), the specified x's do not constitue a set?
- From: Dave Seaman <dseaman@xxxxxxxxxxxx>
- Date: Sat, 21 May 2005 21:35:15 +0000 (UTC)
On 21 May 2005 13:03:23 -0700, porky_pig_jr@xxxxxxxxxxx wrote:
> I"m reading 'Naive Set Theory' and the end of Section 3 there is a
> following sentence:
> [quote]
> In case S(x) is (x \notin x), or in case S(x) is (x = x), the specified
> x's do not constitute a set.
> [quote]
> The first one, (x \notin x) is Russell's Paradox, that one I
> understand. I"m not clear what's wrong with the second one, (x = x).
> Seems like this is tautology, always true. So if I have something like
> B = { x \in A : x = x }
> then B = A, since (x = x) is valid for all x's in A.
> I assume I'm missing something, but what?
There is a difference between
B = { x in A : x = x } (1)
and
B = { x : x = x } (2)
in that (1) always represents a set (for each choice of A), but (2) would
have to be the set of all sets, which does not exist in the kind of set
theory being discussed here.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
.
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