🚀 go-pugleaf

sa114984@longs.lance.colostate.edu (Steven Arnold) writes:
>	Suppose X is an arbitrary set which contains itself as a
>member.  All other members within the set can be symbolized as
>'g'; therefore,
>
>		X = {g,X}		(1)

As somebody else did, I'll take it Steven meant "X = g U {X}".

>	Eq. 1 can be expanded as follows:
>
>		X = {g, X'}
>		        /\
>		       {g,X''}
>		          /\
>			 {g,X'''}
>...
>	The number of X's below X'' is infinite; nonetheless, it
>is certainly true that X' contains X'' as well as all the X's
>contained by X'';

What does "contains" mean?  If "X contains Y" means "Y is an element
of X", then I'll simply point out that "is an element of" is NOT
transitive (consider the sets {a}, {{a}}, and {{{a}}}).  If on the
other hand he means "there is a finite chain of 'is-an-element-of'
relationships from Y to X" (a nonstandard definition) we can go on.

>therefore X' contains one more X than X''!

This comment only makes sense for "sets with multiplicity" (i.e. the
set {a,b} is different from the set {a,b,b,b,b}).  If this nonstandard
definition is what he means, we can go on.

The statement "X' contains one more X than X''!" is no contradiction,
since both numbers are infinite (as Steven had already pointed out).
So let's go on.

>It
>follows that X' != X'', for X' contains an X not contained in X''. 

But I thought all these X's were assumed to be identical.  To prove
X'!=X by this method, you'd need to be able to point out one object
that's in one but not in the other.

Yes, there are several flaws in the argument.  And each of them is
interesting in a way; they point out several of the ways standard set
theory violates the intuitions one develops by dealing with finite
(not to mention well-founded!) sets.

-- 
					Stephen Bloch
				sbloch@silver.cs.umanitoba.ca