🚀 go-pugleaf

	I am not an expert in set theory or the predicate
calculus, but I was leafing through a book on the subject and the
following informal "proof" occured to me.  Could any of you experts out
there tell me if this contains a major flaw -- and if so, what it
is?  Thanks.

	The proposition to be proven is: A set can never contain
itself as a member.
	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)

(Of course, g may be the empty set.)
	Eq. 1 can be expanded as follows:

		X = {g, X'}
		        /\
		       {g,X''}
		          /\
			 {g,X'''}

...and so on forever.  (The ' symbol is merely meant to
distinguish an X of one iteration from that of another.)  X'',
for example, is equivalent to {g,X'''}; X''' is equivalent to
{g,X''''}, etc etc.
	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''; therefore X' contains one more X than X''!  It
follows that X' != X'', for X' contains an X not contained in X''. 
By the same reasoning, X != X', because X contains one more X
than does X'.  Therefore, for any arbitrary set X, X cannot
contain itself as a member.

Steve