🚀 go-pugleaf

In article  <Oct05.201328.96457@yuma.ACNS.ColoState.EDU> Steven Arnold writes:
]
]	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)

A minor nit-pick: the above notation restricts X to be a two-element
set of which one element is the set g and the other set is X.  I
believe what you want is

  X = g union {X}

](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''!

This can only prove something about the particular representation
you use for the set.  In other words, assuming that your "therefore"
is correct, you have only proven that a set that contains itself
cannot be represented by an infinite expansion of the sort you give
above.

And your "therefore" itself is on shaky ground.  Rather than try to
approach it directly, let me propose a related falacy due to Zeno:

Suppose that you are at point A and want to move in a straight line to
point B, say 10 feet away.  Before you can get to B, you have to get
to C, halfway between A and B.  But before you can get to C, you have
to get to D, halfway between A and C.  You can see that this regress
is infinite, but that each segment is finite and takes a finite amount
of time to cross.  Moving an infinite number of finite distances
requires an infinite amount of time, so movement is impossible.
-- 
					David Gudeman
gudeman@cs.arizona.edu