Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.1 6/24/83; site pucc-i
Path: utzoo!watmath!clyde!burl!ulysses!mhuxl!ihnp4!inuxc!pur-ee!CS-Mordred!Pucc-H:Pucc-I:ags
From: ags@pucc-i (Seaman)
Newsgroups: net.math
Subject: Proof that A < 2**A by diagonalization
Message-ID: <230@pucc-i>
Date: Tue, 6-Mar-84 09:42:32 EST
Article-I.D.: pucc-i.230
Posted: Tue Mar  6 09:42:32 1984
Date-Received: Wed, 7-Mar-84 06:29:48 EST
Organization: Purdue University Computing Center
Lines: 57

First, a clarification.  Someone posted an induction proof that n < 2**n
for all n.  The proof was correct, but it led to some rather vague
statements about numbers with infinitely many digits being greater
than 2**n for any n.  As I pointed out, "greater than" has no meaning
in this context.

What follows is a different proof that n < 2**n, which has the advantage of
extending to arbitrary sets (finite and infinite) as well as the natural
numbers (which are finite sets of a particular form).  This leads to a 
different proof that the real numbers are uncountable.

Quick review:  If A and B are sets, then A**B is the set of all mappings

			f: B --> A.

In particular, what does 2**A mean?  Since 2 is the set {0,1}, 2**A must
be the set of all mappings

			f: A --> {0,1}.

There is a natural correspondence between 2**A and the set of all subsets
of A:  With the mapping f we may associate the subset

			{x in A | f(x) = 1}.

Another way of saying this is that each subset is associated with its
CHARACTERISTIC MAPPING:  1 in the set and 0 outside.  Because of this
natural correspondence, the set of all subsets of A is often written
as 2**A, called the POWER SET OF A.

PROPOSITION.  For any set A, the power set 2**A is larger than A.

PROOF.  We must show that for every map f: A --> 2**A, there is an
element of 2**A which is not in the image of f.  Such an element is

			B = {x in A | not (x in f(x))}.

B is clearly a member of 2**A (i.e. a subset of A).  Suppose B = f(y) for
some y in A.  It follows by the definition of B that y is in B if and only
if y is not in B.  Therefore no such y can be found, and the mapping f is
not onto, Q.E.D.
			-------------------

Exercise:  The above argument is called DIAGONALIZATION.  Why?  How does
	   it relate to the familiar diagonalization proof that the reals
	   are uncountable?

Notice that we have not assumed anywhere that A is a finite set.  This is
the basis for a new proof that the real numbers are uncountable, to follow
in next (and probably last) article.
-- 

Dave Seaman
..!pur-ee!pucc-i:ags

"Against people who give vent to their loquacity 
by extraneous bombastic circumlocution."