Page 370 - DMTH401_REAL ANALYSIS

Real Analysis

Notes Next our aim is to determine the cardinality of ( ). We need some set-theoretic preparation.
d
Definition: An order  on a set X is a partial order if (i) x  x for every x  X, (ii) x  y and y  x
 x = y for every x, y  X, (iii) x  y and y  z  x  z for every x, y, z  X. We say (X, ) is a totally
ordered set if  is a partial order and any two elements of X are comparable. We say (X, ) is a
well-ordered set if (X, ) is totally ordered and any nonempty subset Y  X has a least element
in Y.

Examples:
(i) Let X be the collection of all nonempty subsets of . Define an order  on X as A  B iff the
minimum of A is less than or equal to the minimum of B. Then this is not a partial order
since the second condition fails.
(ii) If X is any nonempty set, then (X) with inclusion as order is partially ordered, but in
general not totally ordered.
(iii)  with the usual order is totally ordered, but not well-ordered since the subset (0,1) does
not contain a least element.

(iv)  with the usual order is well-ordered.
Well-ordering principle (equivalent to the axiom of choice): Any nonempty set admits a well-
ordering.

Now we describe the construction of some ordinal numbers. Start with an uncountable set X
such that card(X) = card(), and let  be a well-ordering on X. Let  denote the least element of
X. By adding one extra element to X if necessary, we may also assume that (X, ) has a largest
element, say . For each   X, let L = {  X :  < } be the left section of  in X. Let Y = {  X

: L is uncountable}. Then Y  Ø since   Y. So Y has a least element, say . Then L is
 
uncountable, but L is countable for every < . Here,  is called the first uncountable ordinal,

and each   L is called a countable ordinal number since each   L represents the type of a
 
countable well-ordered set through L .

Fact: If A  L is a nonempty countable set, then A has a least upper bound in L . [Proof: If B = 
   A
L , then B is countable and hence L \B  Ø. The least element of L \B is the least upper bound of
  
A]
If   L , then the least element of the nonempty set {  L :  < } will be denoted as  +1. Note
 
that there are no elements between  and  + 1 in L . On the other hand, given   L , there may
 
or may not exist   L such that  + 1 = . For example, if   L is the least upper bound of the
 
countable set {,  +1, + 2,…} (where recall that  is the least element of L ), then there is no

  X with  + 1 = . We say   L is a limit ordinal if there is no   L with  + 1 = .
 
card(( )) = card().
d
Proof: We will use transfinite induction (i.e., induction with respect to ordinal numbers) by using
L described above. Recall that we denoted the least element of L by the symbol . To start the
 
induction process, let  = {U   : U is open}. Let   L and assume that we have defined  for
d
  
every   L . If  is a limit ordinal, define  =  < A . If  =  + 1 for some   L , let  
    

d
d
= {A   :  \ A   }, and  = {A   : A is a countable union of members from    }.
d
   
This defines  for every   L . Finally, put  =  <  . From our construction, it is clear that
  
  ( ).
d
We verify that  is a -algebra on  . It suffices to check only the third property. So consider
d
A ,A ,…. Then there are  , ,…  L such that A   for every n  . By the Fact
1 2 1 2  n  n
mentioned above, the countable set { : n  } has a least upper bound, say  in L . Then A  
n  n 
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