Page 98 - DMTH401_REAL ANALYSIS

Real Analysis

Notes As an application of the above result, we will see that perfect sets are closed sets that contain lots
of points:

8.3.1 Perfect Sets are Uncountable

Every non-empty perfect set must be uncountable.

Proof: If S is perfect, it consists of accumulation points, and therefore can not be finite. Therefore
it is either countable or uncountable. Suppose S was countable and could be written as
S = {x , x , x , ...}
1 2 3
The interval U = (x – 1, x + 1) is a neighbourhood of x . Since x must be an accumulation point
1 1 1 1 1
of S, there are infinitely many elements of S contained in U .
1
Take one of those elements, say x and take a neighbourhood U of x such that closure (U ) is
2 2 2 2
contained in U and x is not contained in closure (U ). Again, x is an accumulation point of S, so
1 1 2 2
that the neighbourhood U contains infinitely many elements of S.
2
Select an element, say x , and take a neighbourhood U of x such that closure (U ) is contained in
3 3 3 3
U but x and x are not contained in closure (U ).
2 1 2 3
Continue in that fashion: we can find sets U and points x such that:
n n
 closure ( ) 
n + 1 n
 x is not contained in  for all 0 < j < n
j n
 x is contained in 
n n
Now consider the set
 V = (closure ( )  S)
n
Then each set closure ( )S) is closed and bounded, hence compact. Also, by construction,
n
(closure ( )S) (closure ( )S). Therefore, by the above result, V is not empty. But which
n + 1 n
element of S should be contained in V? It can not be x , because x is not contained in closure (U ).
1 1 2
It can not be x because x is not in closure ( ), and so forth.
2 2 3
Hence, none of the elements {x , x , x , ...} can be contained in V. But V is non-empty, so that it
1 2 3
must contain an element not in this list. That means, however, that S is not countable.
8.4 Cantor Middle Third Set

Start with the unit interval

S = [0, 1]
0
Remove from that set the middle third and set
S = S \(1/3, 2/3)
1 0
Remove from that set the two middle thirds and set

S = S \{(1/9, 2/9) (7/9, 8/9) }
2 1
Continue in this fashion, where
S = S \{middle thirds of subintervals of S }
n+1 n n
Then the Cantor set C is defined as

C = S
n

92 LOVELY PROFESSIONAL UNIVERSITY