Page 101 - DMTH401_REAL ANALYSIS

Unit 8: Completeness and Compactness

Another way to write the Cantor set is to note that each of the sets A can be written as a finite Notes
n
n
n
union of 2 closed intervals, each of which has a length of 1/3 , as follows:
A = [0, 1]
0
A = [0, 1/3]  [2/3, 1]
1
A = [0, 1/9][2/9, 3/9]  [6/9, 7/9]  [8/9, 1]
2
...
Now suppose that there is an open set U contained in C. Then there must be an open interval
(a, b) contained in C. Now pick an integer N such that

1/3 < b – a
N
Then the interval (a, b) can not be contained in the set A , because that set is comprised of
N
intervals of length 1/3 . But if that interval is not contained in A it can not be contained in C.
N
N
Hence, no open set can be contained in the Cantor set C.
8.5 Baire Category Theorem

Definition: S  M is

 Dense in M if S = M.

 Nowhere dense in M if M\ S is dense in M.
 Meagre in M if it is the union of a sequence of nowhere dense sets.

Proposition: S  M nowhere dense in M iff S = 0 /


Proof: S = 0 / = M\(M \S) so if RHS = 0 / then M\ S is dense in M so S is nowhere dense.
Conversely if S is nowhere dense in M then M\ S = M so RHS = 0 / .
Theorem: Baire Category
A complete metric space is not meagre in itself.
I.e. if S are the nowhere dense subsets of non-empty complete M then
n
¥
M \  Sn  0 /
=
n 1
Proof: IDEA: Find decreasing sequence of dense sets with non-empty intersection of their closures
by Cantor. Any point in this intersection cannot be in any nowhere dense set.

G := M\ S dense in M, open.
k k
Then G  0 / . Choose x  G and  > 0 s.t. B(x ,  )  G .
1 1 1 i i i 1
Continue inductively: Having defined x ,  use fact that G dense to find x  G  B
k–i k–i k k k
æ  k 1 ö  k 1
-
-
ç è x k 1 , 2 ø ÷ . Find 0 <  < 2 s.t. B(x ,  )  G .
-
k
k
k
k
 ¾¾¾® 0 and " k, B(x ,  k )  B(x ,  ).
k
k
k–i
k®¥
k–i
LOVELY PROFESSIONAL UNIVERSITY 95