Page 256 - DMTH503_TOPOLOGY

Topology

Notes Then <f > converges pointwise to g :    s.t.
n
 0 if 0  x  1
g(x) = 
 1 if x  1
Evidently g is not continuous.

31.1.3 Baire Category Theorem

Theorem 1: Every complete metric space is of second category.

Proof: Let (X, d) be a complete metric space.
To prove that X is of second category.
Suppose not. Then X is not of second category so that X is of first category. By def., X is expressible
as a countable union of nowhere dense sets arranged in a sequence <A >. Since A is non-dense
n 1
1
and soa closed sphere K of radius r < s.t. K A = .
1 1 1 1
2
Let the open sphere with same centre and radius as r be denoted by S . In S , we can find a closed
1 1 1
2
 1 
sphere K of radius r    s.t.
2 2  2 
K A =  and so K A = 
1 2 2 1
Continuing like this we construct a nested sequence <K > of closed spheres having the following
n
properties:
(i) For each positive integer n, K does not intersect
n
A , A , …, A .
1 2 n
1
(ii) The radius of K tends to zero as n. For 0 as n.
n n
2
Since (X, d) is complete and so by Cantor’s intersection theorem, K contains a single point x .
n o
n

 x  K  x K  n
o n o n
=
n 1
 x A  n (according to (i))
o n

 x  A = X
o n
=
n 1
 x  X. A contradiction
o
For X is universal set.
Hence X is not of first category. A contradiction. Hence the required result follows.
Remarks: The theorem 1 can also be expressed in the following ways:
1. If <A > is a sequence of nowhere dense sets in a complete metric space (X, d), thena point
n
in X, which is not in A ’s.
n
2. If a complete metric space is the union of a sequence of its subsets, then the closure of at
least one set in the sequence must have non-empty interior.
Theorem 2: Let X be a space; let (Y, d) be a metric space. Let f : X  Y be a sequence of continuous
n
functions such that f (x)  f (x) for all x  X, where f : X  Y. If X is a Baire space, the set of points
n
at which f is continuous is dense in X.

250 LOVELY PROFESSIONAL UNIVERSITY