Page 257 - DMTH503_TOPOLOGY

Unit 31: Baire Spaces

Proof: Given a positive integer N and given  > 0, define Notes
A () = {x|d(f (x), f (x))   for all n, m  N}.
N n m
Note that A () is closed in X. For the set of those x for which d (f (x), f (x))   is closed in X, by
N n m
continuity of f and f and A () is the intersection of these sets for all n, m  N.
n m N
For fixed , consider the sets A ()  A ()  …. The union of these sets is all of X. For, given
1 2
x  X, the fact that f (x )  f (x ) implies that the sequence f (x ) is a Cauchy sequence; hence
0 n 0 0 n 0
x  A () for some N.
0 N
Now let
() = IntA ( ).
N
N  
We shall prove two things:
(1) () is open and dense in X.
(2) The function f is continuous at each point of the set

 = (1) (1/2) (1/3) ….
Our theorem then follows from the fact that X is a Baire space. To show that () is dense in X,
it suffices to show that for any non-empty open set V of X, there is an N such that the set V Int
A () is non-empty. For this purpose, we note first that for each N, the set V A () is closed
N N
in V. Because V is a Baire space by the preceding lemma, at least one of these sets, say V A (),
M
must contain a non-empty open set W of V. Because V is open in X, the set W is open in X;
therefore, it is contained in Int A ().
M
Now we show that if x  , then f is continuous at x . Given  > 0, we shall find a neighborhood
0 0
W of x such that d (f (x), f (x )) <  for x  W.
0 0
First, choose K so that 1/K < /3. Since x  , we have x  (1/K) therefore, there is an N such
0 0
that x  Int A (1/K). Finally, continuity of the function f enables us to choose a neighborhood
0 N N
W of x , contained in A (1/K), such that
0 N
(*) d (f (x), f (x ))  /3 for x  W.
N N 0
The fact that W  A (1/K) implies that
N
(**) d (f (x), f (x))  1/K for n  N and x  W.
n N
Letting n  , we obtain the inequality
(***) d (f (x), f (x))  1/K < /3 for x  W.
N
In particular, since x  W, we have
0
d (f (x ), f (x )) < /3
0 N 0
Applying the triangle inequality (*), (**) and (***) gives us our desired result.
Theorem 3: If Y is a first category subset of a Baire space (X, T) then the interior of Y is empty.


Proof: As Y is first category, Y = Y , where each Y , n  N is nowhere dense.
n
n

n 1
  
Let  T be such that  Y. Then  Y  Y . So X\  (X \Y ), and each of the sets
n
n
n


n 1 n 1 n 1


X \Y is open and dense in (X, T). As (X, T) is Baire. (X \Y ) is dense in (X, T). So the closed set
n
n
n 1

n 1

X\ is dense in (X, T). This implies X\ = X . Hence = . This completes the proof.
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