Baire Category Theorems
Definition. A Baire space is a topological space X that has the property that if O ! is { i }i=1 n a sequence of dense open sets, then is dense in X. !Oi i=1
The following theorems hold.
Theorem 1. If X is a complete metric space, then X is a Baire space.
Theorem 2. If X is a compact Hausdorff space, then X is a Baire space.
Theorem 3. If X is a Baire space and A is a dense G! in X, then A is also a Baire space.
Theorem 4. The set of irrationals with the usual metric forms a Baire space.
Definition. A completely regular space is an absolute G! if it is a G! in every space in which it is embedded as a dense subset.
Theorem 5. A completely regular space X is an absolute G! if and only if X is a G! in its Stone-Cech compactification.
Theorem 6. Every complete separable metric space X is an absolute G! .
Proof of Theorem 1. n Suppose that O ! is a sequence of dense open sets in X. We need to show that O { i }i=1 ! i i=1 n is dense in X. Let x X and 0 . We need to show that there is a z O B (x) . ! ! > !! i " # i=1 We will do this by finding a Cauchy sequence x ! converging to the required z. { n }n=1
" Let ! = and let x be any point in B (x) " O . Such a point exists because O is 1 3 1 !1 1 1 dense in X.
# Let 0 < ! " such that B (x ) " O # B (x) . Such a ! exists because O ! B (x) is 2 9 !2 1 1 !1 2 1 "1 open. Let x be any point in B (x ) O . The point x exists because O is dense. 2 !2 1 " 2 2 2
# Let x be a point in B (x ) " O and let 0 < ! " be such that B (x ) " B (x ) # O . 3 !2 2 3 3 33 !3 3 !2 2 3 The point x exists because O is dense. The number exists because B (x ) O is 3 3 ! 3 !2 2 " 3 open.
We continue inductively using the above argument. Suppose that a sequence of points x , x ,…, x has been chosen along with a sequence of positive numbers ! ,! ,…,! { 1 2 n } { 1 2 n } having the following properties.
(1) B (x ) O for1 i n , !i i " i ! !
# (2) 0 < ! " for1 ! i ! n , i 3i
(3) x B (x ) O for1 i n , and i+1 ! "i i # i+1 ! <
(4) B (x ) " B (x ) # O for1 ! i < n . !i+1 i+1 !i i i+1
# Then we let x be a point in B (x ) " O and choose 0 < ! " such n+1 !n n n+1 n+1 3n+1 that B (x ) " B (x ) # O . This increases the sequence of points and numbers by !n+1 n+1 !n n n+1 one more element each to x , x ,…, x and ! ,! ,…,! . The increased sequences { 1 2 n+1} { 1 2 n+1} have the above four properties.
By induction there are infinite sequences x , x ,… and ! ,! ,… such that { 1 2 } { 1 2 }
(1) B (x ) O for1 i , !i i " i ! < " # (2) 0 < ! " for1 ! i < " , i 3i (3) x B (x ) O for1 i , and i+1 ! "i i # i+1 ! < "
(4) B (x ) " B (x ) # O for1 ! i < " . !i+1 i+1 !i i i+1
One can easily show using (1), (2), (3), and (4) that the sequence x ! is Cauchy. { n }n=1
Since X is complete, this Cauchy sequence has a limit point z = lim xn . Since z is the n!" " limit, z !B (x ) . By property (1) above z !O for all n. Thus, z ! O . "n+1 n+1 n ! n n=1
Now d(x, x ) d(x, x ) d(x , x ) d(x , x ) n ! 1 + 1 2 +!+ n"1 n
" " " n 1 $ 1 " d(x, x ) n ! + 2 +!+ n = "# i < "# i = 3 3 3 i=1 3 i=1 3 2
! $ ! As a result, d(x,z) . So, z B (x) O . This completes the proof that O is < ! " # ! i ! i 2 i=1 i=1 dense in X.