Basic Analysis – Gently Done Topological Vector Spaces

10: Fréchet Spaces 115
Definition 10.9 A subset of a metric space is said to be nowhere dense if its closure
has empty interior.
Example 10.10 Consider the metric space R with the usual metric, and let S be
the set S = {1, 1 1 1
1 1
2 , 3 , 4 ,...}. Then S has closure S = {0,1, 2 , 3 ,...} which has
empty interior.
We shall denote the open ball of radius r around the point a in a metric
space by B(a;r). The statement that a set S is nowhere dense is equivalent to
the statement that the closure, S of S, contains no open ball B(a;r) of positive
radius.
The next result, the Baire Category theorem, tells us that countable unions of
nowhere dense sets cannot amount to much.
Theorem 10.11 (Baire Category theorem) The complement of any countable
union of nowhere dense subsets of a complete metric space X is dense in X.
Proof Suppose that A n , n ∈ N, is a countable collection of nowhere dense sets in
the complete metric space X. Set A 0 = X\ �
n∈N A n . We wish to show that A 0 is
dense in X. Now, X\ �
n∈N A n
⊆ X\�
n∈N A n
, and a set is nowhere dense if and
only its closure is. Hence, by taking closures ifnecessary, we may assume that each
A n , n = 1,2,... is closed. Suppose then, by way of contradiction, that A 0 is not
dense in X. Then X \A 0 �= ∅. Now, X \A 0 is open, and non-empty, so there is
x 0 ∈ X\A 0 and r 0 > 0 such that B(x 0 ;r 0 ) ⊆ X\A 0 , that is, B(x 0 ;r 0 )∩A 0 = ∅.
The idea of the proof is to construct a sequence of points in X with a limit which
does not lie in any of the sets A 0 ,A 1 ,.... This will lead to a contradiction, since
X is the union of the A n ’s.
We start by noticing that since A 1 is nowhere dense, the open ball B(x 0 ;r 0 )
is not contained in A 1 . This means that there is some point x 1 ∈ B(x 0 ;r 0 )\A 1 .
Furthermore, since B(x 0 ;r 0 )\A 1 is open, there is 0 < r 1 < 1 such that B(x 1 ;r 1 ) ⊆
B(x 0 ;r 0 ) and also B(x 1 ;r 1 )∩A 1 = ∅.
Now, since A 2 is nowhere dense, the open ball B(x 1 ;r 1 ) is not contained in
A2 . Thus, there is some x2 ∈ B(x1 ;r1 )\A 2 . Since B(x1 ;r1 )\A 2 is open, there is
0 < r2 < 1
2 such that B(x2 ;r2 ) ⊆ B(x1 ;r1 ) and also B(x2 ;r2 )∩A 2 = ∅.
Similarly, we argue that there is some point x3 and 0 < r3 < 1
3 such that
B(x3 ;r3 ) ⊆ B(x2 ;r2 ) and also B(x3 ;r3 )∩A 3 = ∅.
Continuing in this way, we obtain a sequence x0 ,x1 ,x2 ,... in X and positive
real numbers r0 ,r1 ,r2 ,... satisfying 0 < rn < 1
n , for n ∈ N, such that
and B(x n ;r n )∩A n = ∅.
B(x n ;r n ) ⊆ B(x n−1 ;r n−1 )