Basic Analysis – Gently Done Topological Vector Spaces

116 Basic Analysis
For any m,n > N, both x m and x n belong to B(x N ;r N ), and so
d(x m ,x n ) ≤ d(x m ,x N )+d(x n .x n )
< 1
N
+ 1
N .
Hence (x n ) is a Cauchy sequence in X and therefore there is some x ∈ X such
that x n → x. Since x n ∈ B(x n ;r n ) ⊆ B(x N ;r N ), for all n > N, it follows
that x ∈ B(x N ;r N ). But by construction, B(x N ;r N ) ⊆ B(x N−1 ;r N−1 ) and
B(x N−1 ;r N−1 ) ∩ A N−1 = ∅. Hence x /∈ A N−1 for any N. This is our required
contradiction and the result follows.
Remark 10.12 The theorem implies, in particular, that a complete metric space
cannot be given as a countable union of nowhere dense sets. In other words, if
a complete metric space is equal to a countable union of sets, then not all of
these can be nowhere dense; that is, at least one of them has a closure with nonempty
interior. Another corollary to the theorem is that if a metric space can be
expressed as a countable union of nowhere dense sets, then it is not complete.
Corollary 10.13 The intersection of any countable family of dense open sets in
a complete metric space is dense.
Proof Suppose that {Gn : n ∈ N} is a countable family of dense open subsets of
a complete metric space. For each n, Gn is open, so its complement, Gc n , is closed.
Also, a set is dense if and only if its complement has no interior. Therefore each
Gc n is closed and nowhere dense. By the theorem, Theorem 10.11, the complement
of the union �
nGcn is dense, that is,
�
Gn = ��
G c �c n
is dense, as required.
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Example 10.14 For each rational q ∈ Q, let F q = {q}. Then the complement of F q
in R is a dense open set, and the intersection of all these complements, as q runs
over Q, is a countable intersection equal to the set of irrationals and so is dense.
On the other hand, if each F q is considered as a subset of Q, then the intersection
of the complements (in Q) is empty—which is evidently not dense in Q.
Definition 10.15 Suppose that X and Y are topological vector spaces and that E
is a collection of linear mappings from X into Y. The collection E is said to be
equicontinuous iftoeveryneighbourhoodW of0inY thereissomeneighbourhood
V of 0 in X such that T(V) ⊆ W for all T ∈ E.
Department of Mathematics King’s College, London
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