Basic Analysis – Gently Done Topological Vector Spaces

26 Basic Analysis
Remark 3.2 Let G be a non-empty open set in X, equipped with the prod-
uct topology, and let γ ∈ G. Then, by definition of the topology, there exist
α1 ,...,α n ∈ I and open sets Uαi in X , 1 ≤ i ≤ n, such that
αi
γ ∈ p −1
α1 (Uα1 )∩···∩p−1 α1 (U ) ⊆ G. α1
HencethereareopensetsS α , α ∈ I, suchthatγ ∈ �
α S α ⊆ GandwhereS α = X α
except possibly for at most a finite number of values of α in I. This means that
G can differ from X in at most a finite number of components.
Now let us consider the second candidate for a topology on X. Let S be the
topology on X with base given by the sets of the form �
αVα , where Vα ∈ Tα for
α ∈ I. Thus, a non-empty set G in X belongs to S if and only if for any point x
in G there exist Vα ∈ Tα such that
x ∈ �
Vα ⊆ G.
α
Here there is no requirement that all but a finite number of the Vα are equal to
the whole space Xα .
Definition 3.3 The topology on the cartesian product �
αXα constructed in this
way is called the box-topology on X and denoted Tbox .
Evidently, in general, S is strictly finer than the product topology, T prod .
Proposition 3.4 A net (x λ ) converges in ( �
α X α ,T prod ) if and only if (p α (x λ ))
converges in (X α ,T α ) for each α ∈ I.
Proof This is a direct application of Theorem 2.14.
Example 3.5 Let I = N, let Xk be the open interval (−2,2) for each k ∈ N, and
let Tk be the usual (Euclidean) topology on Xk . Let xn ∈ �
kXk be the element
xn = ( 1 1 1
n , n , n ,...), that is, pk (x 1
n ) = n for all k ∈ I = N. Clearly pk (xn ) → 0
as n → ∞, for each k and so the sequence (xn ) converges to z in ( �
kXk ,Tprod )
where z is given by pk (z) = 0 for all k.
However, (xn ) doesnot converge toz withrespect tothe box-topology. Indeed,
to see this, let G = �
kAk where Ak is the open set A 1
k = (−1 k , k ) ∈ Tk . Then G
is open with respect to the box-topology and is a neighbourhood of z but x n /∈ G
for any n ∈ N. It follows that, in fact, (x n ) does not converge at all with respect
to the box-topology—if it did, then the limit would have to be the same as that
for the product topology, namely z.
Department of Mathematics King’s College, London