Basic Analysis – Gently Done Topological Vector Spaces

10 Basic Analysis
Proposition 1.36 Let X be any non-empty set and let S be any collection of
subsets of X which covers X, i.e., for any x ∈ X, there is some A ∈ S such that
x ∈ A. Let B be the collection of intersections of finite families of elements of
S. Then the collection T of subsets of X consisting of ∅ together with arbitrary
unions of elements of members of B is a topology on X, and it is the weakest
topology on X containing the collection of sets S. Moreover, S is a sub-base for T,
and B is a base for T.
Proof Clearly, ∅ ∈ T and X ∈ T, and any union of elements of T is also a member
of T. It remains to show that any finite intersection of elements of T is also an
element of T. It is enough to show that if A,B ∈ T, then A∩B ∈ T. If A or B is
the empty set, there is nothing more to prove, so suppose that A �= ∅ and B �= ∅.
Then we have that A = �
αA �
α and B = βBβ for families of elements {Aα } and
{Bβ } belonging to B. Thus
A∩B = �
Aα ∩ �
Bβ = �
α
β
α,β
(A α ∩B β ).
Now, each A α is an intersection of a finite number of elements of S, and the same
is true of B β . It follows that the same is true of every A α ∩B β , and so we see that
A∩B ∈ T, which completes the proof that T is a topology on X.
It is clear that T contains S. Suppose that T ′ is any topology on X which also
contains thecollectionS. Then certainlyT ′ must also containB. But then T ′ must
contain arbitrary unions of families of subsets of B, that is, T ′ must contain T. It
follows that T is the weakest topology on X containing S. From the definitions, it
is clear that B is a base and that S is a sub-base for T.
Remark 1.37 We can describe the σ(X,F)-topology on X determined by the
family of maps {f α : α ∈ I}, discussed earlier, as the topology with sub-base given
by the collection {f −1
α (V) : α ∈ I, V ∈ S α }.
Example 1.38 Let (X,T) and (Y,S) betopologicalspaces and let
�
pX : X×Y
�
→ X
and
�
pY :
�
X × Y → Y be the projection maps defined by pX (x,y) = x and
−1
pY (x,y) = y for (x,y) ∈ X ×Y. For any U ⊆ X and V ⊆ Y, pX (U) = U ×Y
and p −1
Y (V) = X × V, so that p−1
X (U) ∩ p−1
Y (V) = U × V. Now, the collection
{U × V : U ∈ T, V ∈ S} is a base for the product topology on X × Y and it
(V) : U ∈ T, V ∈ S}. It
is clear it has as a sub-base the collection {p −1
X
(U), p−1
Y
follows that the product topology is the same as the σ(X ×Y,{p X ,p Y })-topology
and therefore the product topology is the weakest topology on X × Y such that
the projection maps are continuous.
We will adopt this later as our definition for the product topology on an arbitrary
cartesian product of topological spaces.
Department of Mathematics King’s College, London