Basic Analysis – Gently Done Topological Vector Spaces

30 Basic Analysis
Proof (version 2) Let (γα ) α∈A be any given net in X = �
i∈I Xi . We shall show
that (γα ) has a cluster point. For each i ∈ I, (γα (i)) is a net in the compact
space Xi and therefore has a cluster point zi , say, in Xi . However, the element
γ ∈ X given by γ(i) = zi need not be a cluster point of (γα ). (For example,
let I = {1,2}, X1 = X2 = [−1,1] with the usual topology and let (γn ) be the
sequence ((xn ,yn )) = � ((−1) n ,(−1) n+1 ) � in X1 × X2 . Then 1 is a cluster point
of both (xn ) and (yn ) but (1,1) is not a cluster point of the sequence ((xn ,yn )).)
The idea of the proof is to consider the set of partial cluster points, that is, cluster
points of the net (γα ) with respect to some subset of components. These are
naturally partially ordered, and an appeal to Zorn’s lemma assures the existence
of a maximal such element. One shows that this is truly a cluster point of (γα ) in
the usual sense.
For given γ ∈ X and J ⊆ I, J �= ∅, let γ ↾ J denote the element of the partial
cartesian product �
j∈J Xj whose jth component is given by γ ↾ J(j) = γ(j), for
j ∈ J. In other words, γ ↾ J is obtained from γ by simply ignoring the components
in each Xj for j /∈ J. Let g ∈ �
j∈J Xj . We shall say that g is a partial cluster
point of (γα ) if g is a cluster point of the net (γα ↾ J) α∈A in the topological space
�
j∈J Xj . Let P denote the collection of partial cluster points of (γα ). Now, for
any j ∈ I, X j is compact, by hypothesis. Hence, (γ α (j)) α∈A has a cluster point,
xj , say, in Xj . Set J = {j} and define g ∈ �
i∈{j} Xi = Xj by g(j) = xj . Then g
is a partial cluster point of (γα ), and therefore P is not empty.
The collection P is partially ordered by extension, that is, if g1 and g2 are
elements of P, where g1 ∈ �
j∈J1 Xj and g �
2 ∈ j∈J2 Xj , we say that g1 � g2 if
J1 ⊆ J2 and g1 (j) = g2 (j) for all j ∈ J1 . Let {gλ ∈ �
j∈Jλ Xj : λ ∈ Λ} be any
totally ordered family in P. Set J = �
λ∈Λ J λ
and define g ∈ �
j∈J X j
by setting
g(j) = g λ (j), j ∈ J, where λ is such that j ∈ J λ . Then g is well-defined because
{g λ : λ ∈ Λ} is totally ordered. It is clear that g � g λ for each λ ∈ Λ. We claim
that g is a partial cluster point of (γα ). Indeed, let G be any neighbourhood of g
in XJ = �
j∈J Xj . Then there is a finite set F in J and open sets Uj ∈ Xj , for
j ∈ F, such that g ∈ ∩j∈Fp −1
j (Uj ) ⊆ G. By definition of the partial order on P, it
follows that there is some λ ∈ Λ such that F ⊆ Jλ , and therefore g(j) = gλ (j), for
j ∈ F. Now, gλ belongs to P and so is a cluster point of the net (γα ↾ Jλ ) α∈A . It
follows that for any α ∈ A there is α ′ � α such that pj (γα ′) ∈ Uj for every j ∈ F.
Thus γα ′ ∈ G, and we deduce that g is a cluster point of (γα ↾ J) α . Hence g is a
partial cluster point of (γα ) and so belongs to P.
We have shown that any totally ordered family in P has an upper bound and
hence, by Zorn’s lemma, P possesses a maximal element, γ, say. We shall show
that the maximality of γ implies that it is, in fact, not just a partial cluster point
but a cluster point of the net (γα ). To see this, suppose that γ ∈ �
j∈J Xj , with
J ⊆ I, so that γ is a cluster point of (γα ↾ J) α∈A . We shall show that J = I. By
way of contradiction, suppose that J �= I and let k ∈ I \J. Since γ is a cluster
Department of Mathematics King’s College, London