Basic Analysis – Gently Done Topological Vector Spaces

24 Basic Analysis
and so � A∩B ∩B 0 ∈ C. But
�A∩B ∩B 0 = (X \(A∩B 0 ))∩(B ∩B 0 )
= ((X \A)∪(X \B 0 ))∩B ∩B 0
= {(X \A)∩B ∩B 0 }∪{(X \B 0 )∩B ∩B 0 }
� �� �
=∅
= (X \A)∩B ∩B 0
and so we see that (x α ) is frequently in (X \A)∩B ∩B 0 and hence is frequently
in (X \ A) ∩ B for any B ∈ C. Again, by the above argument, we deduce that
X \A ∈ C. This proves the claim and completes the proof of the lemma.
Theorem 2.27 Every net has a universal subnet.
Proof To prove the theorem, let (x α ) I be any net in X, and let C be a family
of subsets as given by the lemma. Then, in particular, the conditions of Proposition
2.20 hold, and we deduce that (x α ) I has a subnet (y β ) J such that (y β ) J is
eventually in each member of C. But, for any A ⊆ X, either A ∈ C or X \A ∈ C,
hence the subnet (y β ) J is either eventually in A or eventually in X \ A; that is,
(y β ) J is universal.
Theorem 2.28 A topological space is compact if and only if every universal net
converges.
Proof Suppose that (X,T) is a compact topological space and that (x α ) is a
universal net in X. Since X is compact, (x α ) has a convergent subnet, with limit
x ∈ X, say. But then x is a cluster point of the universal net (x α ) and therefore
the net (x α ) itself converges to x.
Conversely, suppose that every universal net in X converges. Let (x α ) be any
net in X. Then (x α ) has a subnet which is universal and must therefore converge.
In other words, we have argued that (x α ) has a convergent subnet and therefore
X is compact.
Corollary 2.29 A non-empty subset K of a topological space is compact if and
only if every universal net in K converges in K.
Proof The subset K of the topological space (X,T) is compact if and only if it is
compact with respect to the induced topology T K on K. The result now follows
by applying the theorem to (K,T K ).
Department of Mathematics King’s College, London