Basic Analysis – Gently Done Topological Vector Spaces

8 Basic Analysis
Theorem 1.27 Suppose that T 1 ,T 2 are topologies on a set X such that T 1 ⊆ T 2 ,
T 1 is a Hausdorff topology and such that (X,T 2 ) is compact. Then T 1 = T 2 .
Proof Let U ∈ T 2 and set F = X\U. Then F is T 2 -closed and hence T 2 -compact.
But any T 1 -open cover is also a T 2 -open cover, so it follows that F is T 1 -compact.
Since(X,T 1 )isHausdorff, itfollowsthatF isT 1 -closedandthereforeU isT 1 -open.
Thus T 2 ⊆ T 1 and the result follows.
Remark 1.28 Thus we see that if (X,T) is a compact Hausdorff topological space,
then T cannot be enlarged without spoiling compactness or reduced without spoiling
the Hausdorff property. This expresses a rigidity of compact Hausdorff spaces.
Let X be a given (non-empty) set, let (Y,S) be a topological space and let
f : X → Y be a given map. We wish to investigate topologies on X which make
f continuous. Now, if f is to be continuous, then f −1 (V) should be open in X
for all V open in Y. Let T = � T ′ , where the intersection is over all topologies T ′
on X which contain all the sets f −1 (V), for V ∈ S. (The discrete topology on X
is one such.) Then T is a topology on X, since any intersection of topologies is
also a topology. Moreover, T is evidently the weakest topology on X with respect
to which f is continuous. We can generalise this to an arbitrary collection of
mappings. Suppose that {(Y α ,S α ) : α ∈ I} is a collection of topological spaces,
indexed by I, and that F = {f α : X → Y α } is a family of maps from X into the
topological spaces (Y α ,S α ). Let T be the intersection of all those topologies on X
which contain all sets of the form f −1
α (V α ), for f α ∈ F and V α ∈ S α
. Then T is a
topology on X and it is the weakest topology on X with respect to which every
f α ∈ F is continuous.
Definition 1.29 The topology T, described above, is called the σ(X,F)-topology
on X.
Theorem 1.30 Suppose that each (Y α ,S α ) is Hausdorff and that F separates
points of X, i.e., for any a,b ∈ X with a �= b, there is some f α ∈ F such that
f α (a) �= f α (b). Then the σ(X,F)-topology is Hausdorff.
Proof Suppose that a,b ∈ X, with a �= b. By hypothesis, there is some α ∈ I such
that f α (a) �= f α (b). Since (Y α ,S α ) is Hausdorff, there exist elements U,V ∈ S α
such that f α (a) ∈ U, f α (b) ∈ V and U ∩V = ∅. But then f −1
α
are open with respect to the σ(X,F)-topology and a ∈ f −1
α
f−1 α (U)∩f−1 α (V) = ∅.
(U) and f−1
α (V)
(U), b ∈ f−1
α
(V) and
Department of Mathematics King’s College, London