Basic Analysis – Gently Done Topological Vector Spaces

36 Basic Analysis
Q x contains no rationals less than q. It follows that infQ x ≥ q > b, and therefore
f(x) > b. Thus we have shown that
{x ∈ X : f(x) > b} = �
X \Uq whichisanopenset inX. Itfollowsthatf −1 ((a,b)) = f −1 ((−∞,a))∩f −1 ((b,∞))
isan open set inX and weconclude that f iscontinuous and the proof iscomplete.
Remark 4.7 The values 0 and 1 in the theorem are not critical. Indeed, if f is
as in the statement of the theorem and if α < β is any pair of real numbers, put
g = α+(β −α)f. Then g is a continuous map from X into R with values in the
interval [α,β] such that g ↾ A = α and g ↾ B = β.
Note also that the theorem makes no claim as to the value of f outside the
sets A and B, other than it lies in [0,1]. It is quite possible for f to assume either
of the values 0 or 1 outside A or B. The next result sheds some light on this. A
subset of a topological space is said to be a G δ set if it is equal to a countable
intersection of open sets.
Theorem 4.8 Let A and B be non-empty closed disjoint subsets of a normal
space (X,T). There is a continuous map f : X → R with values in [0,1] such that
(i) A ⊆ {x ∈ X : f(x) = 0} and
(ii) B ⊆ {x ∈ X : f(x) = 1}
with equality in (i) if and only if A is a G δ set, and equality in (ii) if and only if
B is a G δ set.
Proof If f : X → R is continuous with values in [0,1], then the closed set
{x ∈ X : f(x) = 0} can be written as
q>b
{x ∈ X : f(x) = 0} = �
{x ∈ X : f(x) < 1
n }
n∈N
which is evidently a G δ set since each term on the right hand side is an open set
in X. Similarly,
{x ∈ X : f(x) = 1} = �
{x ∈ X : f(x) > 1− 1
n }
n∈N
is a G δ set. Thus equality in (i) or (ii) demands that A or B be a G δ set, respectively.
Department of Mathematics King’s College, London