Mathematics 442C
Suggested solutions to exercise sheet 3
1. (a) Since T is a topology, it is closed under finite intersections and arbitrary
unions. Since Sb is the collection of unions of finite intersections of sets
in S and S ⊆ T , we have Sb ⊆ T .
b
The empty set is the empty union (take I = ∅ in the definition of S)
b
and X is the intersections of zero sets, so ∅, X ∈ S.
b say
If A, B ∈ S,
[
[
A=
Ai and B =
Bj
i∈I
j∈J
where I and J are index sets and Ai , Bj are finite intersections of sets
in S for every i ∈ I and j ∈ J, then Ai ∩ Bj is a finite intersection of
sets in S and so
[ [ [
b
A∩B =
Ai ∩
Bj =
(Ai ∩ Bj ) ∈ S.
i∈I
j∈J
i∈I, j∈J
Similarly, since a union of unions can be written as a single union, we
b So Sb is a topology.
see that an arbitrary union of sets in Sb is again in S.
(b) (i) =⇒ (ii): If S is a subbase for T , then T is the smallest topology
b
containing S. Since Sb is a topology containing S, this shows that T ⊆ S.
b
We have seen that S ⊆ T in (a). Hence we have equality.
(ii) =⇒ (iii): Suppose that Sb = T . If G ⊆ X is a subset which satisfies
the given condition, then G is a union of finite intersections
S of sets in S.
b
b
So G is in S = T . Conversely, if G ∈ S = T then G = i∈I Gi where
each Gi is a finite intersection of sets in S. Hence, if x ∈ G then x ∈ Gi
for some i ∈ I. Since Gi = S1 ∩ · · · ∩ Sn for some S1 , . . . , Sn ∈ S, this
establishes (iii).
(iii) =⇒ (i): We must show that T is the weakest topology on X
containing S. Suppose that W is a topology on X with S ⊆ W. Let
G ∈ T . If x ∈ G then, by (iii), there are S1 , . . . , Sn ∈ S such that,
writing S(x) = S1 ∩ · · · ∩ Sn , we have x ∈ S(x) ⊆ G. S
Since W is a
topology containing S, we have S(x) ∈ W. Clearly, G = x∈G S(x); so
G ∈ W. Hence T ⊆ W, so T is the weakest topology containing S.
(c) If f : Y → X is continuous and S ∈ S, then since S is an open subset
of X, f −1 (S) is an open subset of Y .
Conversely, suppose that f : Y → X and f −1 (S) is open for all S ∈ S.
Any open set G ⊆ X is of the form
G=
n
[\
Sij
i∈I j=1
where each Sij is in S. Hence
f
−1
(G) = f
−1
n
[ \
i∈I j=1
Sij =
n
[\
f −1 (Sij )
i∈I j=1
which is an open subset of Y , by hypothesis. Hence f is continuous.
2. (a) The topology T is defined by saying that a set G ⊆ X is open if and
only if, for every x ∈ G there is ε > 0 such that B(x, ε) ⊆ G. By
Exercise 1(b)(iii), {B(x, ε) : x ∈ X, ε > 0} is a subbase for T .
(b) Let W be the weak topology induced by {dx : x ∈ X}. Since d is
continuous with respect to T (or more accurately, with respect to the
product topology from T on X × X), the function dx is continuous with
respect to T for every x ∈ X. Hence W ⊆ T .
Let G ∈ T and let x ∈ G. There is εx > 0 such that
B(x, εx ) = {y ∈ X : d(x, y) < εx } ⊆ G.
Since B(x, εx ) = d−1
(−∞, εx ) and (−∞, εx ) is open in R, and dx : X →
x
R is continuous
S with respect to W, we have B(x, εx ) ∈ W for all x ∈ X.
Now G = x∈X B(x, εx ), so G ∈ W. Hence T ⊆ W. By the previous
paragraph, we have equality.
(c) Let V be the weak topology induced by C(X, T ). We have dx ∈ C(X, T )
for all x ∈ X (when we consider dx as a mapping X → C). Hence
W ⊆ V. Moreover, T = W by (a), and the inclusion V ⊆ T is part of
the definition of V. Hence T = W ⊆ V ⊆ T , so we have equality.
3. (a) By the Hahn-Banach theorem [FA 3.7], there is a functional α ∈ X ∗
z
with α(x) = kxk. Hence ϕ = kxk
α ∈ X ∗ and ϕ(x) = z.
(b) Let ϕ ∈ X ∗ with ϕ(x) = z. We have
Jx−1 (B(ϕ(x), ε)) = {ψ ∈ X ∗ : Jx (ψ) ∈ B(ϕ(x), ε)}
= {ψ ∈ X ∗ : |Jx (ψ) − ϕ(x)| < ε}
= {ψ ∈ X ∗ : |ψ(x) − ϕ(x)| < ε}
= U (ϕ; x; ε).
(c) Let W be the weak* topology on X ∗ . We must show that Sb = W.
First we show that Sb ⊆ W. If S = U (ϕ; x; ε) then S = Jx−1 (G) for some
open set G, by (b). Hence S ∈ W. So S ⊆ W. Now Sb is the smallest
topology containing S, so Sb ⊆ W.
b Let x ∈ X. If x = 0 then Jx = J0 = 0,
We now show that W ⊆ S.
which is continuous with respect to Sb [why?]. Suppose that x 6= 0.
By Exercise 2(a), the sets {B(z, ε) : z ∈ C, ε > 0} form a subbase
for the topology on C. Let z ∈ C, ε > 0 and let ϕ ∈ X ∗ be the
functional from (a) with ϕ(x) = z. Since Jx−1 (B(z, ε)) = U (ϕ; x; ε) is
in S by (b), Jx is continuous with respect to Sb by Exercise 1(c). Since
W is the weakest topology such that every Jx is continuous, this shows
b
that W ⊆ S.
(d) Let δ = ε − |ψ(x) − ϕ(x)| > 0. If τ ∈ U (ψ; x; δ) then
|τ (x) − ϕ(x)| ≤ |τ (x) − ψ(x)| + |ψ(x) − ϕ(x)| < δ + |ψ(x) − ϕ(x)| < ε,
so τ ∈ U (ϕ; x; ε). Hence U (ψ; x; δ) ⊆ U (ϕ; x; ε).
2
(e) By (c) and Exercise 1(b)(iii), a set G ⊆ X ∗ is open if and only if, for
every ϕ ∈ G there are sets S1 , . . . , Sn ∈ S such that ϕ ∈ S1 ∩ · · · ∩ Sn ⊆
G. We have ϕ ∈ Si = U (ϕi ; xi ; εi ) for some ϕi ∈ X ∗ , xi ∈ X and
εi > 0. By (d), there are δi > 0 such that U (ϕ; xi ; δi ) ⊆ Si . Let
ε = min(δ1 , . . . , δn ) > 0. Then U (ϕ; xi ; ε) ⊆ U (ϕ; xi ; δi ) ⊆ Si and so
U (ϕ; x1 , . . . , xn ; ε) =
n
\
i=1
U (ϕ; xi ; ε) ⊆
n
\
Si ⊆ G.
i=1
4. (a) We have ken − em k = 1 for all n, m ≥ 1 with n 6= m. Hence the only
Cauchy sequences in K are eventually constant, so K contains all of its
accumulation points and so K is closed.
P
(b) Let x = (xm )m≥1 ∈ ℓ1 . We have
m≥1 |xm | < ∞, so |xm | → 0 as
m → ∞. Hence en (x) = xn → 0 as n → ∞.
(c) We must show that for any weak* open set G with 0 ∈ G, there is
n0 ∈ N such that en ∈ G for all n ≥ n0 .
Let G be an open set with 0 ∈ G. By Exercise 3, there are x1 , . . . , xm ∈
ℓ1 and ε > 0 such that U (0; x1 , . . . , xm ; ε) ⊆ G. For 1 ≤ i ≤ m we have
en (xi ) → 0 as n → ∞, by (b). Hence there is ni ∈ N with |en (xi )| < ε
for n ≥ ni . Let n0 = max(n1 , . . . , nm ). For n ≥ n0 , we have |en (xi )| < ε
for 1 ≤ i ≤ m, so en ∈ U (0; x1 , . . . , xm ; ε). Hence en ∈ G for n ≥ n0 .
(d) We have shown that K is closed in the norm topology, but it is not
closed in the weak* topology since en converges in the weak* topology
to a point outside K. So the complement of K is open in the norm
topology, but not in the weak* topology. So the weak* topology is
strictly weaker than the norm topology.
3