Download Functional Analysis Exercise Class

Functional Analysis Exercise Class
Week 19 October – 23 October:
Deadline to hand in the homeworks: your exercise class on week 26 October – 30 October
Exercises with solutions
(1) Show that any finite set in a topological space is compact.
Solution: Let M be a finite set in a topological space X , and let {Ui }i∈I be an
open cover. Then for every x ∈ M we can choose an ix such that x ∈ Uix , and
hence {Uix }x∈M is a finite subcover of M .
(2) (Discrete topology) Let X be a set, and
(
1, x 6= y,
d(x, y) :=
0, x = y.
a) Show that d is a metric on X , and it generates the discrete topology τd =
P(X ) := {M ⊂ X }.
b) What are the open/closed/compact sets in this topology?
c) Given a set M ⊂ X , what is its closure/interior/boundary?
d) Describe the open balls Bε (x) := {y ∈ X : d(x, y) < ε} and the closed balls
B̄ε (x) := {y ∈ X : d(x, y) ≤ ε} for every ε ≥ 0. Is it true that
Bε (x) = B̄ε (x)
for all
ε > 0,
where Bε (x) denotes the closure of Bε (x)?
Solution:
a) It is clear that d(x, y) = d(y, x), x, y ∈ X , and d(x, y) = 0 ⇐⇒ x = y. The
inequality d(x, z) ≤ d(x, y) + d(y, z) is trivial when x 6= y or x 6= z; on the
other hand, if x = y and y = z then x = z and thus d(x, z) = 0, in which
case the inequality holds again.
1
b) By definition, all subsets of X are open, and hence all subsets are closed
as well. The compact sets are exactly the finite sets. Indeed, if M is finite
then it is compact by Exercise (1). On the other hand, if M has infinite
cardinality then consider the open cover {{x}}x∈M . Obviously, this open
cover has no finite subcover.
c) Since every set is open and closed, every set is equal to its closure and its
interior. As a consequence, the boundary of every set is empty.
d)


ε = 0,
∅,
Bε (x) = {x}, ε ≤ 1,


X,
ε > 1,
and
(
{x}, ε < 1,
B̄ε (x) =
X,
ε ≥ 1.
6 X = B̄1 (x) unless |X | = 1.
In particular, B1 (x) = {x} =
(3) (Indiscrete/trivial topology) Recall that the indiscrete topology on a set X is
τid = {∅, X }.
a) Is this topology metrizable? Is it Hausdorff?
b) What are the open/closed/compact sets in this topology?
c) Given a set M ⊆ X , what is its closure/interior/boundary?
Solution:
a) If |X | > 1 then any way we choose two different points x, y ∈ X , all
neighbourhoods of x will contain y and vice versa, since their only neighbourhood is the whole space. Hence, X is not Hausdorff, and therefore it
is not metrizable, either.
b) By definition, the only open sets are ∅ and X . The closed sets are their
complements, i.e., X and ∅. Every set is compact in this topology. Indeed,
the empty set is compact in any topological space, and if M ⊆ X is nonempty then its only open cover is {X }, which is already finite.
c) For any non-empty set M ⊆ X , its closure is X . The interior of a set
M is the empty set if M 6= X , and the interior of X is X. Hence, the
boundary of the empty set and the boundary of X are both the empty set;
the boundary of any other set is X .
(4) (Half-line topology) Let τhl := {(a, +∞) : a ∈ R ∪ {−∞, +∞}}.
a) Show that τhl is a topology on R.
2
b) Is this topology metrizable? Is it Hausdorff?
c) What are the open/closed/compact sets in this topology?
d) Given a set M ⊂ X , what is its closure/interior/boundary?
Solution:
a) We have ∅ = (+∞, +∞), and R = (−∞, +∞). Given ai , i ∈ I, where I
is any index set, we have ∪i∈I (ai , +∞) = (inf i ai , +∞) ∈ τhl . If I is finite
then ∩i∈I (ai , +∞) = (maxi ai , +∞) ∈ τhl . Hence, τhl contains the empty
set, the whole space, it is closed under taking the arbitrary union and the
finite intersection of its elements, and therefore it is a topology.
b) Given x, y ∈ R, x < y, we see that all neighbourhoods of x contain y, and
hence they can not be separated by open sets. Thus, R with the half-line
topology is not Hausdorff, and therefore not metrizable, either.
c) The open sets are given by the definition of the topology. The closed sets
are the empty set, R, and (−∞, a] for any a ∈ R. Note that any open cover
of M is of the form {(ai , +∞)}i∈I . If inf M ∈ M then there must exist an
i such that ai < inf M , and hence {(ai , +∞)} is a finite subcover (with a
single open set). On the other hand, if inf M ∈
/ M then we can choose a
strictly decreasing sequence an ∈ R, n ∈ N, that converges to inf M , and
define an open cover of M by {(an , +∞)}n∈N ; this open cover has no finite
subcover. Hence, the compact sets are exactly those that contain their
infimum.
d) By definition, the interior of M is (amin , +∞), where amin := inf{b ∈ R :
(b, +∞) ⊆ M }. Note that in particular, if M doesn’t contain an interval of
the form (b, +∞) with b ∈ R then amin = inf ∅ = +∞, and (+∞, +∞) = ∅
is the interior of M . The closure of M is (−∞, sup M ] if sup M is finite,
otherwise it is R. The boundary of M is (−∞, amin ] if amin 6= +∞, otherwise
it is equal to the closure of M .
1 Definition. Let (X , τ ) be a topological space, and let xn ∈ X , n ∈ N, be a sequence
in it. We say that x ∈ X is a limit of the sequence, or equivalently, that the sequence
converges to x, if for every neighbourhood U of x, there exists an nU ∈ N such that
for all n ≥ nU , xn ∈ U .
We say that an x ∈ X is a cluster point of the sequence, if for every neighbourhood
U of x, and every n ∈ N, there exists an m ∈ N such that m ≥ n and xm ∈ U .
2 Remark. Cluster points are also called accumulation points, condensation points
or limit points. Be careful, the latter terminology is a bit confusing, and easy to mix
with the notion of a limit. From the definition it is clear that every limit of a sequence
is also a limit point, but not every limit point has to be a limit.
3
(5) Show that every sequence in a Hausdorff space has at most one limit. How
about limit points? Construct a sequence in [0, 1] (equipped with the Euclidean
topology) that has continuum many limit points.
Solution: Assume that x 6= y are both limits of a sequence (xn )n∈N . By the
Hausdorff property, there exist disjoint open sets Ux 3 x and Uy 3 y. Since
x = limn xn , there exists an N1 ∈ N such that for all n ≥ N1 , xn ∈ Ux . Likewise,
y = limn xn implies the existence of an N2 ∈ N such that for all n ≥ N2 , xn ∈ Uy .
Thus, for all n ≥ max{N1 , N2 }, xn ∈ Ux ∩ Uy = ∅, a contradiction.
Let {qn }n∈N be an arbitrary ordering of the rational numbers in [0, 1]. Define a
sequence by x1 := q1 ; x2 := q1 , x3 := q2 ; x4 := q1 , x5 := q2 , x6 := q3 , etc. For
any x ∈ [0, 1], we can choose a sequence of rational numbers (rm )m∈N such that
limm rm = x. Then for any ε > 0 and n ∈ N, there exists a k ∈ N such that
rk ∈ (x − ε, x + ε) and an m ∈ N such that m > n and xm = rk ∈ (x − ε, x + ε).
Hence, all the points in [0, 1] are limit points of the sequence (xn )n∈N .
(6) Describe the set of limits of a sequence in a set equipped with the a) discrete b)
indiscrete topology, and c) in the half-line topology on R.
Solution:
Let X be equipped with the discrete topology. Since {x} is a neighbourhood of
any x ∈ X , the only sequences that have a limit are those that are constant after
some index. If X is equipped with the indiscrete topology then every point of X
is a limit of any sequence in X . Finally, in the half-line topology, a point x ∈ R
is a limit of a sequence (xn )n∈N if and only if for all ε > 0, {n : xn < x − ε} is
finite. Equivalently, the set of limits of a sequence (xn )n∈N is (−∞, lim inf n xn ].
(7) Let X be a topological space and V be a normed space over the field K = R or
K = C. Define
kf k∞ := sup kf (x)k ,
f ∈ V X,
and
x∈X
Cb (X , V ) := {f : X → V, f continuous and kf k∞ < +∞}.
a) Show that if fn ∈ Cb (X , V ), n ∈ N, and f ∈ V X is such that
lim kfn − f k∞ = 0 then f ∈ Cb (X , V ).
n→+∞
b) Show that if V is a Banach space then Cb (X , V ) is a Banach space, i.e., it
is a complete metric space with respect to the metric d(f, g) := kf − gk∞ .
Solution:
a) First, we show that f is continuous, which is equivalent to showing that it
is continuous at every x ∈ X . To this end, fix an x ∈ X and an ε > 0;
4
we have to show that there exists a neighbourhood U of x such that for
all y ∈ U , kf (y) − f (x)k < ε. Since limn→+∞ kfn − f k∞ = 0, we have an
n ∈ N such that kfn − f k∞ < ε/3. Since fn is continuous, there exists
a neighbourhood U of x such that for all y ∈ U , kfn (y) − fn (x)k < ε/3.
Using now the triangle inequality, we get
kf (y) − f (x)k = kf (y) − fn (y) + fn (y) − fn (x) + fn (x) − f (x)k
≤ kf (y) − fn (y)k + kfn (y) − fn (x)k + kfn (x) − f (x)k
≤ 2 kfn − f k∞ + kfn (y) − fn (x)k < ε.
Next, we show that f is bounded. This is because for every n ∈ N,
kf (x)k = kf (x) − fn (x) + fn (x)k ≤ kf (x) − fn (x)k + kfn (x)k
≤ kf − fn k∞ + kfn k∞ ,
x ∈ X.
Hence, f is continuous and bounded, and thus an element of Cb (X , V ).
b) Let fn ∈ Cb (X , V ), n ∈ N, be a Cauchy sequence, i.e., for every ε > 0 there
exists an Nε ∈ N such that for all n, m ≥ Nε , kfn − fm k∞ < ε. Since for
any x ∈ X , kfn (x) − fm (x)k ≤ kfn − fm k∞ , we see that (fn (x))n∈N is a
Cauchy sequence in V . Since V is complete, this sequence is convergent,
and we denote its limit by f (x) := limn→+∞ fn (x). We have to show that
kfn − f k∞ converges to 0. To this end, let ε > 0, and Nε be such that
kfn − fm k∞ < ε for all n, m ≥ Nε . Then, for any n ≥ Nε , and any x ∈ X ,
we have
kfn (x) − f (x)k = lim kfn (x) − fm (x)k < ε,
m→+∞
and hence
kfn − f k∞ = sup kfn (x) − f (x)k < ε.
x∈X
This shows that limn→+∞ kfn − f k∞ = 0, and hence, by the previous point,
f ∈ Cb (X , V ).
(8) According to the definition of a topology, the infinite union of open sets and
the finite intersection of open sets are open. Give an example when the infinite
intersection of open sets is not open.
Solution: For n ∈ N, let Un = (− n1 , n1 ) be an open set in R with a usual metric.
Then ∩n∈N Un = {0}, a single point. Since no ball with positive radius around 0
is contained in {0}, it is not an open set.
5
(9) In the lecture it was proved that a compact set K in a Hausdorff space (X , τ ) is
closed. Show that this result need not be true if the space X is not Hausdorff.
Solution 1: As we have seen in Exercise (3), and is indeed trivial to verify, any
set in an indiscrete topological space is compact; however, if the space has at
least two points then no proper non-empty subset of it is closed.
Solution 2: Consider the Sierpinski two-point topological space - a space that
consists of two points, X := {a, b}, and the topology is τ := {∅, {a}, {a, b}}.
Then the set {a} is compact since it is finite (see Exercise (1)). But it is not
closed, since it is not the complement of an open set.
(10) Show that a set in a topological space is closed if and only if it contains its
boundary, and open if and only if it is disjoint from its boundary.
Solution:
By the definition of the boundary, ∂A = A \ int(A) and hence A = int(A) ∪ ∂A
with int(A) ∩ ∂A = ∅. Note that A is closed if and only if A = A, and since any
set contains its interior by definition, we see that indeed A = A if and only if
∂A ⊆ A. Similarly, A is open if and only if A = int(A), which is equivalent to
A being disjoint from its bounday.
6