Exercises for the lecture

MA3008 – Algebraic Topology

Solutions to Sheet 2 Spring Semester 2017

Exercise 1 (Properties of closed sets).

Let (X, TX ) be a topological space. Show the following facts about closed sets in X:
a) The empty set ∅ and the whole space X are closed.

b) If A ⊂ X and B ⊂ X are closed, so is their union A ∪ B.

T
c) Let I be a set and let (Ai )i∈I be a family of closed subsets of X. Then i∈I Ai is closed
as well.
Hint: Use De Morgan’s laws.

Solution to 1: De Morgan’s laws refer to the following facts about taking complements:
Let X, I be sets and let Ui ⊂ X for each i ∈ I be a subset. Then we have
!
[ \
X\ Ui = (X \ Ui ) ,
i∈I i∈I
!
\ [
X\ Ui = (X \ Ui ) .
i∈I i∈I

To show that the empty set is closed, we have to see that X \ ∅ is open, but X = X \ ∅ is
open in any topology by definition. Likewise, X \ X = ∅ is open, hence X is also closed.
If A and B are closed sets, then they are complements of open sets U and V . Say we have
A = X \ U and B = X \ V . Then
A ∪ B = (X \ U ) ∪ (X \ V ) = X \ (U ∩ V ) ,
where we used De Morgan’s laws for the second equality. But since U and V are open, so
is U ∩ V . Hence, A ∪ B is the complement of an open set and therefore closed. Now let I
be a set and let (Ai )i∈I be a family of closed subsets of X. Each Ai is the complement of
an open set Ui ⊂ X. By De Morgan’s laws we have
!
\ \ [
Ai = (X \ Ui ) = X \ Ui .
i∈I i∈I i∈I
T
Since each Ui is open, their union is also open. Thus, i∈I Ai is the complement of an
open set and therefore closed.
Exercise 2 (The indiscrete topology).
Let X be a set and consider the topological space (X, Tind ), where Tind is the indiscrete
topology from Example 2.2.8 in the lecture notes. Let (Y, dY ) be a metric space and
consider it as a topological space equipped with the metric topology T (dY ). Show that
any continuous map f : X → Y has to be constant.

Solution to 2: Since f : X → Y is continuous, f −1 (U ) is open for any open subset

U ⊂ Y . By the definition of the indiscrete topology this means that f −1 (U ) has to be
either empty or equal to X, since these are the only two open subsets in the indiscrete
topology.
Now let x1 , x2 ∈ X and assume that f (x1 ) 6= f (x2 ). Consider U = Y \ {f (x1 )} ⊂ Y . We
claim that this set is open in Y . In fact, let y ∈ U . Then we have r = d(y,f2(x1 )) > 0 and
y ∈ Br (y). Suppose that f (x1 ) ∈ Br (y). This would mean that d(y, f (x1 )) < r = d(y,f2(x1 )) ,
which is a contradiction. Hence, Br (y) ⊂ U and U is open.
Since U ⊂ Y is open, we have that f −1 (U ) is either empty or all of X. It can not be empty,
since x2 ∈ f −1 (U ), hence we must have f −1 (U ) = X. But this means that all points of X
are mapped into U , which can not be the case, since f (x1 ) ∈ / U . This contradiction proves
that there can be no two points x1 , x2 with f (x1 ) 6= f (x2 ). Therefore f must be constant.

Exercise 3 (The discrete topology).

Let X be a set and let d : X × X → R be the following map

1 if x 6= y ,
d(x, y) =
0 if x = y .

This defines a metric on X. Show that the metric topology T (d) of d agrees with the
discrete topology Tdis on X (see Example 2.2.10 in the lecture notes) for this particular
choice of metric.

Solution to 3: Let U ⊂ X. We want to show that U ∈ T (d). If U = ∅ we are done.

Suppose that x ∈ U . We have
 
1
B 1 (x) = x0 ∈ X | d(x, x0 ) < = {x} ⊂ U
2 2
Since x ∈ B 1 (x) ⊂ U for all x ∈ U , we see that U is open with respect to the metric
2
topology.
Exercise 4 (Basis).
Let X = Rn let B = {X} be the family that contains only the element X.

a) Check that this family B is the basis of a topology TB on X.

b) We already know this topology. Which topology TB on X = Rn do you get this way?

Solution to 4: To prove a) we need to see first that for any x ∈ X we have a B ∈ B, such
that x ∈ B. However, since X ∈ B we can always choose B = X. The second property B
needs to satisfy to be a basis is the following: If B1 , B2 ∈ B are such that x ∈ B1 ∩ B2 ,
then there is a B3 ∈ B with x ∈ B3 ⊂ B1 ∩ B2 . In our case, the intersection B1 ∩ B2 is
only non-empty if it agrees with X, hence B3 = X will have the desired property.
The topology TB is defined as follows. A subset U ⊂ X is in TB if and only if for any point
x ∈ U we can find a subset B ∈ B with x ∈ B ⊂ U . This is only true for the empty set,
since there is no point to check there, and the whole space, in which case we can choose
B = X. Hence, TB = {∅, X} = Tind agrees with the indiscrete topology.

Exercise 5 (Convergent sequences).

Consider the set N+ = N ∪ {∞}, i.e. the set of all natural numbers plus an additional
point, which we call ∞. We define a topology T+ on N+ in the following way: All subsets
of N are in T+ and all sets that contain ∞ and a complement of a finite set are in T+ . For
example, the set {n ∈ N |n > 5} ∪ {∞} is in T+ , while the sets {2k | k ∈ N} ∪ {∞} and
{1, 2, 3, ∞} are not.

a) Show that T+ is in fact a topology on N+ .

b) Show that the inclusion ι : N → N+ is continuous.

c) Let (Y, TY ) be another topological space and let (an )n∈N be a sequence of points in Y .
It gives rise to a continuous map f : N → Y with f (n) = an . Show that the sequence
converges if and only if there is a continuous map F : N+ → Y , such that F ◦ ι = f .

The space N+ is also called the one-point compactification of N and we will meet it again
in the section about compactness.

Solution to 5: To prove a) we have to check that T+ has the properties of a topology.

Note that ∅ ⊂ N ⊂ N+ and since T+ contains all subsets of N, we have ∅ ∈ T+ . Note
that the empty set is a finite set (it has 0 elements, which is finite). Therefore N is the
complement of a finite set and N+ = N ∪ {∞} is in T+ .
Let U, V ∈ T+ . If at least one of the two sets does not contain ∞, then their intersection
U ∩ V is a subset of N and therefore in T+ . If both of them contain ∞, then we can write
U = U 0 ∪ {∞}, where U 0 does not contain ∞, and N \ U 0 is finite. We have a similar
decomposition of V = V 0 ∪ {∞}. Now note that N \ (U 0 ∩ V 0 ) = (N \ U 0 ) ∪ (N \ V 0 ) is
finite. Thus, we obtain that U ∩ V = (U 0 ∩ V 0 ) ∪ {∞} is in T+ .
S
Now let (Ui )i∈I be a family of subsets with Ui ∈ T+ for all i ∈ I. Let U = i∈I Ui .
Suppose that at least one of the Ui contains ∞. Otherwise, U is a subset of N, which
implies U ∈ T+ . So, assume that ∞ ∈ Uj for some j ∈ I. Then we know that N \ (Uj ∩ N)
is finite, but since (Uj ∩ N) ⊂ (U ∩ N) we have N \ (U ∩ N) ⊂ N \ (Uj ∩ N) and since the
subset of a finite set is again finite, it follows that U ∈ T+ .
This finishes the proof that T+ is a topology.
The proof of b) is actually very simple: Since N has the discrete topology, every subset
of N is open. Therefore ι−1 (U ) ⊂ N is open for any open subset U ⊂ N+ and hence ι is
continuous. (alternatively: We have also shown that ι is always continuous if its source is
equipped with the subspace topology.)
To answer c) we first prove that a sequence (an )n∈N in Y that converges to a ∈ Y gives rise
to a continuous map F : N+ → Y . We define F as follows: F (n) = an for all n ∈ N and
F (∞) = a. We certainly have F ◦ ι = f then. We have to check that this is continuous.
Let U ⊂ Y be an open subset. If U ⊂ Y does not contain a ∈ Y , then F −1 (U ) is a subset
of N and hence open. So, suppose a ∈ U . Since (an )n∈N converges to a, there is an N ∈ N,
such that F (m) = am ∈ U for all m > N . Therefore, {m ∈ N | m > N } ∪ {∞} ⊂ F −1 (U ),
which proves that the complement of F −1 (U ) ∩ N in N is finite and therefore that F −1 (U )
is open.
Next we show that if we have a continuous extension F : N+ → Y such that F ◦ ι = f ,
then (an )n∈N converges to a = F (∞). Let U ⊂ Y be an open subset with a ∈ U . Since F
is continuous, V = F −1 (U ) ⊂ N+ is open. We have ∞ ∈ V , since F (∞) = a ∈ U . Since
V is open and ∞ ∈ V , it is of the form V = V 0 ∪ {∞} with V 0 ⊂ N and N \ V 0 finite. In
particular, there is N ∈ N, such that m ∈ V 0 ⊂ F −1 (U ) for all m > N . (Otherwise N \ V 0
would not be finite.) But this means am = F (m) ∈ U for all m ∈ N with m > N . In other
words: (an )n∈N converges to a ∈ Y .
Exercise 6 (Basis).
Let X be a set and let B = {{x} | x ∈ X} be the family of subsets of X, which contains
only one-point sets.

a) Check that this family B is the basis of a topology TB on X.

b) We also know this topology already. Which one is it?

Solution to 6: To prove a) we need to see that for any x ∈ X we have a B ∈ B, such

that x ∈ B. We can choose B = {x}. Then we have x ∈ B. To see that B also satisfies
the second property, let B1 , B2 ∈ B. In particular, we have B1 = {x1 } and B2 = {x2 }. If
x ∈ B1 ∩B2 , then we must have x = x1 = x2 . With B3 = {x} we obtain x ∈ B3 ⊂ B1 ∩B2 .
Therefore B defines a topology TB on X given by

TB = {U ⊂ X | for each x ∈ X there is B ∈ B such that x ∈ B ⊂ U } .

We claim that TB is the discrete topology on X. To see this, we have to check that any
subset U ⊂ X is open with respect to this topology. If U is empty we are done. Otherwise,
let x ∈ U . Then we have x ∈ {x} ⊂ U and {x} ∈ B. Hence, U ∈ TB . Therefore TB contains
all subsets of X and is the discrete topology.

Exercise 7 (Subspace topology on Z).

Let R be the real line equipped with the metric d : R × R → R given by d(x, y) = |x − y|.
Consider the topological space (R, T (d)) and let Z ⊂ R be the subset of integers. Show
that the subspace topology TZ⊂R on Z induced by R agrees with the discrete topology Tdis
on Z.

Solution to 7: The subspace topology induced by R on Z is given by

TZ⊂R = {U ⊂ Z | there is V ⊂ R open with U = V ∩ Z} .

To show that this is the same as the discrete topology we need to show that it contains
all subsets of Z. Let U ⊂ Z be an arbitrary subset and define
[ 1

V = x ∈ R | |x − y| < .
10
y∈U

1
These are all x ∈ R, such that the distance to one of the elements of U is less than 10 .
An example for such a set V ⊂ R is shown in Figure 1. It is the (disjoint) union of open
 −2 −1 0 1 2 3 4 5 6 7 8 9 10

Figure 1: For U = {1, 2, 5, 6, 8} we have the picture above. The set V is sketched in blue.

intervals, hence open in R with the metric topology. We have U = V ∩ Z by definition.

Since U was arbitrary, TZ⊂R contains all subsets of Z.

Exercise 8 (Subspace topology and restriction of the metric).

Let (X, dX ) be a metric space and let Y ⊂ X be a subset. Then Y is a metric space by
restricting the metric, i.e. we define dY = dX |Y ×Y and consider the metric space (Y, dY ).
Show that the induced metric topology T (dY ) on Y agrees with the subspace topology
TY ⊂X on Y induced by the topological space (X, T (dX )).

Solution to 8: The two topologies that are involved are defined as

TY ⊂X = {U ⊂ Y | there is V ∈ T (dX ) such that V ∩ Y = U } ,

n o
T (dY ) = U ⊂ Y | for all y ∈ U there is r > 0 such that BrdY (y) ⊂ U .

To show that the two topologies agree, we show that TY ⊂X ⊂ T (dY ) and T (dY ) ⊂ TY ⊂X .
The key observation in both cases is the following: For r > 0 and y ∈ Y ⊂ X we have

BrdY (y) = y 0 ∈ Y | dY (y, y 0 ) < r = y 0 ∈ Y | dX (y, y 0 ) < r = BrdX (y) ∩ Y ,

 
(1)

where we used dY = dX |Y ×Y to get the second equality.

Now let U ∈ TY ⊂X . If U = ∅, then U ∈ T (dY ). Otherwise, let y ∈ U . We have V ∈ T (dX )
with V ∩ Y = U by the definition of TY ⊂X . Since V is open with respect to T (dX ) and
y ∈ U ⊂ V , we have an r > 0 such that BrdX (y) ⊂ V . Using (1) this implies

BrdY (y) = BrdX (y) ∩ Y ⊂ V ∩Y =U .

But this means that U ∈ T (dY ).

For the other inclusion let U ∈ T (dY ). To show that U ∈ TY ⊂X we have to find V ∈ T (dX )
dY
such that V ∩ Y = U . For each y ∈ U we have a radius r(y) > 0 such that Br(y) (y) ⊂ U
S dY
by the definition of T (dY ). In particular, we have U = y∈U Br(y) (y). Let
[ dX
V = Br(y) (y) ⊂ X
y∈U
and note that we are using the metric dX to define this subset! This is an open subset of
X with respect to the topology T (dX ) and we have (using (1) again):
[ dX
[ dY
V ∩Y = (Br(y) (y) ∩ Y ) = Br(y) (y) = U .
y∈U y∈U

This proves that U ∈ TY ⊂X and therefore the two topologies are the same.