Math 535 - General Topology

Math 535 - General TopologyFall 2012

Homework 1, Lecture 8/27

Definition. Let V be a (real or complex) vector space. A norm on V is a function ‖·‖ : V → Rsatisfying:

1. Positivity: ‖x‖ ≥ 0 for all x ∈ V and moreover ‖x‖ = 0 holds if and only if x = 0.

2. Homogeneity: ‖αx‖ = |α|‖x‖ for any scalar α and x ∈ V .

3. Triangle inequality: ‖x+ y‖ ≤ ‖x‖+ ‖y‖ for all x, y ∈ V .

A normed vector space is the data (V, ‖·‖) of a vector space V equipped with a norm ‖·‖.

Problem 1. Let (V, ‖·‖) be a normed vector space. Define a function d : V × V → R by

d(x, y) := ‖x− y‖.Show that d is a metric on V , called the metric induced by the norm ‖·‖.

Problem 2. Denote by ‖·‖2 the standard (Euclidean) norm on Rn, defined by

‖x‖2 :=

(n∑

i=1

x2i

) 12

.

Now consider the function ‖·‖1 : Rn → R defined by

‖x‖1 :=n∑

i=1

|xi|.

a. Show that ‖·‖1 is a norm on Rn.

Remark. The norms ‖·‖1 and ‖·‖2 are special cases of the so-called p-norm, for any real numberp ≥ 1 or p =∞. See:

http://en.wikipedia.org/wiki/Lp_spaces#The_p-norm_in_finite_dimensions.

b. Find constants C,D > 0 satisfying

‖x‖2 ≤ C‖x‖1

‖x‖1 ≤ D‖x‖2for all x ∈ Rn.

Definition. Two norms ‖·‖1 and ‖·‖2 on a vector space V are equivalent if they can becompared as in Problem 2b.

Definition. Two metrics d1 and d2 on a set X are topologically equivalent if for everyx ∈ X and ε > 0, there is a δ > 0 satisfying

d1(x, y) < δ ⇒ d2(x, y) < ε

d2(x, y) < δ ⇒ d1(x, y) < ε.

In other words, the identity function (X, d1)→ (X, d2) is a homeomorphism.

1

Problem 3. Show that equivalent norms on a vector space V induce topologically equivalentmetrics.

Problem 4. (Bredon Prop. I.1.3) Show that topologically equivalent metrics induce thesame topology (which explains the terminology). In other words, if d1 and d2 are topologicallyequivalent metrics on X, then a subset U ⊆ X is open with respect to d1 if and only if it isopen with respect to d2.

2



Definition. Let (X, T ) be a topological space. A subset C ⊆ X is closed (with respect to T )if its complement Cc := X \ C is open (with respect to T ).

Problem 5. Show that the collection of closed subsets of X satisfies the following properties.

1. The empty subset ∅ and X itself are closed.

2. An arbitrary intersection of closed subsets is closed: Cα closed for all α implies⋂αCα is

closed.

3. A finite union of closed subsets is closed: C,C ′ closed implies C ∪ C ′ is closed.

Remark. In fact, a collection of subsets satisfies these properties if and only if their comple-ments form a topology. Moreover, open subsets and closed subsets determine each other.

Upshot: One might as well define a topology via a collection of “closed subsets” satisfying thethree properties above. Their complements then form the topology in question.

Problem 6. Let X be a set. Consider the collection of cofinite subsets of X together withthe empty subset:

Tcofin := {U ⊆ X | X \ U is finite} ∪ {∅}.

a. Show that Tcofin is a topology on X, called the cofinite topology.

b. Assuming X is infinite, show that the cofinite topology on X cannot be induced by ametric on X.

Definition. Let X be a set.

• The discrete topology on X is the one where all subsets are open:

Tdisc = P(X) = {U ⊆ X}.

• The anti-discrete (or trivial) topology on X is the one where only the empty subsetand X itself are open:

Tanti = {∅, X}.

Problem 7. Let D be a discrete topological space and A an anti-discrete topological space.

a. Describe all continuous maps f : D → X, where X is an arbitrary topological space.

b. Describe all continuous maps f : X → A, where X is an arbitrary topological space.

c. Describe all continuous maps f : A→ X, where X is a metric space.

Remark. We will come back to the question of mapping into a discrete space when discussingthe notion of connectedness.

1



Problem 8. Let f : X → Y be a function between topological spaces, and let x ∈ X.

a. Show that the following conditions (defining continuity of f at x) are equivalent.

1. For all neighborhood N of f(x), there is a neighborhood M of x such that f(M) ⊆ N .

2. For all open neighborhood V of f(x), there is an open neighborhood U of x such thatf(U) ⊆ V .

3. For all neighborhood N of f(x), the preimage f−1(N) is a neighborhood of x.

b. Find an example of function f : X → Y between metric spaces which is continuous at apoint x ∈ X, but there is an open neighborhood V of f(x) such that the preimage f−1(V ) isnot an open neighborhood of x.

Upshot: The description “preimage of open is open” is really about global continuity, notpointwise continuity (or even local continuity).

Problem 9. Let X be a topological space and B a collection of open subsets of X.

a. Show that B is a basis for the topology of X if and only if for every open subset U ⊆ Xand x ∈ U , there is a B ∈ B satisfying x ∈ B ⊆ U .

b. Assuming X is a metric space, show that the collection of open balls

B = {B 1n(x) | x ∈ X,n ∈ N}

is a basis for the topology of X.

Problem 10. Let X be a set and S a collection of subsets of X.

a. Show that the collection

T :=

{⋃α

nα⋂i=1

Sα,i | Sα,i ∈ S

}

of (arbitrary) unions of finite intersections of members of S is a topology on X.

b. Show that T is the topology TS generated by S. In other words: T contains S and anyother topology T ′ containing S must satisfy T ≤ T ′.

1

Math 535 - General TopologyAdditional notes

Martin Frankland

August 31, 2012

1 Homeomorphisms

Definition 1.1. A map f : X → Y between topological spaces is a homeomorphism if it iscontinuous, invertible (i.e. bijective), and its inverse f−1 : Y → X is also continuous.

2 Neighborhoods

Definition 2.1. Let X be a topological space. A neighborhood of a point x ∈ X is a subsetN ⊆ X such that there is an open U satisfying x ∈ U ⊆ N .

3 Bases and subbases

Definition 3.1. Let (X, T ) be a topological space. A basis for the topology T of X is acollection B of subsets of X satisfying

T =

{⋃α

Bα | Bα ∈ B

}i.e. open sets are precisely unions of members of B.

Exercise 3.2. Let X be a set. Show that a collection B of subsets of X is a basis for sometopology on X if and only if B satisfies the following conditions:

1. B covers X, i.e.⋃B∈B B = X.

2. Finite intersections are unions: For any B,B′ ∈ B, we have B ∩ B′ =⋃αBα for some

family {Bα} of members of B.

Definition 3.3. Let (X, T ) be a topological space. A subbasis for the topology T of X is acollection S of subsets of X satisfying

T :=

{⋃α

nα⋂i=1

Sα,i | Sα,i ∈ S

}i.e. finite intersections of members of S form a basis for the topology.

1

4 Comparing topologies

For a given set X, topologies on X can be partially ordered by inclusion.

Definition 4.1. Let X be a set, and T1 and T2 two topologies on X. We say T1 is smallerthan T2, denoted T1 ≤ T2, if the inclusion T1 ⊆ T2 holds, viewed as subsets of the power setP(X). In other words, every T1-open is also T2-open.

One can also say that T2 is larger than T1.Some references say that T1 is coarser than T2, while T2 is finer than T1.

Remark 4.2. The anti-discrete topology Tanti = {∅, X} is the least element in that partialorder, whereas the discrete topology Tdis = P(X) is the greatest element. In other words, theinequalities

Tanti ≤ T ≤ Tdishold for any topology T on X.

Remark 4.3. By definition, the inequality T1 ≤ T2 holds if and only if the identity function

id : (X, T2)→ (X, T1)

is continuous. Note the reversal, mapping “from fine to coarse”.

The poset of topologies on X has arbitrary meets (infima), described explicitly in the followingproposition.

Proposition 4.4. Let {Tβ} be a family of topologies on X. Then the intersection⋂β Tβ is a

topology on X, and therefore the infimum of the family {Tβ}.

Proof. Exercise.

Remark 4.5. If we consider an empty family of topologies, then their intersection is⋂β

Tβ = P(X) = Tdis

which is a topology on X. Thus the proposition also holds in that case.

Definition 4.6. Let X be a set and S be a collection of subsets of X. The topology generatedby S (if it exists) is the smallest topology TS containing S. In other words, it satisfies S ⊆ TSand for any other topology T ′ containing S, we have TS ≤ T ′.Note that this universal property makes TS unique, if it exists.

Proposition 4.7. For any collection of subsets S, the topology TS exists.

Proof. The topology

TS =⋂

topologies Tsuch that S⊆T

T

has the required properties.

The following proposition provides an explicit description of TS .

2

Proposition 4.8. The topology generated by S is

TS =

{⋃α

nα⋂i=1

Sα,i | Sα,i ∈ S

}

i.e. the topology for which S is a subbasis.

Proof. Homework 1 Problem 10.

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Homework 1 Solutions

Definition. Let V be a (real or complex) vector space. A norm on V is a function ‖·‖ : V → Rsatisfying:

1. Positivity: ‖x‖ ≥ 0 for all x ∈ V and moreover ‖x‖ = 0 holds if and only if x = 0.

2. Homogeneity: ‖αx‖ = |α|‖x‖ for any scalar α and x ∈ V .

3. Triangle inequality: ‖x+ y‖ ≤ ‖x‖+ ‖y‖ for all x, y ∈ V .

A normed vector space is the data (V, ‖·‖) of a vector space V equipped with a norm ‖·‖.

Problem 1. Let (V, ‖·‖) be a normed vector space. Define a function d : V × V → R by

d(x, y) := ‖x− y‖.

Show that d is a metric on V , called the metric induced by the norm ‖·‖.

Solution. We check the three properties of a metric.

1. Positivity:d(x, y) = ‖x− y‖ ≥ 0 for all x, y ∈ V,

d(x, y) = 0⇔ ‖x− y‖ = 0

⇔ x− y = 0

⇔ x = y.

2. Symmetry:

d(y, x) = ‖y − x‖

= ‖(−1)(x− y)‖

= |−1|‖x− y‖

= ‖x− y‖

= d(x, y).

3. Triangle inequality:

d(x, y) = ‖x− y‖

= ‖x− z + z − y‖

≤ ‖x− z‖+ ‖z − y‖

= d(x, z) + d(z, y).

1

Problem 2. Denote by ‖·‖2 the standard (Euclidean) norm on Rn, defined by

‖x‖2 :=

(n∑i=1

x2i

) 12

.

Now consider the function ‖·‖1 : Rn → R defined by

‖x‖1 :=n∑i=1

|xi|.

a. Show that ‖·‖1 is a norm on Rn.

Solution. We check the three properties of a norm.

1. Positivity:

‖x‖1 =n∑i=1

|xi| ≥ 0 for all x ∈ Rn,

‖x‖1 = 0⇔n∑i=1

|xi| = 0

⇔ |xi| = 0 for 1 ≤ i ≤ n

⇔ xi = 0 for 1 ≤ i ≤ n

⇔ x = 0.

2. Homogeneity:

‖αx‖1 =n∑i=1

|αxi|

=n∑i=1

|α||xi|

= |α|n∑i=1

|xi|

= |α|‖x‖1.

3. Triangle inequality:

‖x+ y‖1 =n∑i=1

|xi + yi|

≤n∑i=1

(|xi|+ |yi|)

=n∑i=1

|xi|+n∑i=1

|yi|

= ‖x‖1 + ‖y‖1.

2

Remark. The norms ‖·‖1 and ‖·‖2 are special cases of the so-called p-norm, for any real numberp ≥ 1 or p =∞. See:

http://en.wikipedia.org/wiki/Lp_spaces#The_p-norm_in_finite_dimensions.

3

b. Find constants C,D > 0 satisfying

‖x‖2 ≤ C‖x‖1

‖x‖1 ≤ D‖x‖2

for all x ∈ Rn.

Solution. Claim: ‖x‖2 ≤ ‖x‖1 for all x ∈ Rn. In other words, we can take the constantC = 1.

Proof 1. The claim is equivalent to

‖x‖22 ≤ ‖x‖2

1

n∑i=1

x2i ≤

(n∑i=1

|xi|

)2

n∑i=1

x2i ≤

n∑i=1

x2i +

∑i 6=j

|xi||xj|

0 ≤∑i 6=j

|xi||xj|

which holds for all x ∈ Rn.

Proof 2. Write ei for the standard basis vector ei = (0, . . . , 0,

ith︷︸︸︷1 , 0, . . . , 0). Then we have

‖x‖2 = ‖n∑i=1

xiei‖2

≤n∑i=1

‖xiei‖2

=n∑i=1

|xi|‖ei‖2

=n∑i=1

|xi|

= ‖x‖1.

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For the bound ‖x‖1 ≤ D‖x‖2, here are two solutions.

Solution 1: Crude bound. Noting |xi| ≤ ‖x‖2, we obtain

‖x‖1 =n∑i=1

|xi|

≤n∑i=1

‖x‖2

= n‖x‖2

so that we can take the constant D = n.

Solution 2: Better bound. The 1-norm can be expressed as a dot product

‖x‖1 =n∑i=1

|xi|

=n∑i=1

sign(xi)xi

= s · x

where s ∈ Rn is the vector with entries ±1 given by si = sign(xi). Let’s say sign(0) = +1 byconvention.

The Cauchy-Schwarz inequality yields

‖x‖1 = s · x

≤ ‖s‖2‖x‖2

=√n ‖x‖2

so that we can take the constant D =√n.

Remark. In fact, this is the best possible bound, because equality is achieved by the vectoru = (1, 1, . . . , 1). Indeed, we have ‖u‖1 = n whereas ‖u‖2 =

√n.

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Definition. Two norms ‖·‖1 and ‖·‖2 on a vector space V are equivalent if they can becompared as in Problem 2b.

Definition. Two metrics d1 and d2 on a set X are topologically equivalent if for everyx ∈ X and ε > 0, there is a δ > 0 satisfying

d1(x, y) < δ ⇒ d2(x, y) < ε

d2(x, y) < δ ⇒ d1(x, y) < ε.

In other words, the identity function (X, d1)→ (X, d2) is a homeomorphism.

6

Problem 3. Show that equivalent norms on a vector space V induce topologically equivalentmetrics.

Solution. Let ‖·‖1 and ‖·‖2 be two norms on V , and let C,D > 0 be constants satisfying

‖x‖2 ≤ C‖x‖1

‖x‖1 ≤ D‖x‖2

for all x ∈ V . Thus the induced metrics satisfy

d2(x, y) = ‖x− y‖2

≤ C‖x− y‖1

= Cd1(x, y)

and likewise d1(x, y) ≤ Dd2(x, y) for all x, y ∈ V .

Take K := max{C,D}. Let x ∈ V and ε > 0 be given. Pick δ := εK

(which in this case can bechosen uniformly, independently of x). Then we have

d1(x, y) < δ ⇒ d2(x, y) ≤ Cd1(x, y)

< Cδ

≤ Kδ

= ε

and likewise

d2(x, y) < δ ⇒ d1(x, y) ≤ Dd2(x, y)

< Dδ

≤ Kδ

= ε.

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Problem 4. (Bredon Prop. I.1.3) Show that topologically equivalent metrics induce thesame topology (which explains the terminology). In other words, if d1 and d2 are topologicallyequivalent metrics on X, then a subset U ⊆ X is open with respect to d1 if and only if it isopen with respect to d2.

Solution. By symmetry of the situation, it suffices to show one side of the equivalence. LetU ⊆ X be a subset which is open with respect to d1. We want to show that U is open withrespect to d2.

Let x ∈ U . Because U is open with respect to d1, there is an ε-ball (in the d1 metric) aroundx entirely contained in U , i.e. B1

ε (x) ⊆ U , for some ε > 0.

Because d2 is topologically equivalent to d1, there is some δ > 0 satisfying B2δ (x) ⊆ B1

ε (x). Inparticular we have

B2δ (x) ⊆ B1

ε (x) ⊆ U

so that U is open with respect to d2.

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Definition. Let (X, T ) be a topological space. A subset C ⊆ X is closed (with respect to T )if its complement Cc := X \ C is open (with respect to T ).

Problem 5. Show that the collection of closed subsets of X satisfies the following properties.

1. The empty subset ∅ and X itself are closed.

2. An arbitrary intersection of closed subsets is closed: Cα closed for all α implies⋂αCα is

closed.

3. A finite union of closed subsets is closed: C,C ′ closed implies C ∪ C ′ is closed.

Solution.

1. The empty set ∅ is closed because its complement ∅c = X is open.

The entire set X is closed because its complement Xc = ∅ is open.

2. Let Cα be closed for all α and consider the intersection⋂αCα. Its complement(⋂

α

Cα

)c

=⋃α

Ccα

is a union of open sets, therefore open. Thus⋂αCα is closed.

3. Let C,C ′ be closed and consider the union C ∪ C ′. Its complement

(C ∪ C ′)c = Cc ∩ C ′c

is a finite intersection of open sets, therefore open. Thus C ∪ C ′ is closed.

Remark. In fact, a collection of subsets satisfies these properties if and only if their comple-ments form a topology. Moreover, open subsets and closed subsets determine each other.

Upshot: One might as well define a topology via a collection of “closed subsets” satisfying thethree properties above. Their complements then form the topology in question.

9

Problem 6. Let X be a set. Consider the collection of cofinite subsets of X together withthe empty subset:

Tcofin := {U ⊆ X | X \ U is finite} ∪ {∅}.

a. Show that Tcofin is a topology on X, called the cofinite topology.

Solution. By problem 5, it suffices to check that the collection of subsets

{F ⊆ X | F is finite} ∪ {X}.

satisfies the axioms of closed subsets.

1. The empty set ∅ is finite, and therefore belongs to the collection, whereas X belongs tothe collection by definition.

2. Let Cα be a family of subsets that are either finite or all of X. Then the intersection⋂αCα is finite (if at least one Cα is finite) or all of X (if all Cα are X).

3. Let C,C ′ be subsets that are either finite or all of X. Then the union C ∪ C ′ is finite (ifboth C and C ′ are finite) or all of X (if at least one of C or C ′ is X).

b. Assuming X is infinite, show that the cofinite topology on X cannot be induced by ametric on X.

Solution. Let Br(x) be an open ball not containing all of X. If the complement Br(x)c isinfinite, then we are done: we have found an open subset which is not cofinite.

If the complement Br(x)c is finite, then pick a point y ∈ Br(x)c and choose a radius ε > 0small enough so that the open ball Bε(y) does not intersect Br(x); any value ε ≤ d(x, y) − rwill do. Then Bε(y) is finite, and thus its complement Bε(y)c is infinite, since X is infinite. Weare done: we have found an open subset Bε(y) which is not cofinite.

Slightly different solution. If X is infinite, then:

• Any cofinite subset of X is infinite;

• Any cofinite subset and any infinite subset must intersect.

In particular, any two non-empty open subsets of X intersect.

However, in any metric space containing at least two points, we can find non-empty open subsetsthat do not intersect. Indeed, pick distinct points x and y, and take small enough open ballsBr(x) and Br(y) around them; any radius r ≤ d(x,y)

2will guarantee Br(x) ∩Br(y) = ∅.

Therefore the cofinite topology on X cannot be induced by a metric.

Remark. In a few lectures, we will say that such a topology is not Hausdorff, hence not metri-zable.

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Definition. Let X be a set.

• The discrete topology on X is the one where all subsets are open:

Tdisc = P(X) = {U ⊆ X}.

• The anti-discrete (or trivial) topology on X is the one where only the empty subsetand X itself are open:

Tanti = {∅, X}.

Problem 7. Let D be a discrete topological space and A an anti-discrete topological space.

a. Describe all continuous maps f : D → X, where X is an arbitrary topological space.

Solution. The condition that f−1(V ) be open in D for any open V ⊆ X is automaticallysatisfied, since every subset of D is open. Thus every function f : D → X is continuous.

Remark. We will come back to the question of mapping into a discrete space when discussingthe notion of connectedness.

b. Describe all continuous maps f : X → A, where X is an arbitrary topological space.

Solution. For any function f : X → A, we have f−1(∅) = ∅ and f−1(A) = X, both of whichare open in X. Since ∅ and A are the only open subsets of A, every function f : X → A iscontinuous.

c. Describe all continuous maps f : A→ X, where X is a metric space.

Solution. Let f : A→ X be a continuous. We claim that f is a constant function.

Pick some a ∈ A and look at its value x := f(a) ∈ X. For any other point y ∈ X, pick an opensubset V ⊆ X satisfying x ∈ V but y /∈ V . Since f is continuous, the preimage f−1(V ) is openin A. The condition f(a) = x ∈ V yields a ∈ f−1(V ) so that f−1(V ) is non-empty and musttherefore be all of A, the only non-empty open subset of A.

Now the condition f(A) ⊆ V implies that f never takes the value y /∈ V . Since y was anarbitrary point distinct from x, we conclude that f is the constant function with value x.

Remark. The proof still holds whenever X is a T1 space, a property that will be discussed in afew lectures.

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Problem 8. Let f : X → Y be a function between topological spaces, and let x ∈ X.

a. Show that the following conditions (defining continuity of f at x) are equivalent.

1. For all neighborhood N of f(x), there is a neighborhood M of x such that f(M) ⊆ N .

2. For all open neighborhood V of f(x), there is an open neighborhood U of x such thatf(U) ⊆ V .

3. For all neighborhood N of f(x), the preimage f−1(N) is a neighborhood of x.

Solution. (1 ⇒ 2) Let V be an open neighborhood of f(x). Since V is in particular aneighborhood of f(x), the assumption (1) guarantees that there is a neighborhood M of xsuch that f(M) ⊆ V . Let U be an open of X satisfying x ∈ U ⊆ M . Then U is an openneighborhood of x satisfying f(U) ⊆ f(M) ⊆ V .

(2 ⇒ 3) Let N be a neighborhood of f(x). Let V be an open of Y satisfying f(x) ∈ V ⊆ N .Then we have x ∈ f−1(V ) ⊆ f−1(N). By the assumption (2), f−1(V ) is an open neighborhoodof x, so that f−1(N) is a neighborhood of x.

(3 ⇒ 1) Let N be a neighborhood of f(x) and take M := f−1(N). By the assumption (3),f−1(N) is a neighborhood of x, and moreover it satisfies f (f−1(N)) ⊆ N .

b. Find an example of function f : X → Y between metric spaces which is continuous at apoint x ∈ X, but there is an open neighborhood V of f(x) such that the preimage f−1(V ) isnot an open neighborhood of x.

Upshot: The description “preimage of open is open” is really about global continuity, notpointwise continuity (or even local continuity).

Solution. Consider the “step” function f : R→ R defined by

f(x) =

{0 if x ≤ 43

1 if x > 43.

Then f is continuous at x = 0 (in fact everywhere except at 43). However, take the openneighborhood (−1

2, 1

2) of f(0) = 0. Its preimage under f is

f−1

((−1

2,1

2)

)= (−∞, 43]

which is not open in R.

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Problem 9. Let X be a topological space and B a collection of open subsets of X.

a. Show that B is a basis for the topology of X if and only if for every open subset U ⊆ Xand x ∈ U , there is a B ∈ B satisfying x ∈ B ⊆ U .

Solution. (⇒) Assume B is a basis for the topology. Let U ⊆ X be open (WLOG non-empty)and x ∈ U . Because B is a basis, U can be written as a union U =

⋃αBα for some family of

subsets Bα ∈ B. Thus x is in at least one of those subsets Bαx , yielding x ∈ Bαx ⊆ U .

(⇐) To show that B is a basis, there are two things to check.

1) Every union of members of B is open in X. This is automatic, because each B ∈ B wasassumed to be open.

2) Let U ⊆ X be open (WLOG non-empty). We want to show that U is a union of subsetsin the collection B. By assumption on B, for each x ∈ U , there is some Bx ∈ B satisfyingx ∈ Bx ⊆ U . Thus we have U =

⋃x∈U Bx.

b. Assuming X is a metric space, show that the collection of open balls

B = {B 1n(x) | x ∈ X,n ∈ N}

is a basis for the topology of X.

Solution. We use the criterion from part (a).

Let U ⊆ X be open and x ∈ U . Then there is some radius r > 0 such that the open ball ofradius r centered at x is contained within U , i.e. Br(x) ⊆ U .

Pick n large enough so that 1n≤ r. Then we have x ∈ B 1

n(x) ⊆ Br(x) ⊆ U .

13

Problem 10. Let X be a set and S a collection of subsets of X.

a. Show that the collection

T :=

{⋃α

nα⋂i=1

Sα,i | Sα,i ∈ S

}

of (arbitrary) unions of finite intersections of members of S is a topology on X.

Solution. We check the properties of a topology.

1. Because unions indexed by the empty family are allowed, the empty set ∅ =⋃∅ Sα is in

T .

Because intersections indexed by the empty family are allowed, the entire set X =⋂∅ Sα

is in T .

2. Finite intersections of members of T are in T . Let U =⋃α

⋂nαi=1 Sα,i and V =

⋃β

⋂nβj=1 Sβ,j

be members of T . Their intersection is

U ∩ V =

(⋃α

nα⋂i=1

Sα,i

)∩

(⋃β

nβ⋂j=1

Sβ,j

)

=⋃α,β

(nα⋂i=1

Sα,i ∩nβ⋂j=1

Sβ,j

)

which is in T since each Sα,i and Sβ,j is in S.

3. Arbitrary unions of members of T are in T . Let {Uγ =⋃α

⋂n(γ)α

i=1 Sγα,i}γ be a family of

members of T . Then their union is

⋃γ

Uγ =⋃γ

⋃α

n(γ)α⋂i=1

Sγα,i

=⋃γ,α

n(γ)α⋂i=1

Sγα,i

which is in T .

14

b. Show that T is the topology TS generated by S. In other words: T contains S and anyother topology T ′ containing S must satisfy T ≤ T ′.

Solution.

1. T contains S: Any S ∈ S can be viewed as the union of one set which is the intersectionof one set, namely S itself, which is in S. Therefore we have S ⊆ T .

2. Let T ′ be a topology containing S. For any family of subsets Sα,i ∈ S, the finite inter-section

⋂nαi=1 Sα,i is in T ′, since T ′ is a topology. Moreover, the union

⋃α

nα⋂i=1

Sα,i

is also in T ′, since T ′ is a topology. Thus we have T ⊆ T ′, as claimed.

15


Martin Frankland

September 5, 2012

1 Subspaces

Definition 1.1. Let X be a topological space and A ⊆ X any subset. The subspace topologyon A is the smallest topology T sub

A making the inclusion map i : A ↪→ X continuous.

In other words, T subA is generated by subsets V ⊆ A of the form

V = i−1(U) = U ∩ A

for any open U ⊆ X.

Proposition 1.2. The subspace topology on A is

T subA = {V ⊆ A | V = U ∩ A for some open U ⊆ X}.

In other words, the collection of subsets of the form U ∩ A already forms a topology on A.

2 Products

Before discussing the product of spaces, let us review the notion of product of sets.

2.1 Product of sets

Let X and Y be sets. The Cartesian product of X and Y is the set of pairs

X × Y = {(x, y) | x ∈ X, y ∈ Y }.

It comes equipped with the two projection maps pX : X × Y → X and pY : X × Y → Y ontoeach factor, defined by

pX(x, y) = x

pY (x, y) = y.

This explicit description of X × Y is made more meaningful by the following proposition.

1

Proposition 2.1. The Cartesian product of sets satisfies the following universal property. Forany set Z along with maps fX : Z → X and fY : Z → Y , there is a unique map f : Z → X × Ysatisfying pX ◦ f = fX and pY ◦ f = fY , in other words making the diagram

Z

fX

��

∃!f��

fY

X × YpX

{{

pY

""X Y

commute.

Proof. Given fX and fY , define f : Z → X × Y by

f(z) := (fX(z), fY (z))

which clearly satisfies pX ◦ f = fX and pY ◦ f = fY .

To prove uniqueness, note that any pair (x, y) ∈ X × Y can be written as

(x, y) = (pX(x, y), pY (x, y))

i.e. the projections give us each individual component of the pair. Therefore, any functiong : Z → X × Y can be written as

g(z) = (pX(g(z)), pY (g(z)))

= ((pX ◦ g)(z), (pY ◦ g)(z))

so that g is determined by its components pX ◦ g and pY ◦ g.

In slogans: “A map into X × Y is the same data as a map into X and a map into Y ”.

Yet another slogan: “X × Y is the closest set equipped with a map to X and a map to Y .”

As usual with universal properties, this characterizes X × Y up to unique isomorphism. Thisstatement is made precise in the following proposition.

Proposition 2.2. Let W be a set equipped with maps πX : W → X and πY : W → Y satisfying

the universal property of the product. Then there is a unique isomorphism ϕ : W∼=−→ X × Y

commuting with the projections, i.e. making the diagrams

W

πX ##

ϕ// X × Y

pX��

W

πY ##

ϕ// X × Y

pY��

X Y

commute.

2

Proof. Starting from the data of the maps πX : W → X and πY : W → Y , the universal propertyof X × Y provides a unique map ϕ : W → X × Y commuting with the projections.

Likewise, starting from the data of the maps pX : X×Y → X and pY : X×Y → Y , the universalproperty of W provides a unique map ψ : X × Y → W commuting with the projections.

We claim that ϕ is an isomorphism, with inverse ψ.

The composite ψ ◦ϕ : W → W is a map into W commuting with the projections. But so is theidentity map idW : W → W . By uniqueness (guaranteed in the universal property of W ), weobtain ψ ◦ ϕ = idW .

Likewise, the composite ϕ ◦ ψ : X × Y → X × Y is a map into X × Y commuting with theprojections. But so is the identity map idX×Y : X × Y → X × Y . By uniqueness (guaranteedin the universal property of X × Y ), we obtain ϕ ◦ ψ = idX×Y .

2.2 Product topology

The next goal is to define the product X×Y of topological spaces X and Y such that it satisfiesthe analogous universal property in the category of topological spaces.

In other words, we want to find a topology on X×Y such that the projection maps pX : X×Y →X and pY : X × Y → Y are continuous, and such that for any topological space Z along withcontinuous maps fX : Z → X and fY : Z → Y , there is a unique continuous map f : Z → X×Ysatisfying pX ◦ f = fX and pY ◦ f = fY .

Definition 2.3. Let X and Y be topological spaces. The product topology TX×Y on X ×Yis the smallest topology on X×Y making the projections pX : X×Y → X and pY : X×Y → Ycontinuous.

In other words, TX×Y is generated by “strips” of the form

p−1X (U) = U × Y

p−1Y (V ) = X × V

for some open U ⊆ X or some open V ⊆ Y .

Proposition 2.4. The collection of “rectangles”

{U × V | U ⊆ X is open and V ⊆ Y is open}

is a basis for the product topology on X × Y .

Proof. Finite intersections of strips

(U × Y ) ∩ (X × V ) = U × V

provide all rectangles. However a finite intersection of rectangles

(U1 × V1) ∩ (U2 × V2) = (U1 ∩ U2)× (V1 ∩ V2)

is again a rectangle, since U1 ∩ U2 ⊆ X is open and V1 ∩ V2 ⊆ Y is open.

Proposition 2.5. The topological space (X×Y, TX×Y ) along with the projections pX : X×Y →X and pY : X × Y → Y satisfies the universal property of a product.

3

Proof. Let Z be a topological space along with continuous maps fX : Z → X and fY : Z → Y .In particular, these continuous maps are functions, so that there is a unique function f : Z →X × Y satisfying pX ◦ f = fX and pY ◦ f = fY . In other words, f is given by

f(z) = (fX(z), fY (z)) .

It remains to check that f is continuous. For any rectangle U × V ⊆ X × Y where U ⊆ X isopen and V ⊆ Y is open, its preimage is

f−1(U × V ) = {z ∈ Z | f(z) ∈ U × V }

= {z ∈ Z | fX(z) ∈ U and fY (z) ∈ V }

= f−1X (U) ∩ f−1Y (V ).

Since fX and fY are continuous, the subsets f−1X (U) and f−1Y (V ) are open in Z, and so istheir intersection f−1X (U) ∩ f−1Y (V ). Since those rectangles U × V form a basis for the producttopology on X × Y , the function f : Z → X × Y is continuous.

Remark 2.6. Why did we choose the smallest topology making the projections pX and pYcontinuous?

If there is a product topology TX×Y satisfying the universal property, consider any other topol-ogy T on X × Y making the projections pX and pY continuous. Then the universal propertyof TX×Y provides a unique continuous map f making the diagram

(X × Y, T )

pX

��

∃!f��

pY

��

(X × Y, TX×Y )pX

yy

pY

%%X Y

commute. As a function, f : X × Y → X × Y must be the identity:

f(x, y) = (pX(x, y), pY (x, y))

= (x, y).

The identity id : (X × Y, T ) → (X × Y, TX×Y ) being continuous means precisely the inequal-ity TX×Y ≤ T . That is why TX×Y had to be the smallest topology making the projectionscontinuous.

Exercise 2.7. Let (X, dX) and (Y, dY ) be metric spaces.

1. For points (x, y) and (x′, y′) in X × Y , define their distance as the sum

d ((x, y), (x′, y′)) := dX(x, x′) + dY (y, y′).

Show that d is a metric on X × Y .

2. Show that the metric d induces the product topology on X × Y .

4



Problem 1. (Brown Exercise 2.4.5) Consider X = [0, 2] \ {1} as a subspace of the real lineR. Show that the subset [0, 1) ⊂ X is both open and closed in X.

Problem 2. (Bredon Exercise I.3.8) Let X be a topological space that can be written as aunion X = A ∪ B where A and B are closed subsets of X. Let f : X → Y be a function,where Y is any topological space. Assume that the restrictions of f to A and to B are bothcontinuous. Show that f is continuous.

Problem 3. A map between topological spaces f : X → Y is called an open map if for everyopen subset U ⊆ X, its image f(U) ⊆ Y is open in Y .

a. (Munkres Exercise 2.16.4) Let X and Y be topological spaces. Show that the projectionmaps pX : X × Y → X and pY : X × Y → Y are open maps.

b. Find an example of metric spaces X and Y , and a closed subset C ⊆ X × Y such that theprojection pX(C) ⊆ X is not closed in X.

In other words, the projection maps are (usually) not closed maps.

1


Martin Frankland

September 7, 2012

1 Infinite products

Definition 1.1. Let {Xα}α∈A be a family of topological spaces. The product topologyTprod on the Cartesian product

∏αXα is the smallest topology making all projection maps

pβ :∏

αXα → Xβ continuous.

In other words, the product topology is generated by subsets of the form p−1β (Uβ) for Uβ ⊆ Xβ

open.

A basis for Tprod is the collection of “large boxes”

{∏α

Uα | Uα ⊆ Xα is open, and Uα = Xα except for at most finitely many α}

Proposition 1.2. The topological space (∏

αXα, Tprod) along with the projections pβ :∏

αXα →Xβ satisfies the universal property of a product.

Proof. Let Z be a topological space along with continuous maps fα : Z → Xα for all α ∈ A.In particular, these continuous maps are functions, so that there is a unique function f : Z →∏

αXα whose components are pα ◦ f = fα. In other words, f is given by

f(z) = (fα(z))α∈A .

It remains to check that f is continuous. The product topology is generated by subsets of theform p−1β (Uβ) for Uβ ⊆ Xβ open. Its preimage under f is

f−1(p−1β (Uβ)

)= (pβ ◦ f)−1(Uβ)

= f−1β (Uβ)

which is open in Z since fβ : Z → Xβ is continuous.

Definition 1.3. The box topology Tbox on the Cartesian product∏

αXα is the topology forwhich the collection of “boxes”

{∏α

Uα | Uα ⊆ Xα is open}

is a basis.

1

Note that we always have Tprod ≤ Tbox, and equality holds for finite products. For an infiniteproduct, the inequality is usually strict.

Exercise 1.4. Show that the projection maps pβ :∏

αXα → Xβ are open maps in the boxtopology (and therefore also in the product topology).

2 Disjoint unions

In this section, we describe a construction which is dual to the product. The discussion will beeerily similar to that of products, because the ideas are the same, and because of copy-paste.

2.1 Disjoint union of sets

Let X and Y be sets. The disjoint union of X and Y is the set

X q Y = {w | w ∈ X or x ∈ Y }.

It comes equipped with the inclusion maps iX : X → X q Y and iY : Y → X q Y fromeach summand. This explicit description of X q Y is made more meaningful by the followingproposition.

Proposition 2.1. The disjoint union of sets XqY , along with inclusion maps iX and iY , is thecoproduct of sets, i.e. it satisfies the following universal property. For any set Z along withmaps fX : X → Z and fY : Y → Z, there is a unique map f : X q Y → Z whose restrictionsare f ◦ iX = fX and f ◦ iY = fY , in other words making the diagram

X

fX

++

iX ##

Y

fY

tt

iY{{

X q Y

∃!f��

Z

commute.

Proof. Given fX and fY , define f : X q Y → Z by

f(w) :=

{fX(w) if w ∈ XfY (w) if w ∈ Y

which clearly satisfies f ◦ iX = fX and f ◦ iY = fY .

To prove uniqueness, note that any element w ∈ X q Y is in one of the summands:

w =

{iX(w) if w ∈ XiY (w) if w ∈ Y.

2

Therefore, any function g : X q Y → Z can be written as

g(w) =

{g(iX(w)) = (g ◦ iX)(w) if w ∈ Xg(iY (w)) = (g ◦ iY )(w) if w ∈ Y

so that g is determined by its restrictions g ◦ iX and g ◦ iY .

In slogans: “A map out of X q Y is the same data as a map out of X and a map out of Y ”.

Yet another slogan: “X q Y is the closest set equipped with a map from X and a map fromY .”

As usual with universal properties, this characterizes X q Y up to unique isomorphism.

2.2 Coproduct topology

The next goal is to define the coproduct X q Y of topological spaces X and Y such that itsatisfies the analogous universal property in the category of topological spaces.

In other words, we want to find a topology onXqY such that the inclusion maps iX : X → XqYand iY : Y → X q Y are continuous, and such that for any topological space Z along withcontinuous maps fX : X → Z and fY : Y → Z, there is a unique continuous map f : XqY → Zwhose restrictions are f ◦ iX = fX and f ◦ iY = fY .

Definition 2.2. Let X and Y be topological spaces. The coproduct topology is the largesttopology on X q Y making the inclusions iX : X → X q Y and iY : Y → X q Y continuous.

This means that a subset U ⊆ X q Y is open if and only if i−1X (U) is open in X and i−1Y (U) isopen in Y .

More concretely, noting i−1X (U) = U ∩ X and i−1Y (U) = U ∩ Y , open sets can be described asU = UX q UY where UX = U ∩X is open in X and UY = U ∩ Y is open in Y .

This definition works for an infinite disjoint union as well.

Definition 2.3. Let {Xα}α∈A be a family of topological spaces. The coproduct topologyTcoprod on the disjoint union

∐αXα is the largest topology making all inclusion maps iβ : Xβ →∐

αXα continuous.

This means that a subset U ⊆∐

αXα is open if and only if i−1α (U) is open in Xα for all α ∈ A.

More concretely, noting i−1α (U) = U ∩ Xα, open sets can be described as U =∐

α Uα whereUα = U ∩ Xα is open in Xα. That is, open subsets are disjoint unions of open subsets fromeach of the summands.

Proposition 2.4. Each summand Xβ ⊆∐

αXα is open in the coproduct topology.

Proof. Write Xβ =∐

α Uα where

Uα =

{Xβ if α = β

∅ if α 6= β

is open in Xα for all α.

3

Remark 2.5. More generally, the same proof shows that each inclusion map iβ : Xβ →∐

αXα

is an open map.

Proposition 2.6. The topological space (∐

αXα, Tcoprod) along with the inclusions iβ : Xβ →∐αXα is a coproduct of topological spaces.

Proof. We verify the universal property of a coproduct.

Let Z be a topological space along with continuous maps fα : Xα → Z for all α ∈ A. Inparticular, these continuous maps are functions, so that there is a unique function f :

∐αXα →

Z whose restrictions are f ◦ iα = fα. In other words, f is given by

f(w) = f(iα(w)) = fα(w)

where α is the unique index for which w ∈ Xα.

It remains to check that f is continuous. Let U ⊆ Z be open and consider its preimagef−1(U) ⊆

∐αXα. To show that this subset is open, it suffices to check that its restriction to

each summand is open:

i−1α(f−1(U)

)= (f ◦ iα)−1(U)

= f−1α (U)

is indeed open in Xα since fα : Xα → Z is continuous.

Upshot: A map f :∐

αXα → Z is continuous if and only if its restriction f ◦ iα : Xα → Z toeach summand is continuous.

4



Problem 4. (Munkres Exercise 2.19.7) Consider the set of sequences of real numbers

RN = {(x1, x2, . . .) | xn ∈ R for all n ∈ N} ∼=∏n∈N

R

and consider the subset of sequences that are “eventually zero”

R∞ := {x ∈ RN | xn 6= 0 for at most finitely many n}.

a. In the box topology on RN, is R∞ a closed subset?

b. In the product topology on RN, is R∞ a closed subset?

Problem 5. Let X be a topological space, S a set, and f : X → S a function. Consider thecollection of subsets of S

T := {U ⊆ S | f−1(U) is open in X}.

a. Show that T is a topology on S.

b. Show that T is the largest topology on S making f continuous.

c. Let Y be a topological space. Show that a map g : S → Y is continuous if and only if thecomposite g ◦ f : X → Y is continuous.

d. Show that T is the smallest topology on S with the property that a map g : S → Y iscontinuous whenever g ◦ f is continuous.

Problem 6. Consider the subset X = {0}∪{ 1n| n ∈ N} ⊂ R viewed as a subspace of the real

line R. As a set, X is the disjoint union of the singletons {0} and { 1n} for all n ∈ N. However,

show that X does not have the coproduct topology on {0} q∐

n∈N{1n}.

1


Homework 2 Solutions

Problem 1. (Brown Exercise 2.4.5) Consider X = [0, 2] \ {1} as a subspace of the real lineR. Show that the subset [0, 1) ⊂ X is both open and closed in X.

Solution. [0, 1) is open in X because we can write

[0, 1) = (−8, 1) ∩X

and (−8, 1) is open in R.

On the other hand, [0, 1) is closed in X because we can write

[0, 1) = [0, 1] ∩X

and [0, 1] is closed in R.

1

Problem 2. (Bredon Exercise I.3.8) Let X be a topological space that can be written as aunion X = A ∪ B where A and B are closed subsets of X. Let f : X → Y be a function,where Y is any topological space. Assume that the restrictions of f to A and to B are bothcontinuous. Show that f is continuous.

Solution.

Lemma. Let A ⊆ X be a closed subset. If C ⊆ A is closed in A, then C is also closed in X.

Proof. Since C is closed in A, it can be written as C = C̃ ∩ A for some closed subset C̃ ⊆ X.Therefore C is an intersection of closed subsets of X, and thus is closed in X.

Let C ⊂ Y be a closed subset. Its preimage under f is the union

f−1(C) =(f−1(C) ∩ A

)∪(f−1(C) ∩B

)= (f |A)−1(C) ∪ (f |B)−1(C).

Since the restriction f |A : A → Y is continuous, (f |A)−1(C) is closed in A, and thus closed inX by the lemma. Likewise, (f |B)−1(C) is closed in X. Therefore their union

f−1(C) = (f |A)−1(C) ∪ (f |B)−1(C).

is closed in X, so that f in continuous.

Remark. The same proof shows that the statement still holds if A and B are both open in X.

2

Problem 3. A map between topological spaces f : X → Y is called an open map if for everyopen subset U ⊆ X, its image f(U) ⊆ Y is open in Y .

a. (Munkres Exercise 2.16.4) Let X and Y be topological spaces. Show that the projectionmaps pX : X × Y → X and pY : X × Y → Y are open maps.

Solution.

Lemma. A map f : X → Y is open if and only if f(B) ⊆ Y is open in Y for every B ∈ Bbelonging to some basis B of the topology on X.

Proof. (⇒) Each member B ∈ B is open in X.

(⇐) Let U ⊆ X be open in X. Then U is a union U =⋃αBα of basic open subsets Bα ∈ B.

Its image under f is

f(U) = f

(⋃α

Bα

)

=⋃α

f(Bα)

where each f(Bα) is open in Y by assumption. Thus f(U) is a union of open subsets and henceopen.

Take an “open box” U × V ⊆ X × Y , where U ⊆ X is open and V ⊆ Y is open. Its projectiononto the first factor is

pX(U × V ) = U ⊆ X

which is open in X. Since open boxes form a basis of the topology on X × Y , the lemmaguarantees that pX is an open map, and likewise for pY .

3

b. Find an example of metric spaces X and Y , and a closed subset C ⊆ X × Y such that theprojection pX(C) ⊆ X is not closed in X.

In other words, the projection maps are (usually) not closed maps.

Solution. Take X = Y = R and consider the hyperbola in R× R

C = {(x, 1

x) | x 6= 0} = {(x, y) ∈ R× R | xy = 1}.

Its projection onto the first factor is

pX(C) = R \ {0}

which is not closed in R.

To show that C is closed in R×R, note that the function f : R×R→ R defined by f(x, y) = xyis continuous, and C is the preimage C = f−1({1}). Since the singleton {1} is closed in R, Cis closed in R× R.

4

Problem 4. (Munkres Exercise 2.19.7) Consider the set of sequences of real numbers

RN = {(x1, x2, . . .) | xn ∈ R for all n ∈ N} ∼=∏n∈N

R

and consider the subset of sequences that are “eventually zero”

R∞ := {x ∈ RN | xn 6= 0 for at most finitely many n}.

a. In the box topology on RN, is R∞ a closed subset?

Solution. Yes, R∞ is closed in the box topology.

Let x ∈ RN \R∞, which means that the sequence x has infinitely many non-zero entries xn 6= 0.For all those indices n, pick an open neighborhood Un of xn ∈ R which does not contain 0. Forother values of n, take Un = R. Then the open box

∏n Un is an open neighborhood of x which

does not intersect R∞.

Indeed, for any y ∈∏

n Un and every index n such that xn 6= 0, we have yn ∈ Un so that yn 6= 0by construction. Because there are infinitely many such indices, we conclude y /∈ R∞.

b. In the product topology on RN, is R∞ a closed subset?

Solution. No, R∞ is not closed in the product topology.

Let x ∈ RN \R∞ and consider any open neighborhood U =∏

n Un of x which is a “large box”,i.e. Un ⊆ R is open for all n and Un = R except for finitely many n. In particular, there is anumber N such that Un = R for all n ≥ N . Consider a sequence y with yn = 0 for all n ≥ Nand yn ∈ Un for 1 ≤ n < N . Then we have y ∈ U ∩ R∞.

Because “large boxes” form a basis of the product topology, every open neighborhood of xintersects R∞. Therefore R∞ is not closed.

Remark. In fact, the argument shows that x is not an interior point of RN \ R∞, so that theinterior of RN \ R∞ is empty. Equivalently, the closure of R∞ is all of RN, i.e. R∞ is dense inRN.

5

Problem 5. Let X be a topological space, S a set, and f : X → S a function. Consider thecollection of subsets of S

T := {U ⊆ S | f−1(U) is open in X}.

a. Show that T is a topology on S.

Solution.

1. The preimage f−1(S) = X is open in X, so that the entire set S is in T . Likewise,f−1(∅) = ∅ is open in X, so that the empty set ∅ is in T .

2. Let Uα be a family of members of T . Then we have

f−1

(⋃α

Uα

)=⋃α

f−1(Uα)

where each f−1(Uα) is open in X by assumption. Thus f−1 (⋃α Uα) is also open in X, so

that the union⋃α Uα is in T .

3. Let U and U ′ be members of T . Then we have

f−1 (U ∩ U ′) = f−1(U) ∩ f−1(U ′)

where f−1(U) and f−1(U ′) are open in X by assumption. Thus f−1 (U ∩ U ′) is also openin X, so that the finite intersection U ∩ U ′ is in T .

b. Show that T is the largest topology on S making f continuous.

Solution. Note that T makes f continuous by construction: for all U ∈ T , the preimagef−1(U) ⊆ X is open in X.

Let T ′ be a topology on S making f continuous. Then for every U ∈ T ′, the preimage f−1(U)is open in X, which means U ∈ T . This proves T ′ ≤ T .

c. Let Y be a topological space. Show that a map g : S → Y is continuous if and only if thecomposite g ◦ f : X → Y is continuous.

Solution. (⇒) The maps f and g are continuous, hence so is their composite g ◦ f .

(⇐) Assume g ◦ f is continuous; we want to show that g is continuous. Let U ⊆ Y be openand take its preimage g−1(U) ⊆ S. To check that this subset is open, consider its preimage

f−1(g−1(U)

)= (g ◦ f)−1(U) ⊆ X

which is open in X since g◦f is continuous. By definition of T , g−1(U) is indeed open in S.

6

d. Show that T is the smallest topology on S with the property that a map g : S → Y iscontinuous whenever g ◦ f is continuous.

Solution. Let T ′ be a topology on S with said property. We know that f : X → (S, T ) iscontinuous, but it can be written as the composite

Xf−→ (S, T ′) id−→ (S, T ).

By the property of T ′, the composite id ◦ f being continuous guarantees that the identityid : (S, T ′)→ (S, T ) is continuous, i.e. T ≤ T ′.

7

Problem 6. Consider the subset X = {0}∪{ 1n| n ∈ N} ⊂ R viewed as a subspace of the real

line R. As a set, X is the disjoint union of the singletons {0} and { 1n} for all n ∈ N. However,

show that X does not have the coproduct topology on {0} q∐

n∈N{1n}.

Solution. In the coproduct topology on {0} q∐

n∈N{1n} (which happens to be the discrete

topology), the summand {0} is open.

However, in the subspace topology on X, the singleton {0} is not open. Indeed, any open ballBr(0) around 0 will contain other points 1

n∈ Br(0), for all n such that 1

n< r.

8

Math 535 - General Topology

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