Math 54: TopologySyllabus, problems and solutions

Pierre Clare

Dartmouth College, Summer 2015

1

Textbook

[M] Topology (2nd ed.), by J. Munkres.

Effective syllabus

O. Introduction - 4 lectures

O.1 Elementary set theoryO.2 Basics on metric spaces

I. Topological spaces and continuous functions - 16 lectures

I.1. TopologiesI.2. Bases and subbasesI.3. Closed setsI.4. Continuous maps and the category Top.I.5. Topologies on cartesian productsI.6. Metrizable topologies

II. Connectedness - 2 lectures

II.1. Connected spacesII.2. Path connectednessII.3. Connected components

III. Compactness - 5 lectures

III.1. Compact spacesIII.2. Frechet and sequential compactnessIII.3. Local compactness, Alexandrov Compactification

IV. Other topics - 3 lectures

IV.1. Separation axiomsIV.2. The Urysohn Lemma and the Urysohn Metrization TheoremIV.3. Normed linear spacesIV.4. Topological properties of GL(n,R) and GL(n,C)

2

Contents

Content of the lectures 4

Problem sets and examinations 10Problem set 1: review on sets and maps 10Problem set 2: metric spaces 11Problem set 3: topological spaces 12Midterm 1 13Problem set 4: closed sets and limit points 15Problem set 5: continuous maps, the product topology 16Problem set 6: metrizable spaces 17Midterm 2: in-class examination 19Midterm 2: take-home examination 21Problem set 7: connectedness and compactness 23Final examination 24

Elements of solutions for the homework sets and examinations 26

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Week 1

Lecture 1. General introduction. (Why) should we study Topology? What is it?Poincare (1895): Analysis situs.

There is a science called analysis situs and which has for its objectthe study of the positional relations of the different elements ofa figure, apart from their sizes. This geometry is purely qual-itative; its theorems would remain true if the figures, insteadof being exact, were roughly imitated by a child. [...] The impor-tance of analysis situs is enormous and can not be too much emphasized[...].

H. Poincare, The value of Science, 1905.

Intuitive notions of neighborhood and deformation.Fun: example of topological problem: the seven bridges of Konigsberg.

Week 2

Lecture 2. [M, §6-7]Finite sets and cardinality. Examples and comparison of infinite subsets of Z+ (evens,odds, squares, primes). Bijection between two line segments.Countably infinite and countable sets. Examples: Z and Z+ × Z+ are countably infinite.Fun: the set S = n2 , n ∈ Z+ is in bijection with Z. However, there are ‘more’ integers than

squares in the sense that∑k∈Z+

1k =

∞∑n=1

1n diverges while

∑k∈S

1k =

∞∑n=1

1n2 converges.

What about∑

p prime

1p?

Lecture 3. [M, §7]

Characterization of countable sets: B is countable iff Z+surj.−−→ B iff B

inj.−→ Z+.Infinite subset of Z+ are countable. Examples of countable sets:

- Z+ × Z+, Q,...- subsets, countable unions, finite products of countable sets.

There exist uncountable sets: 0, 1ω (diagonal extraction argument) and R for instance.

Week 3

Lecture 4. Notion of distance, metrics. Examples of metric spaces: (R2,Euclidean),(R2,Manhattan), Z with the ordinary metric, (Z, 2-adic). Balls in metric spaces.Continuous functions on R: definition and expression in terms of balls and inverse images.Fun: p-adic metric on Q.

Lecture 5. Open sets in metric spaces. Examples in (R,Euclidean).Theorem [MT]: let (E, d) be a metric space. Then,

- E and ∅ are open;- arbitrary unions of open sets are open;- finite intersections of open sets are open.

4

Continuous maps between metric spaces: definition in terms of balls.Theorem [MC]: a map between metric spaces is continuous if and only if the inverseimage of any open set is an open set.

Lecture 6. [M, §12-13]Topology on a set. Discrete and trivial topologies. Finite complement topology.Reformulation of Theorem [MT]: open sets defined by a metric constitute a topology.Comparison between topologies: notion of finer (stronger) and coarser (weaker) topology.Basis for a topology. Examples: disks and rectangles in R2.

Lecture 7. [M, §13]Topology T (B) generated by a basis B. Basis elements are open (B ⊂ T (B)).Description of T (B) (Lemma 13.1): the open sets of T (B) are the unions of elements of B.Criterion to find a basis of a given topology T on a set X (Lemma 13.2):if a subset C of T is a finer covering1 of X, then C is a basis and generates T , i.e. T (C) = T .Topologies can be compared by comparing bases (Lemma 13.1):T (B′) is finer than T (B) if and only if B′ is a finer covering of X than B.

Week 4

Lecture 8. [M, §13-14]Topology generated by a subbasis.Order topology: definition, examples: R2 and 1, 2 × Z+ with the lexicographical order.Comparison with the Euclidean and the discrete topology respectively.

Lecture 9. [M, §15]The product topology: definition, bases. Projections and cylinders.

Lecture 10. [M, §16]The subspace topology: definition, bases. Restriction commutes to products.The topology of the restricted order may differ from the restricted order topology: casesof X = R and Y1 = [0, 1], Y2 = [0, 1) ∪ 2. They coincide in the case of a convex subset.

Lecture 11. [M, §17]Closed sets: definition, examples. Properties: stability under arbitrary intersections andfinite unions. A topology can be defined by its closed sets. Closed sets in the subspacetopology. Closure and interior of a set, closure in the subspace topology.A topology can be defined by its closure operation:Theorem [Closure] Let X be a set and γ : P(X)→ P(X) a map such that,

- γ(∅) = ∅;- A ⊂ γ(A);- γ(γ(A)) = γ(A);- γ(A ∪B) = γ(A) ∪ γ(B).

Then the family X \ γ(A) , A ∈ P(X) is a topology in which A = γ(A).

1If E and F are families of subsets of X, we say that F is a finer covering of X than E if for every‘x ∈ E ∈ E ’ there exists F ∈ F such that x ∈ F ⊂ E

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Week 5

Lecture 12. [M, §17]Characterization of the closure and accumulation points.

Lecture 13. [M, §17]Theorem [MUL] In a metric space, if a sequence converges, it has a unique limit.Convergent sequences in topological spaces. Uniqueness of limits in Hausdorff (T2) spaces.Characterization of accumulation points in T1 spaces.

Midterm 1. Divisor topology on Z+. Equivalent metrics generate the same topology.Topology generated by the union of two topologies. Examples of subspaces of R, [−1, 1]and R` × Ru.

Post-exam fun: Sperner’s Lemma and consequences, by J. Bass.

Guest lecture. The problem with topology, by James Binkoski (Dartmouth College), anintroduction to T. Maudlin’s New foundations for physical geometry.

Week 6

Lecture 14. [M, §18]Continuous maps between general topological spaces. Any function f : X → Y can bemade continuous by equipping X with the discrete topology or Y with the trivial topology.Characterization at the level of a (sub)basis for the topology on the target space.Characterization in terms of the direct image of the closure, inverse image of closed sets,in terms of neighborhoods.

Lecture 15. (D. Freund) [M, §18]Construction of continuous functions, the Pasting Lemma.

Lecture 16. Homeomorphisms: definition, examples, topological properties.Categories, definitions and examples: Set, Mod(R), Top, ordered sets: N and OpX .Isomorphisms in a category. The isomorphisms in Top are the homeomorphisms.

Lecture 17. [M, §19]Cartesian products of arbitrary indexed families of sets.The box topology and the product topology: definition, comparison, bases.A product of Hausdorff spaces is Hausdorff. The closure of a product is the product ofthe closures. Continuous maps into products.

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Week 7

Lecture 18. [M, §20]Metrizable topologies. Equivalent metrics generate the same topology. Example: theEuclidean and L∞ metrics are equivalent and generate the product topology on Rn.Topologically equivalent metrics need not be equivalent: any metric is topologically equiv-alent to its associated standard bounded metric.Generalization of the L∞ topology: the uniform metric and topology on RJ .

Lecture 19. [M, §20]The uniform topology on RJ is intermediate between the product and box topologies.It is metrizable if J is countable.

Lecture 20. [M, §21]Sequential characterization of closure points and continuous maps in metric spaces.Pointwise and uniform convergence a uniform limit of continuous functions is continuous.Uniform convergence is equivalent to convergence in the uniform topology.Functors between categories. Definition. Examples: For : Top −→ Set, the ‘matrixfunctor’ N −→Modf (R).

Lecture 21. (D. Freund) [M, §23-24]Definition: separations and connected spaces. Examples of connected spaces: R`, Q.Examples of disconnected spaces: (Z, Tf.c.), any space with the trivial topology.Linear continua in the order topology and their intervals are connected.A space is connected if and only if it does not have non-trivial open and closed subsets. Aconnected subspace lies entirely in one component of any separation of the ambient space.The union of connected subspaces with a common point is connected.If A is connected and A ⊆ B ⊆ A, then B is connected.

Week 8

Lecture 22. [M, §23-25]The continuous image of a connected space is connected. Finite products of connectedspaces are connected, R∞ is not connected in the box topology.The Intermediate Value Theorem.Path connectedness is a strictly stronger property than connectedness.Connected components, every space is the disjoint union of its connected components.

Lecture 23. [M, §26]Covers, Borel-Lebesgue definition of compact spaces. Examples: R and (0, 1] are notcompact, 0 ∪ 1/Z+ is.Closed subsets of compacts are compact. Compact subspaces of Hausdorff sets are closed.

Midterm 2. Characterization of T1 spaces. Interior and boundary of (x, y) ∈ R2 , 0 ≤ y < x2 + 1.The metric topology is the coarsest topology making the distance continuous.The box topology on Rω is not metrizable.

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Convergence in the uniform topology is equivalent to uniform convergence.Closure of bounded and finitely supported sequences in the uniform topology.

Post-exam fun: Furstenberg’s proof of the infinitude of primes, by N. Ezroura.Topological groups: translation are homeomorphisms, open subgroups are closed.

Lecture 24. [M, §26]Compact subsets of Hausdorff spaces are compact. Points can be separated from compactsby disjoint open sets in Hausdorff spaces. Continuous images of compact sets are compact.The Tube Lemma. Finite products of compacts are compacts.

Week 9

Lecture 25. [M, §27]Segments in a totally ordered set with the least upper bound property (such as R) arecompact in the order topology. Compact sets of Rn in any topology associated with ametric equivalent to the `∞ metric are exactly the closed and bounded subsets.The Extreme Value Theorem.

Lecture 26. [M, §28]Fechet (limit point) compactness. Compact spaces are Frechet compact.Sequential compactness (Bolzano-Weierstraß property). The three notions are equivalentin metric spaces.

Lecture 27. [M, §29]Locally compact Hausdorff spaces. A punctured compact Hausdorff space is locally com-pact and Hausdorff. Alexandrov compactification of locally compact Hausdorff spaces.

Lecture 28. [M, §30-34]Generalization of sequential characterizations: nets in topological spaces.Regular and normal spaces. First and second countability: definitions, examples andcharacterization in terms of closure of neighborhoods. Sequential characterizations ofclosure points and continuity in first countable spaces. Regular second countable spaces,metrizable spaces and compact Hausdorff spaces are normal.The Urysohn Lemma and the Urysohn Metrization Theorem.

Week 10

Lecture 29. Topology of matrix spaces I.Normed linear spaces over k = R or C, metrics associated with norms. Equivalent normsinduce equivalent metrics. In finite dimension, all norms are equivalent.Exercise: characterizations of continuity for linear maps.Case of Mn(k) ⊂ kn

2: operator norms are submultiplicative. Continuity of operations: lin-

ear combinations, products, determinants, transposition, comatrices. The group GL(n, k)of invertible elements in Mn(k) is a topological group. It is open in Mn(k).

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Lecture 30. Topology of matrix spaces II.Density of GL(n, k) in Mn(k) and applications: AB and BA have the same characteristicpolynomial hence the same eigenvalues for any A, B in Mn(k); there exists a basis ofMn(k) that consists only of invertible matrices.Connectedness properties: GL(n,R) is not connected but GL(n,C) is path connected.

- End of the course -

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Problem set 1: review on sets and maps

Solution p.26

(1) Inverse maps.

a. Show that a map with a left (resp. right) inverse is injective (resp. surjec-tive).

b. Give an example of a function that has a left inverse but no right inverse.

c. Give an example of a function that has a right inverse but no left inverse.

d. Can a function have more than one right inverse? More than one left inverse?

e. Show that if f has both a left inverse g and a right inverse h, then f isbijective and g = f−1 = h.

(2) Let X be a non-empty set and m,n ∈ Z+.

a. If m ≤ n, find an injective map f : Xm −→ Xn.

b. Find a bijective map g : Xm ×Xn −→ Xm+n.

c. Find an injective map h : Xn −→ Xω.

d. Find a bijective map i : Xn ×Xω −→ Xω.

e. Find a bijective map j : Xω ×Xω −→ Xω.

f. If A ⊂ B, find an injective map k : (Aω)n −→ Bω.

(3) If A×B, does it follow that A and B are finite?

(4) If A and B are finite, show that the set F of all functions from A to B is finite.

10

Problem set 2: metric spaces

Solution p.28

(1) Balls. No proof is required for this problem.a. Consider Z× Z equipped with the Euclidean metric.

Describe B((3, 2),

√2)

and Bc((3, 2),

√2).

b. Let X be a set equipped with the discrete metric and x a point in X.Describe the balls B(x, r) for all r > 0.

(2) Continuous maps.a. Prove that the map f defined on R by f(x) = x2 + 1 is continuous.

b. Let (E1, d1), (E2, d2), (E3, d3) be metric spaces and u : E2 → E3, v : E1 → E2

be continuous maps. Prove that u v is continuous.

(3) Let (E, d) be a metric space. Prove that a subset Ω ⊂ E is open if and only iffor every point x ∈ Ω, there exists an open ball containing x and included in Ω.

(4) Let (E, d) be a metric space and A ⊂ E a subset. A point a in A is called interiorif there exists r > 0 such that any point x in E such that d(a, x) < r is in A.

The set of interior points of A is called the interior of A and denoted byo

A.

a. Prove thato

A is the union of all the open balls contained in A.

b. Prove thato

A is the largest open subset contained in A.

c. Cano

A be empty if A is not?

11

Problem set 3: topological spaces

Solution p.30

(1) Let Tαα∈A be a family of topologies on a non-empty set X.

a. Prove that I =⋂α∈A

Tα is a topology on X.

b. Prove that I is the finest topology that is coarser than each Tα.

(2) Let p be a prime number. Consider for n ∈ Z and a a positive integer,

Ba(n) = n+ λpa , λ ∈ Z.a. Show that the family B = Ba(n) , n ∈ Z , a ∈ Z+ is a basis for a topology.

b. Is the topology generated by B discrete?

(3) Compare the following topologies on R:- T1: the standard topology;- T2: the K-topology, with basis elements of the form

(a, b) and (a, b) \

1

n, n ∈ Z+

;

- T3: the finite complement topology;- T4: the upper limit topology, with basis elements of the form (a, b];- T5: the topology with basis elements of the form (−∞, a).

(4) Let L be a straight line in R2. Describe the topology L inherits as a subspace ofR` × R and as a subspace of R` × R`.

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Midterm 1

Solution p.32

This exam consists of 4 independent problems. Treat them in the order of your choosing,starting each problem on a new page.

Every claim you make must be fully justified or quoted as a result from the textbook.

Problem 1

Let T be the family of subsets U of Z+ satisfying the following property:

If n is in U , then any divisor of n belongs to U .

1. Give two different examples of elements of T containing 24 (not including Z+).

2. Verify that T is a topology on Z+.

3. Is T the discrete topology?

Problem 2

Let (E, d) be a metric space.

1. Recall the definition of the metric topology and prove that open balls form a basis.

2. Assume that ρ is a second metric on E such that, for every x, y ∈ E,

1

2d(x, y) ≤ ρ(x, y) ≤ 2d(x, y).

Compare the topologies generated by d and ρ.

13

Problem 3

Let T1 and T2 be topologies on a set X.

1. Verify that T1 ∪ T2 is a subbasis for a topology.

From now on, T1 ∨ T2 denotes the topology generated by T1 ∪ T2.

2. Describe T1 ∨ T2 when T1 is coarser than T2.

3. Compare T1 ∨ T2 with T1 and T2 in general.

4. Let T be a topology on X that is finer than T1 and T2.Prove that T is finer than T1 ∨ T2.

Problem 4

1. Consider the set Y = [−1, 1] as a subspace of R. Which of the following sets are openin Y ? Which are open in R?

A =

x ,

1

2< |x| < 1

B =

x ,

1

2< |x| ≤ 1

C =

x ,

1

2≤ |x| < 1

D =

x , 0 < |x| < 1 and

1

x∈ Z+

2. Let X = R` × Ru where R` denotes the topology with basis consisting of all intervalsof the form [a, b) and Ru denotes the topology with basis consisting of all intervals of theform (c, d].

Describe the topology induced on the plane curve Γ with equation y = ex.

14

Problem set 4: closed sets and limit points

Solution p.35

(1) Prove the following result:Theorem Let X be a set and γ : P(X) → P(X) a map such that, for anyA,B ⊂ X,(i) γ(∅) = ∅;

(ii) A ⊂ γ(A);(iii) γ(γ(A)) = γ(A);(iv) γ(A ∪B) = γ(A) ∪ γ(B).Then the family X \ γ(A) , A ∈ P(X) is a topology in which A = γ(A).

Hint: it might be useful to prove that A ⊂ B ⇒ γ(A) ⊂ γ(B).

(2) The questions in this problem are independent.(a) Show that a topological space X is Hausdorff if and only if the diagonal

∆ = (x, x) , x ∈ X is closed in X ×X.

(b) Determine the accumulation points of the subset

1m

+ 1n, m, n ∈ Z+

of R.

(3) The boundary of a subset A in a topological space X is defined by

∂A = A ∩X \ A.(a) Show that A = A t ∂A2.

(b) Show that ∂A = ∅ if and only if A is open and closed.

(c) Show that U is open if and only if ∂U = U \ U .

(d) If U is open, is it true that U = ˚U?

(4) Find the boundary and interior of each of the following subsets of R2.(a) A = (x, y) , y = 0(b) B = (x, y) , x > 0 and y 6= 0(c) C = A ∪B(d) D = Q× R(e) E = (x, y) , 0 < x2 − y2 ≤ 1(f) F = (x, y) , x 6= 0 and y ≤ 1

x

2The disjoint union symbol t is used to indicate that the sets in the union have empty intersection.15

Problem set 5: continuous maps, the product topology

Solution p.40

(1) (a) Consider Z+ equipped with the topology in which open sets are the subsetsU such that if n is in U , then any divisor of n belongs to U . Give a necessaryand sufficient condition for a function f : Z+ −→ Z+ to be continuous.

(b) Let χQ be the indicator of Q. Prove that the map ϕ : R −→ R defined byϕ(x) = x · χQ(x) is continuous at exactly one point.

(2) Let X and Y be topological spaces. If A is a subset of either, we denote by A′

the sets of accumulation points of A and by ∂A its boundary.Let f : X −→ Y be a map. Determine the implications between the followingstatements:

(i) f is continuous.

(ii) f(A′) ⊂ (f(A))′ for any A ⊂ X.

(iii) ∂(f−1(B)) ⊂ f−1(∂B) for any B ⊂ Y .

(3) Let X and Y be topological spaces, and assume Y Hausdorff. Let A be a subsetof X and f1, f2 continuous maps from the closure A to Y .Prove that if f1 and f2 restrict to the same function f : A→ Y , then f1 = f2.

(4) Let Xαα∈J be a family of topological spaces and X =∏α∈J

Xα.

(a) Give a necessary and sufficient condition for a sequence unn∈Z+ to convergein X equipped with the product topology.

(b) Does the result hold if X is equipped with the box topology?

16

Problem set 6: metrizable spaces

Solution p.43

(1) Let ρ be the uniform metric on Rω. For x = (xn)n∈Z+ ∈ Rω and 0 < ε < 1, let

P (x, ε) =∏n∈Z+

(xn − ε, xn + ε).

(a) Compare P (x, ε) with Bρ(x, ε).(b) Is P (x, ε) open in the uniform topology?

(c) Show that Bρ(x, ε) =⋃δ<ε

P (x, δ).

(2) We denote by `2(Z+) the set of square-summable real-valued sequences, that is,

`2(Z+) =

x = (xn)n∈Z+ ∈ Rω ,

∑n≥1

x2n converges

.

We admit that the formula

d(x, y) =

(∑n≥1

(xn − yn)2

)1/2

defines a metric on `2(Z+).

(a) Compare the metric topology induced by d on `2(Z+) with the restrictionsof the box and uniform topologies from Rω.

(b) Let R∞ denote the subset of `2(Z+) consisting of sequences that have finitelymany non-zero terms. Determine the closure of R∞ in `2(Z+).

(3) Let X be a topological space, Y a metric space and assume that (fn)n≥0 is asequence of continuous functions that converges uniformly to f : X −→ Y .Let (xn)n≥0 be a sequence in X such that lim

n→∞xn = x. Prove that

limn→∞

fn(xn) = f(x).

17

(4) Ultrametric spaces (non-mandatory).

Let X be a set equipped with a map d : X×X −→ R such that for all x, y, z ∈ X,(1) d(x, y) ≥ 0(2) d(x, y) = d(y, x)(3) d(x, y) = 0⇔ x = y(4) d(x, z) ≤ max (d(x, y), d(y, z))

(a) Prove that d is a distance.(b) Let B be an open ball for d. Prove that B = B(y, r) for every element y ∈ B

for some r > 0.(c) Prove that closed balls are open and open balls are closed in the topology

induced by d.

18

Midterm 2: in-class examination

Solution p.47

This exam consists of 5 independent problems. You may treat them in the order of yourchoosing, starting each problem on a new page.

Every claim you make must be fully justified or quoted as a result from the textbook.

Problem 1

1. Show that a topological space is T1 if and only if for any pair of distinct points, eachhas a neighborhood that does not contain the other3.

2. Determine the interior and the boundary of the set

Ξ =

(x, y) ∈ R2 , 0 ≤ y < x2 + 1

where R2 is equipped with its ordinary Euclidean topology.

Problem 2

Let E be a set with a metric d and Td the corresponding metric topology on E.

1. Prove that the map d : (E, Td)× (E, Td) −→ R is continuous.

2. Let T be a topology on E, such that d : (E, T )× (E, T ) −→ R is continuous.Prove that T is finer than Td.

Hint: it might be helpful to consider sets of the form d−1 ((−∞, r)) for r > 0.

3A topological space is said T1 if all its singletons are closed.19

Problem 3

The purpose of this problem is to prove that the box topology on Rω is not metrizable.

1. Recall the definition of the box topology on Rω.

Denote by 0 the sequence constantly equal to 0 and let

P = (0,+∞)ω =∏n≥1

(0,+∞)

be the subset of positive sequences.

2. Verify that 0 belongs to P .

3. Prove that no sequence (pn)n≥1 ∈ P ω converges to 0 in the box topology.

4. Conclude.

Problem 4

1. Let X be a set.

(a) Recall the definition of the uniform topology on RX .

(b) Recall the definition of uniform convergence for a sequence of functions fn in RX .

2. Prove that a sequence in RX converges uniformly if and only if it converges for theuniform topology.

Problem 5

Consider the space Rω of real-valued sequences, equipped with the uniform topology.

1. Prove that the subset B of bounded sequences is closed in Rω for the uniform topology.

2. Let R∞ denote the subset of sequences with finitely many non-zero terms.Determine the closure of R∞ in Rω for the uniform topology.

20

Midterm 2: take-home examination

Solution p.51

This is an individual assignment. You may use the text and class notes but no othersource or outside help.

Problem 1

1. Determine the connected components of Rω in the product topology.

2. Consider Rω equipped with the uniform topology.

(a) Prove that x is in the same connected component as 0 if and only if x is bounded.

(b) Deduce a necessary and sufficient condition for x and y in Rω to lie in the sameconnected component for the uniform topology.

3. Consider Rω equipped with the box topology.

(a) Let x, y ∈ Rω be such that x− y ∈ Rω \ R∞.Prove that there exists a homeomorphism

ϕ : Rω −→ Rω

such that (ϕ(x)n)n∈Z+ is a bounded sequence and (ϕ(y)n)n∈Z+ is unbounded.

Hint: given u ∈ Rω, it might be helpful to consider the sequence v defined by

vn =

un − xn if xn = ynun−xnyn−xn if xn 6= yn

.

(b) Deduce a necessary and sufficient condition for x and y in Rω to lie in the sameconnected component for the box topology.

21

Problem 2

Let F be a functor between categories C and C ′. A functor G : C ′ −→ C is said to be aleft adjoint for F if there is a natural isomorphism

HomC(G(X), Y ) ∼= HomC′(X,F (Y ))

for all objects X ∈ C ′ and Y ∈ C. Similarly, G is called a right adjoint for F if there is anatural isomorphism

HomC(X,G(Y )) ∼= HomC′(F (X), Y )

for all objects X ∈ C and Y ∈ C ′.

Recall that the forgetful functor F : Top −→ Set is defined by

- F ((X, T )) = X for any set X equipped with a topology T ;

- F(f) = f for any continuous map f : X −→ Y .

If X is a set, let G(X) denote the topological space obtained by endowing X with thetrivial topology Ttriv. = X, ∅:

G(X) = (X, Ttriv.).

If f is a map between sets, define in addition G(f) = f .

1. Verify that G is a functor.

2. Prove that G is a right adjoint to F.

3. Find a left adjoint for F.

22

Problem set 7: connectedness and compactness

Solution p.55

(1) Let U be an open connected subspace of R2 and a ∈ U .

(a) Prove that the set of points x ∈ U such that there is a path γ : [0, 1] −→ Uwith γ(0) = a and γ(1) = x is open and closed in U .

(b) What can you conclude?

(2) Let X be a topological space and Y ⊂ X a connected subspace.

(a) Are Y and ∂Y necessarily connected?

(b) Does the converse hold?

(3) Let (E, d) be a metric space.

(a) Prove that every compact subspace of E is closed and bounded.

(b) Give an example of metric space in which closed bounded sets are not nec-essarily compact.

(4) (a) Prove that the Alexandrov compactification of R is homeomorphic to theunit circle

S1 = (x, y) ∈ R2 , x2 + y2 = 1.

(b) Verify that Z+ ⊂ R is a locally compact Hausdorff space.

(c) Prove that the Alexandrov compactification of Z+ is homeomorphic to1

n, n ∈ Z+

∪ 0.

23

Final examination

Solution p.58

This exam consists of 6 independent problems. Your may treat them in the order of yourchoosing, starting each problem on a new page.

Problem 1

1. Let X be a Hausdorff space and K1, K2 disjoint compact subsets of X.Prove that there exist disjoint open sets U1 and U2 such that K1 ⊂ U1 and K2 ⊂ U2.

2. Let X be a discrete topological space. Describe the compact subsets of X.

Problem 2

A topological space is said totally disconnected if its only connected subspaces are single-tons.

1. Prove that a discrete space is totally disconnected.

2. Does the converse hold?

Problem 3

Let Xαα∈J be a family of topological spaces; let Aα ⊂ Xα for each α ∈ J .

1. In∏

α∈J Xα equipped with the product topology, prove that∏α∈J

Aα =∏α∈J

Aα.

2. Does the result hold if∏

α∈J Xα carries the box topology?

24

Problem 4

Is R homeomorphic to R2 ?

Problem 5

Let (E, d) be a metric space. An isometry of E is a map f : E −→ E such that

d (f(x), f(y)) = d(x, y)

for all x, y ∈ E.

1. Prove that any isometry is continuous and injective.

Assume from now on that E is compact and f an isometry. We want to prove that f issurjective. Assume to the contrary the existence of a /∈ f(E).

2. Prove that there exists ε > 0 such that B(a, ε) ⊂ E \ f(E).

3. Consider the sequence defined by x1 = a and xn+1 = f(xn). Prove that

d(xn, xm) ≥ ε

for n 6= m and derive a contradiction.

4. Prove that an isometry of a compact metric space is a homeomorphism.

Problem 6

Let X be a set, P(X) the set of subsets of X and ι : P(X) −→P(X) a map satisfying:

(1) ι(X) = X

(2) ι(A) ⊂ A

(3) ι ι(A) = ι(A)

(4) ι(A ∩B) = ι(A) ∩ ι(B)

for all A,B ⊂ X.

1. Check that A ⊂ B ⇒ ι(A) ⊂ ι(B) for A,B ⊂ X.

2. Prove that the family T = ι(A) , A ∈P(X) is a topology on X.

3. Prove that, in this topology, A = ι(A) for all A ⊂ X.

25

Problem set 1: review on sets and maps - Elements of solution

This is supposedly well-known material. We simply indicate solutions and hints.

(1) Inverse maps.

a. Show that a map with a left (resp. right) inverse is injective (resp.surjective).

If f g = Id, then g(x1) = g(x2) ⇒ x1 = f(g(x1)) = f(g(x2)) = x2 so g isinjective and y = f(g(y)) for all y so f is surjective.

b. Give an example of function with a left inverse but no right inverse.c. Give an example of function with a right inverse but no left inverse.

Consider a two-point set a, b and a singleton x. Then ϕa : x 7→ a andϕb : x 7→ b are both right inverses to the only map ψ from a, b to x,which is not injective hence cannot have a left inverse.Since, ψ ϕa = ψ ϕb = Idx, both ϕa and ϕb have a left inverse, but theyhave no right inverse as the only candidate would be ψ and ϕa ψ 6= Ida,b.

d. Can a function have more than one right inverse? More than oneleft inverse?

The function ψ defined above has multiple right inverses.Consider the map λ : a, b −→ u, v, w defined by λ(a) = u and λ(b) = v.The maps µa, µb : u, v, w −→ a, b respectively defined by

µa(u) = µb(u) = a , µa(v) = µb(v) = b

andµa(w) = a , µb(w) = b

are both left inverses of λ.

e. Show that if f has both a left inverse g and a right inverse h, thenf is bijective and g = f−1 = h.

It follows from a. that f is bijective. Therefore, g f = Id implies g = f−1

while f h = Id implies h = f−1.

26

(2) Let X be a non-empty set and m,n ∈ Z+.a. If m ≤ n, find an injective map f : Xm −→ Xn.

Consider f(x,1 , . . . , xm) = (x1, . . . , xm,

n−m︷ ︸︸ ︷x0, . . . , x0) for some x0 ∈ X.

b. Find a bijective map g : Xm ×Xn −→ Xm+n.

Consider g ((x1, . . . , xm) , (y1, . . . , yn)) = (x1, . . . , xm, y1, . . . , yn)

c. Find an injective map h : Xm −→ Xω.d. Find a bijective map i : Xm ×Xω −→ Xω.

Consider ‘infinite versions’ of the functions f and g defined above.

e. Find a bijective map j : Xω ×Xω −→ Xω.

Consider j :((x`)`∈Z+ , (y`)`∈Z+

)7−→ (z`)`∈Z+ with z2`−1 = x` and z2` = y`.

f. If A ⊂ B, find an injective map k : (Aω)n −→ Bω.

Consider k :((1x`)`∈Z+ , . . . , (

nx`)`∈Z+

)7→ (1x1, . . . ,

nx1, , . . . ,1x`, . . . ,

nx`, . . .).

(3) If A×B, does it follow that A and B are finite?

No: ∅× Z+ = ∅. However, if both A and B are non-empty, and one of them isinfinite, say B, then A×B contains some a ×B which is in bijection with B,hence infinite.

(4) If A and B are finite, show that the set F = f : A −→ B is finite.Let A = a1, . . . , an. Then f 7−→ (f(a1), . . . , f(an)) is a bijection from F to Bn,which has cardinal |B|n, hence is finite.

27

Problem set 2: metric spaces - Elements of solution

(1) a. Consider Z× Z equipped with the Euclidean metric.Describe B

((3, 2),

√2)

and Bc((3, 2),

√2).

One can enumerate the elements:

B(

(3, 2),√

2)

= (2, 2); (3, 1); (3, 2); (3, 3); (4, 2).

and

Bc(

(3, 2),√

2)

= B(

(3, 2),√

2)∪ (2, 1); (2, 3); (4, 1); (4, 3)

b. Let X be a set equipped with the discrete metric and x in X.Describe the balls B(x, r) for all r > 0.

By definition, B(x, r) = x for 0 < r ≤ 1 and B(x, r) = X for r > 1.

(2) Continuous maps.a. Prove that the map f defined on R by f(x) = x2 + 1 is continuous.

Let a ∈ R and ε > 0. Note that if a− 1 ≤ x ≤ a+ 1, then |x+ a| ≤ 2|a|+ 1.Therefore, since |f(x)− f(a)| = |x− a||x+ a|, we get, for x ∈ [a− 1, a+ 1],

|f(x)− f(a)| ≤ |x− a|(2|a|+ 1)

and it suffices to choose |x−a| < min ε2|a|+1

, 1 to guarantee |f(x)−f(a)| < ε.

b. Let E1, E2, E3 be metric spaces and u : E2 → E3, v : E1 → E2 becontinuous maps. Prove that u v is continuous.

Let Ω be open in E3 and apply Theorem [MC] twice: u−1(Ω) is open in E2

by continuity of u and (u v)−1(Ω) = v−1(u−1(Ω)) is open by continuity ofv.

(3) Let (E, d) be a metric space. Prove that a subset Ω ⊂ E is open if andonly if for every point x ∈ Ω, there exists an open ball containing x andincluded in Ω.

The definition seen in class for open sets in a metric space differs only by the factthat it requires the ball to be centered at the point considered. Therefore, opensets trivially satisfy the property.Observe that if a point x is included in a ball B(a, r), the triangle inequalityimplies that B(x, r − d(a, x)) is included in B(a, r). The converse follows.

28

(4) Let (E, d) be a metric space and A ⊂ E. A point a in A is called interiorif there exists r > 0 such that any point x in E such that d(a, x) < r is

in A. The seto

A of interior points of A is called the interior of A.

a. Prove thato

A is the union of all the open balls contained in A.

Let A be the union of all the open balls contained in A and let a be in A.By definition of A and the argument used in (3), there exists a ball B(a, r)

included in A, so A ⊂o

A. Conversely, let a be ino

A. By definition, thereexists r > 0 such that B(a, r) ⊂ A so a ∈ A, hence the result.

b. Prove thato

A is the largest open subset contained in A.

First,o

A is open as the union of open subsets, as proved in a. We argue by

contradiction: assume the existence of Ω open such thato

A Ω ⊂ A. Let

x ∈ Ω\o

A. Since x /∈o

A, no ball B(x, r) with r > 0 is contained in A. Since Ωis open, it must contain such a ball, which contradicts the assumption thatΩ ⊂ A.

c. Cano

A be empty if A is not?

Yes: consider for instance A = Z in E = R, or any strict linear subspace ofRn and observe that every ball centered at the origin must contain a basis.

29

Problem set 3: topological spaces - Elements of solution

(1) Let Tαα∈A be a family of topologies on a non-empty set X.

a. Prove that I =⋂α∈A

Tα is a topology on X.

Direct verification, using the definition and the fact that I ⊂ Tα for each α.

b. Prove that I is the finest topology that is coarser than each Tα.By construction, I is coarser than (included in) each Tα. Let J be a finertopology than I and consider U ∈ J \ I. Since U /∈ I, there exists someα0 ∈ A such that U /∈ Tα0 , so that J cannot be coarser than Tα0 .

(2) Let p be a prime number. Consider for n ∈ Z and a a positive integer,

Ba(n) = n+ λpa , λ ∈ Z.

a. Show that B = Ba(n) , n ∈ Z , a ∈ Z+ is a basis for a topology.

Every n ∈ Z is in Ba(n) so the covering property is satisfied. The otherproperty follows from the observation that if Ba(n)∩Bb(m) is either emptyof of the form Bmax(a,b)(q).

b. Is the topology generated by B discrete?

No: no finite subset of Z is open in this topology.

(3) Compare the following topologies on R:- T1: the standard topology;- T2: the K-topology, with basis elements of the form (a, b) and (a, b)\

1n, n ∈ Z+

;

- T3: the finite complement topology;- T4: the upper limit topology, with basis elements of the form (a, b];- T5: the topology with basis elements of the form (−∞, a).

Comparison of T1 with T2, T3, T4, T5

It is proved in Lemma 13.4 that T1 T2.To compare T1 with T3, observe that any subset of R with finite complement isof the form (−∞, x1)∪ (x1, x2)∪ . . .∪ (xn,+∞), that is a union of basis elementsfor T1. Therefore, T3 ⊂ T1 and the inclusion is strict since any union of finitelycomplemented set is finitely complemented: T3 T1.The argument given in Lemma 13.4 for the lower limit topology can be adaptedto show that T1 T4.Finally, T5 ⊂ T1 follows from the observation that (−∞, a) =

⋃z<a(z, a). The

inclusion is strict since no interval of the form (a, b) with a finite lies in T5.30

Comparison of T2 with T3, T4, T5

By transitivity of inclusions, T3 T2 and T5 T2.We shall now prove that T2 T4 (compare to Lemma 13.4 and Problem 13.6).The inclusion relies on the observation that (a, b) =

⋃a<β<b(a, β], which implies

that both types of basis elements for T2 are union of basis elements of T4. Theinclusion is strict: consider for instance (0, 2] ∈ T4. No basis element of T2 cancontain 2 and be included in (0, 2].

Comparison of T3 with T4, T5

By transitivity again, T3 T4.There is no inclusion relation between T3 and T5: as observed before, no unionof basis elements of T3 can have infinite complement as (−∞, 0) ∈ T5 does, soT5 6⊂ T3. Conversely, R \ 0 = (−∞, 0) ∪ (0,+∞) is a basis element for T3 thatdoes not belong to T5 in which all open sets are convex.

Comparison of T4 with T5

By transitivity, T5 T4.

The relations are summed up in the following diagram, in which all the inclusions

are strict:T5

∩T3 ⊂ T1 ⊂ T2 ⊂ T4

(4) Let L be a straight line in R2. Describe the topology L inherits as asubspace of R` × R and as a subspace of R` × R`.Lemma 16.1 states that a basis for the topology in question is given by consideringintersections of L with basis elements of the chosen topology on R2.

- In the case of L ⊂ R`×R, they are of the form L∩ [a, b)× (c, d). Identifying4

L with R, we see that these subsets are of the form (α, β) if L is verticaland include those of the form [α, β) otherwise. Therefore, if L is vertical,the subspace topology is the standard topology on the real line, while if L isslanted, it carries the lower limit topology.

- In the case of L ⊂ R` × R, the same line of reasoning shows that L receivesthe lower limit topology if it is vertical or has a non-negative slope. Ifthe slope is negative, the sets L ∩ [a, b) × (c, d) include those of the form[α, β] and conversely, any segment can be obtained as such an intersection.Since singletons are special cases (α = [α, α]), it follows that the inducedtopology on L is discrete.

4A rigorous way to do this will be explained later in the course.31

Midterm 1 - Elements of solution

[M] refers to Topology, 2nd ed. by J. Munkres.

Problem 1

Let T be the family of subsets U of Z+ satisfying the following property:

If n is in U , then any divisor of n belongs to U .

1. Give two different examples of elements of T containing 24.

The prime factorization of 24 is 24 = 23 · 3 so divisors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.Any element of T containing 24 must contain all of these. If we adjoin another number tothis list, we must also include all of its divisors. For instance, any open set containing 10will also contain 5. Examples are 1, 2, 3, 4, 5, 6, 8, 10, 12, 24 or 1, 2, 3, 4, 6, 8, 12, 17, 24.

2. Verify that T is a topology on Z+.

We verify the three axioms. (O1) The empty set trivially belongs to T and Z+ containsthe divisors of any integer so it is open too.(O2) Let Uαα∈A be a family of elements of T and U =

⋃α∈A Uα. Let n ∈ U . There

exists α0 ∈ A such that n ∈ Uα0 so all the divisors of n belong to Uα0 open, hence to U .This means that U is open by definition.

(O3) Let Ui1≤i≤p be a family of elements of T and U =p⋂i=1

Ui. Let n ∈ U . For every

1 ≤ i ≤ p, the integer n belongs to Ui so every divisor of n belongs to all the Ui’s, henceto U which is therefore open. This proves that T is a topology.

3. Is T the discrete topology?

Since 1 is a divisor of any integer, every non-empty open set must contain 1. Therefore,there exists subsets of Z+ that are not open and T is not discrete.

Problem 2

Let (E, d) be a metric space.

1. Recall the definition of the metric topology and prove that open balls forma basis for that topology.

A subset Ω of E is open if any point of Ω is contained in an open ball included in Ω. By[M, Lemma 13.2], this implies that balls form a basis for the metric topology.

32

2. Assume that ρ is a second metric on E such that, for every x, y ∈ E,

1

2d(x, y) ≤ ρ(x, y) ≤ 2d(x, y).

Compare the topologies generated by d and ρ.

Denote by Td and Tρ the topologies associated with the metrics d and ρ. We shall provethat Tρ is finer than Td.Let Bd(a, r) be a ball and x ∈ Bd(a, r). Since Bd(a, r) is open for Td, there exists a radiusr′ > 0 such that Bd(x, r

′) ⊂ Bd(a, r). The inequalities satisfied by d and ρ imply that

x ∈ Bρ(x,r′

2) ⊂ Bd(x, r

′).

Indeed, if ρ(x, y) <r′

2, then

1

2d(x, y) <

r′

2so y ∈ Bd(x, r

′).

Lemma 13.3 in [M] then implies that Td ⊂ Tρ. The converse inclusion can be proved bythe same argument, using the other inequality satisfied by ρ and d and we can concludethat Td = Tρ.

Problem 3

Let T1 and T2 be topologies on a set X.

1. Verify that T1 ∪ T2 is a subbasis for a topology.

Any x ∈ X is included in an element of T1 (e.g X) hence of T1 ∪ T2, which is therefore asubbasis for a topology.

From now on, T1 ∨ T2 denotes the topology generated by T1 ∪ T2.

2. Describe T1 ∨ T2 when T1 is coarser than T2.

If T1 ⊂ T2, then T1 ∪ T2 = T2 so T1 ∨ T2 is the topology generated by T2, that is, T2 itself.

3. Compare T1 ∨ T2 with T1 and T2 in general.

By definition, T1 ∨ T2 contains T1 and T2 so it is finer than both.

4. Let T be a finer topology than T1 and T2. Prove that T is finer than T1 ∨T2.

Any element of T1∨T2 is the union of subsets of X of the form U =n⋂i=1

Ui with Ui ∈ T1∪T2

for each i. Such a U is an intersection of elements of T1 and of T2, all of which belongto T assumed finer so U belongs to T . Since T is stable under unions, it follows thatT1 ∨ T2 ⊂ T . In other words, T1 ∨ T2 is the coarsest topology containing T1 and T2

33

Problem 4

1. Consider the set Y = [−1, 1] as a subspace of R. Which of the following setsare open in Y ? Which are open in R?

A =x , 1

2< |x| < 1

= (−1,−1

2)∪(1

2, 1) is open in R as the union of two basis elements.

It is also open in Y by definition.

B =x , 1

2< |x| ≤ 1

= [−1,−1

2) ∪ (1

2, 1] is not open in R since no open interval con-

taining 1 is included in B. It is open in Y as the intersection of Y with (−2,−12)∪ (1

2, 2),

open in R.

C =x , 1

2≤ |x| < 1

= (−1,−1

2] ∪ [1

2, 1) is not open in R since no open interval con-

taining 12

is included in C. It is not open in Y for the same reason: no intersection of Y

with an open interval containing 12

is included in C.

D =x , 0 < |x| < 1 and 1

x∈ Z+

=

1n, n ∈ Z+ , n ≥ 2

is neither open in R nor in

Y . The argument used for C can be used without modification.

2. Let X = R` × Ru where R` denotes the topology with basis consisting of allintervals of the form [a, b) and Ru denotes the topology with basis consistingof all intervals of the form (c, d]. Describe the topology induced on the planecurve Γ with equation y = ex.

A basis for the subspace topology on Γ is given by the sets [a, b) × (c, d] ∩ Γ. Singletonsare of this form: (x, ex) = [x, x+1)× (ex−1, ex]∩Γ so the induced topology is discrete.

34

Problem set 4: closed sets and limit points - Elements of solution

(1) Prove the following result:Theorem Let X be a set and γ : P(X)→ P(X) a map such that(i) γ(∅) = ∅;(ii) A ⊂ γ(A);

(iii) γ(γ(A)) = γ(A);(iv) γ(A ∪B) = γ(A) ∪ γ(B).Then the family T = X \γ(A) , A ⊂ X is a topology in which A = γ(A).

First, we prove that A ⊂ B ⇒ γ(A) ⊂ γ(B). To do so, observe that A ⊂ B is

equivalent to A ∪B = B. Therefore, γ(B) = γ(A ∪B)(iv)= γ(A) ∪ γ(B) ⊃ γ(A).

(O1) The subset X = X \ ∅ (i)= X \ γ(∅) is in T . Moreover, (ii) implies that

X = γ(X) so ∅ = X \ γ(X) is also in T .

(O2) Let Uαα∈J be a family such that Uα = X \ γ(Aα) for each α ∈ J , andU =

⋃α∈J Aα. De Morgan’s Laws imply that

X \ U =⋂α∈J

Aα

and we want to prove that this set is of the form γ(B) for some subset B of X.Since

⋂α∈J γ(Aα) ⊂ γ(Aα) for all α ∈ J , and γ preserves inclusions, we get, for

all α ∈ J ,

γ(X \ U) ⊂ γ(γ(Aα))(iii)= γ(Aα)

so that γ(X \ U) ⊂⋂α∈J γ(Aα) = X \ U , the reverse inclusion is guaranteed by

(ii), hence X \ U = γ(X \ U), that is,

U = X \ γ(X \ U)

and T is stable under arbitrary unions.

(O3) Let Ui = X \γ(Ai)1≤i≤n be a finite family of elements of T . De Morgan’sLaws imply that

X \n⋂i=1

Ui = X \n⋂i=1

γ(Ai) = X \ γ

(n⋂i=1

Ai

)where the last equality follows from (iv) by induction. This shows that T isstable under finite intersections, which concludes the proof that it is a topologyon X.

Let A be a subset of X. Then γ(A) is closed by definition of T and A ⊂ γ(A)by (ii) so A ⊂ γ(A). Conversely, observe that X \ A, being open, is of the formX \ γ(B), that is, A = γ(B) for some B ⊂ X. Since A ⊂ A, and γ preservesinclusions, it follows that

γ(A) ⊂ γ(A) = γ(γ(B))(iii)= γ(B) = A,

hence γ(A) = A.35

(2) (a) Show that a topological space X is Hausdorff if and only if thediagonal ∆ = (x, x) , x ∈ X is closed in X ×X.

A key observation is that for A and B subsets of X, the condition A∩B = ∅is equivalent to (A×B) ∩∆ = ∅.Now, assume X Hausdorff and let (x, y) ∈ (X ×X) \∆. Since x 6= y, thereexist disjoint open sets Ux 3 x and Uy 3 y. By definition of the producttopology, U = Ux × Uy is a neighborhood of (x, y) and by the preliminaryobservation, U ∩∆ = ∅ so X ×X \∆ is open hence ∆ is closed.Conversely, assume that ∆ is closed and let x 6= y in X. Since (x, y) belongsto (X ×X) \∆ assumed open, there exists a neighborhood V of (x, y) suchthat V ∩ ∆ = ∅. Product of open sets form a basis for the topology ofX×X, so there exist open sets U1 and U2 such that (x, y) ∈ U1×U2 ⊂ V so(U1 × U2) ∩∆ = ∅ which, by the preliminary observation again, guaranteesthat U1 and U2 are disjoint neighborhoods of x and y respectively.

(b) Determine the accumulation points of A =

1m

+ 1n, m, n ∈ Z+

⊂ R.

Let A′ denote the set of accumulation points of A. The fact that limn→∞1n

=

0 implies that

1p, p ∈ Z+

∪0 ⊂ A′. Let us prove the converse inclusion.

First, observe that if an interval (a, b) with a > 0 contains infinitely manyelements of the form 1

m+ 1

n, then one of the variables m and n must take

only finitely many values, while the other takes infinitely many values. Nowlet x ∈ A′ with x > 0. For any ε > 0, the set Bε = (x − ε, x + ε) ∩ A mustbe infinite. Without loss of generality, we can assume that

Bε =

1

m+

1

n, m ∈ F , n ∈ Im

with F finite and at least one Im infinite, say Im0 . For all n ∈ Im0 , we have∣∣∣∣∣∣∣∣x− 1

m0

∣∣∣∣− 1

n

∣∣∣∣ ≤ ∣∣∣∣x− 1

m0

− 1

n

∣∣∣∣ < ε.

For n large enough, the left-hand side can be made arbitrarily close to∣∣∣x− 1m0

∣∣∣, in particular, we get that 12

∣∣∣x− 1m0

∣∣∣ < ε. If x > 0 is not of

the form 1m0

for any m0 ∈ Z+, then there exists a positive minimum value

for the numbers 12

∣∣∣x− 1m0

∣∣∣ and Bε cannot be infinite for arbitrarily small

values of ε.

36

(3) The boundary of a subset A in a topological space X is defined by

∂A = A ∩X \ A.

(a) Show that A = A t ∂A5.

If x ∈ A, there exists a neighborhood of A that is included in A. If x ∈ ∂A,in particular x ∈ X \ A so every neighborhood of x intersects X \ A. This

is a contradiction so A ∩ ∂A = ∅.The interior and boundary of A are included in A by definition so the inclu-sion A ⊃ At∂A is trivial. Conversely, let x ∈ A. If x has a neighborhood Usuch that U ⊂ A, then x ∈ A. The alternative is that every neighborhood ofx has non-empty intersection with X \A, that is x ∈ X \ A so that x ∈ ∂A.

Therefore, A ⊂ A t ∂A, which concludes the proof.

(b) Show that ∂A = ∅ if and only if A is open and closed.

By definition of the interior and the closure, A ⊆ A ⊆ A and A is open andclosed if and only if A = A. By (a), this is equivalent to ∂A = ∅.

(c) Show that U is open if and only if ∂U = U \ U .

The result of (a) states that U and U are complements in U , so U = U \ ∂Uand U is equal to U , that is, U is open if and only if U = U \ ∂U , which isequivalent to the condition ∂U = U \ U .

(d) If U is open, is it true that U = ˚U?

If U is open, the inclusion U ⊂ U implies that U ⊂ ˚U . However, the reverseinclusion may fail: consider for instance U = R \ 0 in R. It is open as the

union of open intervals and U = R so that ˚U = R ! U .

(4) Find the boundary and interior of each of the following subsets of R2.(a) A = (x, y) , y = 0(b) B = (x, y) , x > 0 and y 6= 0(c) C = A ∪B(d) D = Q× R(e) E = (x, y) , 0 < x2 − y2 ≤ 1(f) F = (x, y) , x 6= 0 and y ≤ 1

x

Note that, except for (d), a picture is very helpful to determine the boundaryand interior of the subsets at hand before rigorously justifying the intuition, usingwhat is known about the (metric) topology of R2.

5The disjoint union symbol t is used to indicate that the sets in the union have empty intersection.37

(a) Observe that A is closed, as the complement of R× (−∞, 0)∪ (0,+∞) whichis open as a product of open sets. Another way to see this is to remark that everyelement of R2 \ A is of the form (x, y) with y 6= 0, and for any x ∈ R, the basiselement

V = (x− 1, x+ 1)×(y − |y|

2, y +

|y|2

)satisfies (x, y) ∈ V ⊂ R2 \ A.Moreover, the interior of A is empty: every element of A is of the form (x, 0), anyneighbourhood of which contains a basis element (a, b) × (c, d) with c < 0 < d,which in turn cannot be included in A, for it contains (x, d

2) /∈ A.

We conclude that A = ∅ and ∂A = A.

(b) Note that B = (0,+∞) × (−∞, 0) ∪ (0,+∞) is open as a product of opensets. Another way to see this is to consider (x, y) ∈ B, that is, x > 0 and y 6= 0.Then

V =

(x

2,3x

2

)×(y − |y|

2, y +

|y|2

)is a neighborhood of (x, y) that is contained in B, which is therefore open.Finally, B is open because it is the inverse image of R2 \ A open under thecontinuous map (x, y) 7→ (lnx, y).Let us prove that the closure of B is the closed half-plane R defined by x ≥ 0.Let V be a neighborhood of (x, y) ∈ R. If (x, y) ∈ B, there is nothing to prove. Ifxy = 0, then V contains a subset of the form (a, b)× (c, d) with 0 < b and cd 6= 0so(x+b

2, y+d

2

)or(x+b

2, y+c

2

)belongs to V ∩ B, which is therefore not empty. We

have proved that R ⊂ B. The converse inclusion follows from the same argumentinvoked to prove that R2 \ A is open.Since B is open, it follows from (c) in the previous problem that ∂B = B \ B,that is ∂B is the union of the vertical axis and the positive horizontal axis.

(c) Since A ∪B = A ∪ B, it follows form (a) and (b) that C = R ∪A consists ofthe points (x, y) such that x ≥ 0 or y = 0.

Next, C is the right half-plane (0,+∞) × R: this set is open as the product ofopen sets and it is maximal. Indeed, if x ≤ 0, then any neighborhood of (x, y)contains a subset of the form (a, b) × (c, d) with a < 0 and cd 6= 0 so

(x+a

2, y+d

2

)or(x+a

2, y+c

2

)belongs to V ∩ (R2 \ C), which is therefore not empty.

It follows from the result proved in (a) of the previous problem that ∂C = C \ Cis the union of the vertical axis and the negative horizontal axis.

(d) Every non-empty open interval of R contains infinitely many rational andirrational numbers, so every product of intervals contains infinitely many elementsof D and R2 \ D. Therefore, ∂D = R2 and, since ∂D = D \ D, it follows

immediately that D = ∅.38

(e) First, observe that the set Ω = (x, y) , 0 < x2−y2 < 1 is open, for instanceas the inverse image of the open set (0, 1) under the map (x, y) 7→ x2− y2, whichis polynomial, hence continuous.A similar argument, shows that Γ = (x, y) , 0 ≤ x2 − y2 ≤ 1 is closed. Since

Ω ⊂ E ⊂ Γ, we get the chain of inclusions Ω ⊂ E ⊂ E ⊂ Γ, hence

∂E = E \ E ⊂ Γ \ Ω.

In other words, a boundary point (x, y) of E satisfies either x2 = y2 or x2−y2 = 1.Conversely, assume that x2− y2 = 1. Every neighbourhood of (x, y) contains thepoints Pδ = (x+ δ, y) for δ ∈ (−δ0, δ0) with δ0 > 0. Since

(x+ δ)2 − y2 = 1 + 2δ(x+ δ),

and the quantity 2δ(x+δ) takes arbitrarily small positive values when δ runs over(−δ0, δ0), we see that there are points Pδ in R2 \E and E so (x, y) is a boundarypoint of E. One can proceed in the same way to verify that the two lines givenby the equation x2 = y2 are also included in ∂E, which concludes the proof that∂E consists exactly of the union of the hyperbola with equation x2 − y2 = 1 andthe lines with equations y = ±x.It also follows that E = Ω. We have already obtained the inclusion Ω ⊂ E.Conversely, assume that (x, y) is a point in E not in Ω. Then x2 − y2 = 1 so

(x, y) belongs to ∂E which is disjoint from E. This proves that E ⊂ Ω and theequality.

(f) No new technique is needed to prove that F is the region located strictlybelow the branches of the hyperbola with equation xy = 1, that is,

F =

(x, y) , x 6= 0 and y <

1

x

,

and that ∂F is the union of the hyperbola and the vertical axis:

F =

(x, y) , x = 0 or y =

1

x

.

39

Problem set 5: continuous maps, the product topology - Elements of solution

(1) (a) Consider Z+ equipped with the topology in which open sets arethe subsets U such that if n is in U , then any divisor of n belongsto U . Give a necessary and sufficient condition for a function f :Z+ −→ Z+ to be continuous.

Assume f continuous. For n ∈ Z+, let Un be the open set of all divisors ofn. Let a ∈ f−1(Un), assumed non-empty. Since f is continuous, f−1(Un)is open, hence contains all the divisors of a. In other words, if b|a, thenf(b) ∈ Un, that is, f(b)|n. A necessary condition for continuity is thereforethat f preserve divisibility:

b|a⇒ f(b)|f(a).

Let us prove that the condition is also sufficient. Assume that f(b)|f(a)whenever b|a and let U be open in Z+. If f−1(U) =, it is open. Otherwise,let a ∈ f−1(U). To prove that f−1(U) is open, it suffices to show that itcontains all the divisors of a. The property of f implies that f(b)|f(a) forevery such divisor b and, U being open, this implies that f(b) ∈ U , that is,b ∈ f−1(U).

(b) Let χQ be the indicator of Q. Prove that the map ϕ : R −→ Rdefined by ϕ(x) = x · χQ(x) is continuous at exactly one point.

We shall prove that ϕ is continuous at 0 and discontinuous everywhere else.Note that |ϕ(x)| ≤ |x| for every x ∈ R. In particular, let ε > 0 and δ = ε.Then |x| < δ implies |ϕ(x)| < ε, so ϕ is continuous at 0.Now observe that ϕ is odd and let a > 0 be a positive number. Then,

ϕ−1(

(a

2,3a

2

)) =

(a

2,3a

2

)∩Q

which is not open as no subset of Q can contain an open interval of R.Therefore, ϕ is not continuous at a.

40

(2) Let X and Y be topological spaces. If A is a subset of either, we denoteby A′ the sets of accumulation points of A and by ∂A its boundary.Let f : X −→ Y be a map. Determine the implications between thefollowing statements.

(i) f is continuous.

(ii) f(A′) ⊂ (f(A))′ for any A ⊂ X.

(iii) ∂(f−1(B)) ⊂ f−1(∂B) for any B ⊂ Y .

Considering a constant function R −→ R shows (i)6⇒(ii). However, the converseis true: let A be a subset of X and x ∈ A = A ∪ A′.

- If x ∈ A, then f(x) ∈ f(A) ⊂ f(A).

- If x ∈ A′, then (ii) implies that f(x) ∈ (f(A))′ ⊂ f(A).

Therefore, f(A) ⊂ f(A) for any A ⊂ X so f is continuous by [M. Th. 18.1].

Let us prove that (i) ⇒ (iii). Assume f continuous, let B ⊂ Y be a subset andx ∈ ∂(f−1(B)). If x /∈ f−1(∂B), there are two possibilities.

Case 1: f(x) ∈ B. Then x ∈ f−1(B), open by continuity of f . Since f−1(B) ⊂f−1(B), it follows that x is an interior point of f−1(B), which is a contradiction.

Case 2: f(x) ∈ Y \ B. Then x ∈ f−1(Y \ B), open by (i). In particular, thereis a neighborhood U of x such that U ⊂ f−1(Y \ B). Since x ∈ ∂(f−1(B)), itfollows that there exists some y in U such that f(y) ∈ B, which contradicts theassumption that f(x) ∈ Y \ B.Altogether, this proves that x ∈ f−1(∂B), hence the inclusion of (iii).

To establish the converse, we rely on the following characterization of continuity:Lemma: f is continuous if and only if f−1(B) ⊂ f−1(B) for every B ⊂ Y .Proof of the lemma: if f is continuous, the inverse image of the closed set B is aclosed set that contains f−1(B) hence its closure. Conversely, if B is closed, the

condition becomes f−1(B) ⊂ f−1(B). The reverse inclusion holds by definition

of the closure, so f−1(B) = f−1(B), hence f−1(B) is closed and f is continuous.

If f is discontinuous, the lemma implies the existence of some B ⊂ Y such thatf−1(B) 6⊂ f−1(B). Let x be an element of f−1(B) such that f(x) /∈ B, hence

f(x) /∈ B. Since f−1(B) = f−1(B) ∪ ∂f−1(B), it follows that x ∈ ∂f−1(B).

The fact that f(x) /∈ B implies that f(x) /∈ B \ B = ∂B, that is

∂(f−1(B)) 6⊂ f−1(∂B)

and we have proved the contrapositive of (iii) ⇒ (i).

To sum up, conditions (i) and (iii) are equivalent and they are implied by (ii),but the converse does not hold:

(ii)⇒ (i)⇔ (iii).41

(3) Let X and Y be topological spaces, and assume Y Hausdorff. Let A bea subset of X and f1, f2 continuous maps from the closure A to Y .Prove that if f1 and f2 restrict to the same function f : A → Y , thenf1 = f2.

We argue by contradiction: if f1 6= f2, there exists x ∈ A such that f1(x) 6= f2(x)and x /∈ A. Since Y is Hausdorff, there exist disjoint neighborhoods V1 of f1(x)and V2 of f2(x). By continuity of f1 and f2, both f−1

1 (V1) and f−12 (V2) are

neighborhoods of x, and so is U = f−11 (V1) ∩ f−1

2 (V2). Since x ∈ A \ A, theneighborhood U contains some a in A such that f1(a) = f(a) = f2(a). Therefore,f(a) ∈ V1 ∩ V2 which is assumed empty.

(4) Let Xαα∈J be a family of topological spaces and X =∏α∈J

Xα.

(a) Give a necessary and sufficient condition for a sequence unn∈Z+

to converge in X equipped with the product topology.

Assume that limn→∞ un = l. The projection maps πα are continuous so the‘non-necessarily metrizable’ part of the sequential characterization theorem[M. Th. 21.3] implies that

(?) ∀α ∈ J , limn→∞

unα = lα.

Conversely, assume that limn→∞ πα(un) = πα(l) for every α ∈ J and let U bea neighborhood of l. We may assume that U is an intersection of cylinders,that is,

U = π−1α1

(Uα1) ∩ π−1α2

(Uα2) ∩ . . . ∩ π−1αp

(Uαp)

since such elements form a basis for the product topology. With our assump-tion, there exists, for each i ∈ 1, . . . , p, a rank Ni such that παi

(un) ∈ Uαi

for all n ≥ Ni. This implies that un ∈ U for all n ≥ max1≤i≤n

Ni, so limn→∞ un =

l.

(b) Does the result hold if X is equipped with the box topology?

The box topology is finer than the product topology so condition (?) iscertainly necessary. It is not sufficient, however, as the following exampleshows.Let J = Z+ and Xk = R with the standard topology for each k ∈ Z+ so thatX = Rω as a set. Then, consider the sequence (nu)n∈Z+ defined by

nuk =

1 if k = n0 if k 6= n

.

Then limn→∞nuk = limn→∞ πk(

nu) = 0 for every k ∈ Z+ so nu converges tothe zero sequence in the product topology.On the other hand, the open box

∏k∈Z+

(−1, 1) is a neighborhood of the zero

sequence that contains no term of the sequence (nu)n∈Z+ , which thereforecannot converge to zero in the box topology.

42

Problem set 6: metrizable spaces - Elements of solution

(1) Let ρ be the uniform metric on Rω. For x ∈ Rω and 0 < ε < 1, let

P (x, ε) =∏n∈Z+

(xn − ε, xn + ε).

(a) Compare P (x, ε) with Bρ(x, ε).

With 0 < ε < 1, the definition of ρ implies that

Bρ(x, ε) =

y ∈ Rω , sup

n≥1|xn − yn| < ε

.

On the other hand, P (x, ε) = y ∈ Rω , |xn − yn| < ε for every n ≥ 1 sothe inclusion Bρ(x, ε) ⊂ P (x, ε) holds by definition of the supremum. Thisinclusion is strict: the sequence y defined by

yn = xn + ε− 1

n

belongs to P (x, ε) but ρ(x, y) = ε so Bρ(x, ε) P (x, ε).

(b) Is P (x, ε) open in the uniform topology?

No: we prove that P (x, ε) contains no ρ-ball centered at the element yintroduced above. For any η > 0, the sequence z defined by zn = yn + η

2

satisfies ρ(y, z) < η hence belongs to Bρ(y, η).

Observe that xn + ε− yn <η

2for n large enough, so

zn − xn = yn +η

2− xn

= ε− 1

n+η

2> ε

which means that z /∈ P (x, ε), which is therefore not a neighborhood of y,hence not open.

(c) Show that Bρ(x, ε) =⋃δ<ε

P (x, δ).

Let y ∈ P (x, δ) with δ < ε. Then, ρ(x, y) ≤ δ < ε so y ∈ Bρ(x, ε), whichproves that Bρ(x, ε) ⊃ P (x, δ). Since δ was arbitrary, we conclude that

Bρ(x, ε) ⊃⋃δ<ε

P (x, δ).

Conversely, if y ∈ Bρ(x, ε), then y ∈ P (x, δy) with δy = ρ(x, y) < ε, so

Bρ(x, ε) ⊂⋃δ<ε

P (x, δ).

43

(2) We denote by `2(Z+) the set of square-summable real-valued sequences:

`2(Z+) =

x = (xn)n∈Z+ ∈ Rω ,

∑n≥1

x2n converges

,

equipped with the metric

d(x, y) =

(∑n≥1

(xn − yn)2

)1/2

.

(a) Compare the metric topology induced by d on `2(Z+) with therestrictions of the box and uniform topologies from Rω.

The inclusion Tρ ⊂ Td follows from the observation that Bd(x, r) ⊂ Bρ(x, r)for any x ∈ `2(Z+) and r > 0. Indeed, observe that

|xk − yk|2 ≤∑n≥1

|xn − yn|2

for any fixed k, so that ρ(x, y) ≤ d(x, y) for any x, y ∈ `2(Z+)6.

This inclusion is strict, as the following example shows. Denote by 0 thesequence that is constantly equal to 0. We will prove that Bd(0, 1), open inTd by definition, is not open in the uniform topology. More precisely, no ball

Bρ(0, r) with r > 0 is included in Bd(0, 1): let n0 >(

2r

)2and consider the

sequence

ξn =

r2

if n ≤ n0

0 if n > n0.

Then ρ(0, ξ) = r2< r so ξ ∈ Bρ(0, r) but d(0, ξ) =

√n0r

2> 1 so ξ /∈ Bd(0, 1).

Next, we prove that Td ⊂ Tbox: let x ∈ `2(Z+), r > 0 and consider the box

B =∏n≥1

(xn −r

2n, xn +

r

2n).

Then x ∈ B ⊂ Bd(x, r) since y ∈ B implies d(x, y)2 ≤∑n≥1

r2

4n=r2

3.

Again, the inclusion is strict: consider the open box

B =∏n≥1

(− 1

n,

1

n

).

Although 0 ∈ B, no ball Bd(0, r) with r > 0 is included in B.

6Another approach to this result consists in proving that uniform convergence implies `2 convergenceand conclude by the sequential characterization of the closure.

44

Indeed, consider n0 >2r

and let η be the sequence defined by

ηn =

r2

if n = n0

0 if n 6= n0.

Then x /∈ B but d(0, x) = r2< r so x ∈ Bd(0, r). Therefore B /∈ Td.

We have proved that Tρ Td Tbox.

(b) Let R∞ denote the subset of `2(Z+) consisting of sequences thathave finitely many non-zero terms. Determine the closure of R∞in `2(Z+).

We will prove that R∞`2(Z+)

= `2(Z+), using the sequential characterizationof the closure in a metric space. For any x ∈ `2(Z+), let nx be the sequencedefined by

nxk =

xk if k ≤ n0 if k > n

.

Then, d(nx, x)2 =∑

k>n x2k. This quantity converges to 0 as n→∞ because

x is square-summable, so limn→∞

nx = x in `2(Z+).

(3) Let X be a topological space, Y a metric space. Assume that (fn)n≥0

is a sequence of continuous functions that converges uniformly to f :X → Y .Let (xn)n≥0 be a sequence in X such that lim

n→∞xn = x. Prove that

limn→∞

fn(xn) = f(x).

Let ε > 0 and observe that the triangle inequality implies that

d(fn(xn), f(x)) ≤ d(fn(xn), f(xn))︸ ︷︷ ︸An

+ d(f(xn), f(x))︸ ︷︷ ︸Bn

.

The convergence of the sequence is assumed uniform so there exists an integer

NA such that d(fn(ξ), f(ξ)) <ε

2for any n ≥ NA and any ξ ∈ X.

In particular, An <ε

2for n ≥ NA.

Moreover, the Uniform Limit Theorem guarantees that f is continuous. There-

fore, there exists NB such that Bn <ε

2for any n ≥ NB.

It follows that d(fn(xn), f(x)) ≤ ε for any n greater than max(NA, NB).

45

(4) Ultrametric spaces.

Let X be a set equipped with a map d : X ×X −→ R such that(1) d(x, y) ≥ 0 (3) d(x, y) = 0⇔ x = y(2) d(x, y) = d(y, x) (4) d(x, z) ≤ max (d(x, y), d(y, z))

(a) Verify that d is a distance.The only condition to check is the triangle inequality:

d(x, y) + d(y, z) = max (d(x, y), d(y, z)) + min (d(x, y), d(y, z))

≥ max (d(x, y), d(y, z)) by (1)

≥ d(x, z) by (4).

(b) Let B be an open ball for d. Prove that B = B(y, r) for everyelement y ∈ B for some r > 0.Let x ∈ X, r > 0 and B = B(x, r). Let y ∈ B, that is, assume d(x, y) < r.Note that

d(x, z) < r ⇔ max(d(x, y)︸ ︷︷ ︸<r

, d(x, z)) < r,

which implies that d(y, z) < r by (4). It follows that B ⊂ B(y, r). Thereverse inclusion follows from exchanging x and y in the previous argument.Every point in the ball is a center!

(c) Prove that closed balls are open and open balls are closed in thetopology induced by d.

Let B be a closed ball, that is B = Bc(x, r) = y ∈ X , d(x, y) ≤ r forsome x ∈ X and r > 0. Let y ∈ B. The open ball B

(y, r

2

)is a neighborhood

of y contained in B:

d(x, z) ≤(4)

max(d(x, y)︸ ︷︷ ︸≤r

, d(y, z)︸ ︷︷ ︸<r/2

) ≤ r

for any z ∈ B(y, r

2

), so B is open.

To prove that open balls are closed, we used the sequential characterization:let (xn)n≥1 be a sequence in B(a, r) such that lim

n→∞xn = x for some x ∈ X.

Since d(xn, x) can be made arbitrarily small, there exists an integer n0 suchthat d(xn0 , x) < r. The ultrametric property of d implies that

d(a, x) ≤ max(d(a, xn0) , d(xn0 , x)) < r,

so x ∈ B(a, r), which is therefore sequentially closed.

Note: it might seem that a distance with such properties may not be useful in any

reasonable circumstances or even not exist at all. It is easy to check that the discrete

metric on any set is ultrametric. More interestingly the p-adic distances defined on

Q are ultrametric, which gives p-adic analysis a very different flavor from that of real

analysis.46

Midterm 2: in-classElements of solution

Problem 1

1. Show that a topological space is T1 if and only if for any pair of distinctpoints, each has a neighborhood that does not contain the other.

Let x 6= y be elements of X, assumed T1. Then x is closed so X \x is a neighborhoodof y that does not contain x. Similarly, X \ y is a neighborhood of x that does notcontain y. Conversely, assume that distinct points have neighborhoods that does notcontain the other and let x ∈ X. Then if y 6= x, there is a neighborhood of y that doesnot contain x so X \ x is open hence x is closed.

2. Determine the interior and the boundary of the set

Ξ =

(x, y) ∈ R2 , 0 ≤ y < x2 + 1

where R2 is equipped with its ordinary Euclidean topology.

Ξ =

(x, y) ∈ R2 , 0 < y < x2 + 1

∂Ξ = y = 0 ∪ y = x2 + 1

Problem 2

Let E be a set with a metric d and Td the corresponding metric topology.

1. Prove that the map d : (E, Td)× (E, Td) −→ R is continuous.

Let (a, b) be an arbitrary basis element for the topology on R, with b > 0, so thatd−1((a, b)) is not empty. Let (x, y) ∈ d−1((a, b)) and d = d(x, y). Then, by the triangleinequality,

(p, q) ∈ B(x,b− d

2)×B(y,

b− d2

)⇒ d(p, q) < b.

The triangle inequality also implies that d(p, q) ≥ d(x, y)− d(x, p)− d(v, y) so

(p, q) ∈ B(x,d− a

2)×B(y,

d− a2

)⇒ a < d(p, q).

It follows that B(x, r) × B(y, r) with r = minb−d

2, d−a

2

is a neighborhood of (x, y)

contained in d−1((a, b)), which is therefore open.47

2. Let T be a topology on E, such that d : (E, T )× (E, T ) −→ R is continuous.Prove that T is finer than Td.

It suffices to prove that every ball B(x, r) is open for T . If y ∈ B(x, r), then (x, y) belongsto d−1((−∞, r)), assumed open, so there exists a basis element U ×V in T ×T such that

(x, y) ∈ U × V ⊂ d−1((−∞, r)).In particular, V is a neighborhood of y. Moreover, if z ∈ V , then (x, z) ∈ U × V ⊂d−1((−∞, r)) so d(x, z) < r, which proves that V ⊂ B(x, r), hence B(x, r) ∈ T .

We have proved that the metric topology is the coarsest topology on E making d contin-uous.

Problem 3

We prove that the box topology on Rω is not metrizable.

1. Recall the definition of the box topology on Rω.

It is the topology generated by the basis∏

n≥1 Un , Un open in R

.

Denote by 0 the sequence constantly equal to 0 and let

P = (0,+∞)ω =∏n≥1

(0,+∞)

be the subset of positive sequences.

2. Verify that 0 belongs to P .

Let U =∏

n≥1 Un be a neighborhood of 0. Then Un is a neigborhood of 0 in R for every n.Therefore, Un contains an interval (an, bn) with an < 0 < bn for every n so the sequence(bn)n≥1 is an element of U ∩ P . Every neighborhood of 0 meets P so 0 ∈ P .

3. Prove that no sequence (pn)n≥1 ∈ P ω converges to 0 in the box topology.

Let (nu)n≥1 be a sequence of elements of P and consider the open box

B =∏n≥1

(−nun, nun).

Then 0 ∈ B, but no nu belongs to B, since the nth term of nu lies outside the nth intervalin the product defining B.

4. Conclude.

In a metrizable space, closure points of are limits of sequences. Here, 0 is a closure pointof P that is the limit of no sequence of elements of P . Therefore, the box topology on Rωis not metrizable.

48

Problem 4

1. Let X be a set.

(a) Recall the definition of the uniform topology on RX.

It is the metric topology associated with ρ(f, g) = supx∈X min|f(x)− g(x)|, 1.

(b) Recall the definition of uniform convergence for a sequence in RX.

The sequence (fn)n≥1 converges uniformly to f in RX if

∀ε > 0 , ∃Nε ∈ Z+ , ∀n ≥ Nε , ∀x ∈ X , |fn(x)− f(x)| < ε.

2. Prove that a sequence in RX converges uniformly if and only if it convergesfor T∞.

Assume that fn converges uniformly to f and let 0 < ε < 1. Then for n ≥ N ε2

and allx ∈ X,

min|f(x)− g(x)|, 1 = |fn(x)− f(x)| < ε

2,

soρ(fn, f) = sup

x∈Xmin|fn(x)− f(x)|, 1 ≤ ε

2< ε,

which means that fn converges to f in the uniform topology.

Conversely, assume that limn→∞ ρ(fn, f) = 0 and let 0 < ε < 1. For n large enough,supx∈X|fn(x)− f(x)| < ε, so that

|fn(x)− f(x)| < ε for all x ∈ X,so fn converges uniformly to f .

Problem 5

Consider the space Rω of real-valued sequences, with the uniform topology.

1. Prove that the subset B of bounded sequences is closed.

It suffices to prove that a uniform limit of bounded sequences is bounded. Let (nu)n≥1 besuch that each (nuk)k≥1 is bounded:

|nuk| ≤Mn for all k ≥ 1

and assume that limn→∞nu = u, uniformly. Then there exists an integer n0 such that

(?) ∀n ≥ n0 , supk≥1|nuk − uk| < 1.

If u were unbounded, it would admit a subsequence uk` such that lim`→∞ |uk` | = +∞.Since n0u is bounded by Mn0 , this would imply that

lim`→∞|n0uk` − uk` | = +∞,

which contradicts (?). Therefore, u must be bounded and B contains all its limit points.49

2. Let R∞ denote the subset of sequences with finitely many non-zero terms.Determine the closure of R∞ in Rω for the uniform topology.

We will prove that R∞ = c0(Z+), the set of sequences that converge to 0.

Let (nu)n≥1 be a uniformly convergent sequence of elements of R∞ and u = limn→∞nu.

If u does not converge to 0, there exists some η > 0 such that

|uk| > η

for arbitrarily large values of k. It follows that, for any n ≥ 1,

|nu− uk| = |uk| > η for some k

since nu has only finitely many non-zero terms. This implies that supk≥1 |nuk − uk| ≥ ηfor all n ≥ 1, which contradicts the uniform convergence of (nu)n≥1.

Conversely, any sequence u in c0(Z+) is the uniform limit of its truncations: let nu be thesequence defined by

nuk =

uk if k ≤ n0 if k > n

.

Then, nu ∈ c0(Z+) andsupk≥1|nuk − uk| = sup

k>n|uk| −→

n→∞0

so nu converges uniformly to u.

50

Midterm 2: take-homeElements of solution

Problem 1

The purpose of this problem is to study the connected components of Rω in various topolo-gies. In what follows, B and U respectively denote the sets of bounded and unboundedsequences. Note that Rω = B t U .

The sequence whose terms are constantly equal to 0 is denoted by 0. The connectedcomponent of x is denoted by Cx.

Finally, if x is an element in Rω and A a subset of Rω, we denote by x + A the set ofA-translates of x, that is

x+ A = x+ a , a ∈ A.

1. Determine the connected components of Rω in the product topology.

For x, y in Rω, the function

sx,y : t 7−→ (1− t)x+ ty

is continuous from [0, 1] to Rω equipped with the box topology, because each componentmap sxn,yn is polynomial hence continuous from [0, 1] to R. Since each sx,y(t) is a real-valued sequence, it follows that Rω is convex, hence (path) connected.

2. Consider Rω equipped with the uniform topology.

(a) Prove that x is in the same connected component as 0 if and only if xis bounded.

We know that B is closed in Rω for the uniform topology. A similar argumentshows that U is also closed, so that they constitute a separation of Rω in thistopology. Therefore, since the zero sequence is bounded, we see that C0 ⊂ B.Furthermore, for x ∈ B, the function s0,x satisfies

|s0,x(t)| ≤ t · supn≥1|xn|

for every t ∈ [0, 1]. It follows that it is continuous and that every s0,x(t) isbounded so B is connected in the uniform topology. Therefore, C0 = B.

(b) Deduce a necessary and sufficient condition for x and y in Rω to lie inthe same connected component for the uniform topology.

For x fixed in Rω, the map y 7−→ y− x is a homeomorphism from Rω onto itself,which sends x to 0. It follows that x and y are in the same connected componentif and only if x− y ∈ C0. In other words, Cx = x+B.

51

3. Consider Rω equipped with the box topology.

(a) Let x, y ∈ Rω be such that x − y ∈ Rω \ R∞. Prove that there existsa homeomorphism ϕ : Rω −→ Rω such that (ϕ(x)n)n∈Z+ is a boundedsequence and (ϕ(y)n)n∈Z+ is unbounded.

Let us prove the following:Lemma. Let (αn)n∈Z+ and (βn)n∈Z+ be fixed sequences of real numbers suchthat αn 6= 0 for all n. Then the map

ϕ : Rω −→ Rω(un)n∈Z+ 7−→ (αnun + βn)n∈Z+

.

is a homeomorphism in the box topology.

Proof. Since ϕ is bijective with inverse (un)n∈Z+ 7−→(unαn− βn

)n∈Z+

it suffices

to prove that every map from Rω to itself with (non-constant) affine componentsis continuous. Let W =

∏n≥1(sn, tn) be a basis element for the box topology.

Then,

ϕ−1(W ) =∏n≥1

˜(snαn− βn,

tnαn− βn

)is also an open box, hence open. Therefore ϕ is continuous.

Now, with x and y fixed in Rω such that x − y /∈ Rω \ R∞, consider the map ϕdefined by

ϕ(u)n =

un − xn if xn = yn

en · un − xnyn − xn

if xn 6= yn

for u ∈ Rω. Then ϕ is a homeomorphism of Rω by the lemma and ϕ(x) = 0 ∈ Bwhile ϕ(y) has an exponentially growing subsequence, hence ϕ(y) ∈ U .

(b) Deduce a necessary and sufficient condition for x and y in Rω to lie inthe same connected component for the box topology.

Let x, y ∈ Rω. Since B t U is a separation of Rω in the box topology, anyhomeomorphism ϕ of Rω must satisfy ϕ(Cx) ⊂ B or ϕ(Cx) ⊂ U . By the previousquestion, if follows that y ∈ Cx implies x−y ∈ R∞. In other words, Cx ⊂ x+R∞.

The converse inclusion follows from a convexity argument. Assume that x− y ∈R∞. Then, sx,y(t) = x+ t(y − x) ∈ x+ R∞ for every t ∈ [0, 1].Let us prove that this map is continuous between R and Rω equipped with thebox topology. Let W =

∏n≥1 In with In an open interval of R for every n ∈ Z+.

Then, for every n, the set Jn = t ∈ [0, 1] , xn + t(yn − xn) is- empty or equal to [0, 1] if xn = yn;- an open interval of [0, 1] if xn 6= yn.

52

Since the latter occurs only for finitely many values of n, the inverse image of Wunder sx,y, which is

⋂n≥1 In is either empty or a finite intersection of open sets,

hence open. This proves that sx,y is a continuous path, so that x + R∞ is pathconnected, hence equal to Cx.

Problem 2

Let F be a functor between categories C and C ′. A functor G : C ′ −→ C is saidto be a left adjoint for F if there is a natural isomorphism

HomC(G(X), Y ) ∼= HomC′(X,F (Y ))

for all objects X ∈ C ′ and Y ∈ C. Similarly, G is called a right adjoint for F ifthere is a natural isomorphism

HomC(X,G(Y )) ∼= HomC′(F (X), Y )

for all objects X ∈ C and Y ∈ C ′.

Recall that the forgetful functor F : Top −→ Set is defined by

- F ((X, T )) = X for any set X equipped with a topology T ;

- F(f) = f for any continuous map f : X −→ Y .

If X is a set, let G(X) denote the topological space obtained by endowing Xwith the trivial topology Ttriv. = X, ∅:

G(X) = (X, Ttriv.).

If f is a map between sets, define in addition G(f) = f .

This problem is a reformulation in the language of categories of the following basic prop-erties of the trivial and discrete topologies:

(C1) If (X, T ) is a topological space, then any map (X, T ) −→ (Y, Ttriv.) is continuous.

(C2) If (Y, T ) is a topological space, then any map (X, Tdisc.) −→ (Y, T ) is continuous.

1. Verify that G is a functor.

At the level of objects, G sends sets to topological spaces. To verify functoriality, itsuffices to check that if f is a morphism in Set, then G(f) is a morphism in Top andcompatibility with compositions in each categroy. Let X and Y be objects in Set, andf ∈ Hom(X, Y ) that is, f is a map between sets X and Y . Then G(f) = f by definitionand the condition

G(f) ∈ Hom(G(X),G(Y ))53

is equivalent to f being continuous between (X, Ttriv.) and (Y, Ttriv.), which follows from(C1). The composition relation is immediate as G(gf) = gf = G(g)G(f). Finally,G(IdX) = x 7→ x = IdG(X).

2. Prove that G is a right adjoint to F.

To prove that G is a right adjoint to F, we need to compare

HomTop((X, T ),G(Y ))

withHomSet(F ((X, T )) , Y )

for any topological space (X, T ) and every set Y . By definition, HomTop((X, T ),G(Y ))is the set of continuous maps from (X, T ) to (Y, Ttriv.).On the other hand, HomSet(F ((X, T )) , Y ) consists of all maps from X to Y . Therefore,it follows from (C1) that every element of HomSet(F ((X, T )) , Y ) can be seen as an el-ement of HomTop((X, T ),G(Y )). In other words, the natural isomorphism realizing theadjunction is the map

HomSet(F ((X, T )) , Y ) −→ HomTop((X, T ),G(Y ))f 7−→ f

3. Find a left adjoint for F.

Similar arguments and (C2) imply that H defined on Set by H(X) = (X, Tdisc.) andH(f) = f is a functorial and a left adjoint for F.

54

Problem set 7: connectedness and compactness - Elements of solution

(1) Let U be an open connected subspace of R2 and a ∈ U .

(a) Prove that the set Γa of points x ∈ U such that there is a pathγ : [0, 1] −→ U with γ(0) = a and γ(1) = x is open and closed in U .

Let x be an element of Γa, connected to a by a path γ, and r > 0 such thatB(x, r) ⊂ U . Then for y ∈ B(x, r), the map γ : [0, 1] −→ U defined by

γ(t) =

γ(2t) if 0 ≤ t ≤ 1

22(1− t)x+ (2t− 1)y if 1

2≤ t ≤ 1

is a continuous path joining a to y, so B(x, r) ⊂ Γa, which is therefore open.

To prove that it is closed, let x ∈ U be an accumulation point of Γa and r > 0such that B(x, r) ⊂ U . Since x is an accumulation point, there exists y 6= xin B(x, r)∩Γa. Then, as in the proof that Γa is open, one can concatenate apath from a to y and the segment from y to x to get a continuous path in Uthat connects a to x. It follows that x ∈ Γa, which means that Γa containsits accumulation points, hence is closed in U .

(b) What can you conclude?

The set of points that can be connected to a by a path is open and closedin U connected. Since it is not empty (it contains a) it is equal to U ,which is therefore path connected. In other words connectedness and pathconnectedness are equivalent for open subsets of R2.

(2) Let X be a topological space and Y ⊂ X a connected subspace.

(a) Are Y and ∂Y necessarily connected?

The answer is negative in both cases. Let Y1 = L ∪ R ⊂ R2 be the union ofthe half plane L = x ≤ 0 and the half-cone R = x ≥ 0 , |y| ≤ x. ThenY1 is connected because both L and R are, and they intersect at the origin.On the other hand,

Y1 = x < 0 t x > 0 , |y| < xis disconnected, as the two terms in the union are disjoint and open.

To see that a connected set need not have a connected boundary, it sufficesto consider a closed interval of R. Another example is that of a closed washerin R2: let Y2 = 1 ≤ x2 + y2 ≤ 4. It is homeomorphic to the rectangle[1, 2]×[0, 2π) via polar coordinates, hence connected. However, its boundaryconsists of two disjoint circles, closed in R2:

∂Y2 = x2 + y2 = 1 t x2 + y2 = 4which are closed as inverse images of singletons under a continuous map.

55

(b) Does the converse hold?

No: in R, consider the union of negative real number and positive rationals

Y3 = (−∞, 0) ∪Q+.

Then Y3 = (−∞, 0) and ∂Y3 = [0,+∞] are connected, but Y3 is disconnectedas each rational is alone in its connected component.

(3) Let (E, d) be a metric space.

(a) Prove that every compact subspace of E is closed and bounded.

In a metric spaces7, compact sets are closed [M. Th.26.3]. Moreover, assumethat K is compact in E and that K can be covered by open balls of radius1. By compactness, K can be covered by finitely many ball of radius 1, sayN . Then the triangle inequality shows that d(x, y) ≤ 2N , so K is bounded.

(b) Give an example of metric space in which closed bounded sets arenot necessarily compact.

Let X be an infinite set, equipped with the discrete metric. Then X isclosed and bounded for the corresponding metric topology. However, thecover given by all singletons, which are open, since the topology is discrete,does not have any finite subcover.

(4) This problem gives concrete descriptions of the Alexandrov compactifications ofsome locally compact spaces. The Alexandrov compactification is defined up tohomeomorphism and it follows from the universal property that homeomorphicspaces have the same compactification.

(a) Prove that the Alexandrov compactification of R is homeomorphicto the unit circle S1 = (x, y) ∈ R2 , x2 + y2 = 1.

Observe that S1 is compact as a closed and bounded subset of R2. Moreover,S1 \ (−1, 0) can be parametrized by the map r : R −→ R2 defined by

r(t) =

(1− t2

1 + t2,

2t

1 + t2

).

This map is continuous because its components are rational functions withnon-vanishing denominators, and takes values in S1. Its inverse is

(x, y) 7−→ y

x+ 1,

continuous on S1 \ (−1, 0) for the same reason.

We have proved that R ∼= S1 \ (−1, 0), which proves that R ∼= S1.

7This actually holds for Hausdorff spaces in general.56

(b) Verify that Z+ ⊂ R is a locally compact Hausdorff space.

Subspaces inherit the Hausdorff property so Z+ is Haudorff because R is.Moreover, every finite subset of Z+ is compact for the subspace topology,which is discrete, n ∈ Z+ can be seen in x, compact and open.

(c) Prove that the Alexandrov compactification of Z+ is homeomorphicto

1n, n ∈ Z+

∪ 0.

Again, observe that

1n, n ∈ Z+

∪ 0 is closed and bounded in R, hence

compact. Moreover, the map n 7−→ 1n

is a homeomorphism between Z+ and1n, n ∈ Z+

so Z+

∼=

1n, n ∈ Z+

∪ 0.

57

Final examination - Elements of solution

Problem 1

1. Let X be a Hausdorff space and K1, K2 disjoint compact subsets of X. Provethat there exist disjoint open sets U1 and U2 such that K1 ⊂ U1 and K2 ⊂ U2.

We know that points can be separated from compact sets in Hausdorff spaces. In otherwords, for every x ∈ K1, there exist Ux

1 neighborhood of x and Ux2 open containing K2

such that

Ux1 ∩ Ux

2 = ∅.The family Ux

1 , x ∈ K1 is an open cover of K1 compact so we can extract a finitesubcover Ux1

1 . . . , Uxn1 . Then

K1 ⊂n⋃i=1

Uxi1

def= U1,

and U1 is open as the union of open sets. Moreover, K2 ⊂ Uxi2 for every i ∈ 1, . . . , n, so

K2 ⊂n⋂i=1

Uxi2

def= U2,

open as the finite intersection of open sets. To conclude, observe that U1 and U2 aredisjoint because Ux

1 ∩ Ux2 = ∅ for all x.

2. Let X be a discrete space. Describe the compact subsets of X.

Let K be a compact subset of X. Since the topology is assumed discrete, singletons areopen and the family x , x ∈ K is a covering of K. The fact that a finite subcovercan be extracted shows that K must be finite.The converse holds for non-necessarily discrete topologies so the compact subsets of adiscrete space are exactly the finite sets.

Problem 2

A topological space is said totally disconnected if its only connected subspacesare singletons.

1. Prove that a discrete space is totally disconnected.

In a discrete space X, singletons are open and closed. Therefore, the connected componentof x ∈ X is x. Another way to see this is to observe that any non-trivial partition of aset is a separation, since all subsets are open and closed in the discrete topology.

58

2. Does the converse hold?

No, consider the example of Q: any open set of Q contains a subset of the form (a, b)∩Qwith a < b. Such a set contains infinitely many rationals. In particular, singletons are notopen, which means that the topology induced by R is not discrete.It is however totally disconnected: for any subset A of Q containing at least two elementsq and r, there exists an irrational z such that q < z < r, so that A∩(−∞, z)tA∩(z,+∞)is a separation of A.

Problem 3

Let Xαα∈J be a family of topological spaces; let Aα ⊂ Xα for each α ∈ J.

1. In∏

α∈J Xα equipped with the product topology, prove that∏α∈J

Aα =∏α∈J

Aα.

Let (xα)α∈J be a closure point of∏

αAα. And consider, for β ∈ J , a neighborhoodUβ of xβ. Since the projection maps are continuous, π−1

β (Uβ) is open in∏

αXα, hence aneighborhood of (xα)α∈J so it contains a point (yα)α∈J ∈

∏αAα. In particular yβ ∈ Uβ∩Aβ

so xβ ∈ Aβ.

Conversely, let (xα)α∈J ∈∏

α Aα and U =∏

α Uα a neighborhood of (xα)α∈J . Thenevery Uα contains a point yα ∈ Uα ∩ Aα so (yα)α∈J ∈ U ∩

∏Aα, which means that

(xα)α∈J ∈∏

αAα.

2. Does the result hold if∏

α∈J Xα carries the box topology?

Yes. Observe that the continuity of the projection maps used in the previous question stillholds in the box topology. The other part of the proof also carries over without change.

Problem 4

Is R homeomorphic to R2 ?

Assume the existence of a homeomorphism ϕ : R −→ R2 and consider the restricted mapϕ = ϕ

∣∣R\0 : R\0 −→ R2 \f(0). Then ϕ is bijective by construction and continuousas the restriction of a continuous function. Observe that ϕ−1 is continuous for the samereason, which means that ϕ is a homeomorphism between the disconnected space R\0,and the connected connected space R2 \ f(0), which is impossible since connectednessis a topological property.

59

Problem 5

Let (E, d) be a metric space. An isometry of E is a map f : E −→ E such that

d (f(x), f(y)) = d(x, y)

for all x, y ∈ E.

1. Prove that any isometry is continuous and injective.

Let f be an isometry. Then, injectivity follows from the equivalences

f(x1) = f(x2)⇔ d(f(x1), f(x2)) = 0⇔ d(x1, x2) = 0⇔ x1 = x2.

To prove that f is continuous, let (xn)n≥1 be a sequence that converges to x ∈ E. Then

limn→∞

d(f(xn), f(x)) = limn→∞

d(xn, x) = 0

so limn→∞

f(xn) = f(x), which proves that f is sequentially continuous at any x in E metric.

Assume from now on that E is compact and f an isometry. We want to provethat f is surjective. Assume to the contrary the existence of a /∈ f(E).

2. Prove that there exists ε > 0 such that B(a, ε) ⊂ E \ f(E).

It suffices to prove that f(E) is closed, which follows from the fact that f(E) is compactas the continuous image of E compact. Since E is Hausdorff, any compact in E is closed.

3. Consider the sequence defined by x1 = a and xn+1 = f(xn). Prove that

d(xn, xm) ≥ ε

for n 6= m and derive a contradiction.

Assume without loss of generality that 1 < n < m. Then by definition of the sequence,

d(xn, xm) = d(a, xm−n) = d(a, f(xm−n−1)︸ ︷︷ ︸∈f(E)

).

Since no point in f(E) is at distance less than ε of a, it follows that d(xn, xm) ≥ ε.Now since (xn)n≥1 is a sequence in E metric and compact, it admits a subsequence (un)n≥1

with limn→∞ un = ` ∈ K. Then for m and n large enough to guarantee that d(un, k) < ε2

and d(um, k) < ε2, the triangle inequality implies that

d(un, um) ≤ d(un, k) + d(k, um) < ε,

which contradicts the property of (xn)n≥1 established above.

4. Prove that an isometry of a compact metric space is a homeomorphism.

we prove that any continous bijection from a compact space to a Hausdorff space is ahomeomorphism. Let f : X −→ Y be such a map. To prove that g = f−1 is continuous,it suffices to prove that g−1(C) = f(C) is closed for any C closed in X. Closed subsets

60

of compacts are compact so C is compact, therefore f(C) is compact too since f iscontinuous. Finally, compact subsets of Hausdorff spaces are closed so f(C) is closed.

Problem 6

Let X be a set, P(X) the set of subsets of X and ι : P(X) −→ P(X) a mapsatisfying, for all A,B ⊂ X:

(1) ι(X) = X(2) ι(A) ⊂ A(3) ι ι(A) = ι(A)(4) ι(A ∩B) = ι(A) ∩ ι(B).

1. Check that A ⊂ B ⇒ ι(A) ⊂ ι(B).

Note that A ⊂ B ⇔ A ∩B = A, in which case (4) implies ι(A) = ι(A) ∩ ι(B) ⊂ ι(B).

2. Prove that the family T = ι(A) , A ∈P(X) is a topology on X.

Condition (1) says that X ∈ T . Moreover, (2) implies that ι(∅) ⊂ ∅ so that ∅ = ι(∅),which means that ∅ ∈ T .To see that T is stable under finite intersections, it suffices to prove that the intersectionof two elements of T belongs to T , which is guaranteed by (4), and argue by induction.Finally, we prove that T is stable under arbitrary unions. Let Aαα∈J be a family ofsubsets of X; we want to prove that

⋃α∈J ι(Aα) = ι(B) for some B ⊂ X. Observe that

(2) implies that ⋃α∈J

ι(Aα) ⊃ ι

(⋃α∈J

ι(Aα)

).

Moreover, ι(Aα) ⊂⋃α∈J ι(Aα) for every α ∈ J so the result proved in (1) gives

ι(Aα)(3)= ι (ι(Aα)) ⊂ ι

(⋃α∈J

ι(Aα)

).

This holds for every α ∈ J so⋃α∈J ι(Aα) ⊂ ι

(⋃α∈J ι(Aα)

), hence⋃

α∈J

ι(Aα) = ι

(⋃α∈J

ι(Aα)

).

3. Prove that, in this topology, A = ι(A) for all A ⊂ X.

The definition of T and (2) imply that ι(A) is open and a subset of A so ι(A) ⊂ A.

Conversely, (A) is open so it must be of the form ι(B) for some B ⊂ X. Since ι(B) =

A ⊂ A, it follows from (3) and the result of (1) that

A = ι(B) = ι(ι(B)) ⊂ ι(A)

which concludes the proof.61