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Math 54: Topology
Syllabus, 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 theory
O.2 Basics on metric spaces
I. Topological spaces and continuous functions - 16 lectures
I.1.
I.2.
I.3.
I.4.
I.5.
I.6.
Topologies
Bases and subbases
Closed sets
Continuous maps and the category Top.
Topologies on cartesian products
Metrizable topologies
II. Connectedness - 2 lectures
II.1. Connected spaces
II.2. Path connectedness
II.3. Connected components
III. Compactness - 5 lectures
III.1. Compact spaces
III.2. Fréchet and sequential compactness
III.3. Local compactness, Alexandrov Compactification
IV. Other topics - 3 lectures
IV.1.
IV.2.
IV.3.
IV.4.
Separation axioms
The Urysohn Lemma and the Urysohn Metrization Theorem
Normed linear spaces
Topological properties of GL(n, R) and GL(n, C)
2
Contents
Content of the lectures
4
Problem sets and examinations
Problem set 1: review on sets and maps
Problem set 2: metric spaces
Problem set 3: topological spaces
Midterm 1
Problem set 4: closed sets and limit points
Problem set 5: continuous maps, the product topology
Problem set 6: metrizable spaces
Midterm 2: in-class examination
Midterm 2: take-home examination
Problem set 7: connectedness and compactness
Final examination
10
10
11
12
13
15
16
17
19
21
23
24
Elements of solutions for the homework sets and examinations
26
3
Week 1
Lecture 1. General introduction. (Why) should we study Topology? What is it?
Poincaré (1895): Analysis situs.
There is a science called analysis situs and which has for its object
the study of the positional relations of the different elements of
a figure, apart from their sizes. This geometry is purely qualitative; its theorems would remain true if the figures, instead
of being exact, were roughly imitated by a child. [...] The importance of analysis situs is enormous and can not be too much emphasized
[...].
H. Poincaré, The value of Science, 1905.
Intuitive notions of neighborhood and deformation.
Fun: example of topological problem: the seven bridges of Königsberg.
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
∞
∞
P 1
P
P 1
P
1
1
=
diverges
while
=
converges.
squares in the sense that
k
n
k
n2
n=1
n=1
k∈Z+
k∈S
P 1
What about
p?
p prime
Lecture 3. [M, §7]
surj.
inj.
Characterization of countable sets: B is countable iff Z+ −−→ B iff B −→ 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, 1}ω (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 inverse
image 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 0 ) is finer than T (B) if and only if B 0 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: cases
of 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 and
finite unions. A topology can be defined by its closed sets. Closed sets in the subspace
topology. 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
5
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), an
introduction 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 be
made 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 of
the closures. Continuous maps into products.
6
Week 7
Lecture 18. [M, §20]
Metrizable topologies. Equivalent metrics generate the same topology. Example: the
Euclidean and L∞ metrics are equivalent and generate the product topology on Rn .
Topologically equivalent metrics need not be equivalent: any metric is topologically equivalent 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 ‘matrix
functor’ 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. A
connected 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 ⊆ Ā, then B is connected.
Week 8
Lecture 22. [M, §23-25]
The continuous image of a connected space is connected. Finite products of connected
spaces 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 not
compact, {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.
7
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 compacts
by 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) are
compact in the order topology. Compact sets of Rn in any topology associated with a
metric equivalent to the `∞ metric are exactly the closed and bounded subsets.
The Extreme Value Theorem.
Lecture 26. [M, §28]
Féchet (limit point) compactness. Compact spaces are Fréchet compact.
Sequential compactness (Bolzano-Weierstraß property). The three notions are equivalent
in metric spaces.
Lecture 27. [M, §29]
Locally compact Hausdorff spaces. A punctured compact Hausdorff space is locally compact 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 and
characterization in terms of closure of neighborhoods. Sequential characterizations of
closure 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 norms
induce equivalent metrics. In finite dimension, all norms are equivalent.
Exercise: characterizations of continuity for linear maps.
2
Case of Mn (k) ⊂ k n : operator norms are submultiplicative. Continuity of operations: linear 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).
8
Lecture 30. Topology of matrix spaces II.
Density of GL(n, k) in Mn (k) and applications: AB and BA have the same characteristic
polynomial hence the same eigenvalues for any A, B in Mn (k); there exists a basis of
Mn (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 -
9
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. surjective).
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 is
bijective 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 : X m −→ X n .
b. Find a bijective map g : X m × X n −→ X m+n .
c. Find an injective map h : X n −→ X ω .
d. Find a bijective map i : X n × X ω −→ X ω .
e. Find a bijective map j : X ω × X ω −→ X ω .
f. If A ⊂ B, find an injective map k : (Aω )n −→ B ω .
(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
metric.
√ with the Euclidean
√ 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 if
for 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 interior
if there exists r > 0 such that any point x in E such that d(a, x) < r is in A.
o
The set of interior points of A is called the interior of A and denoted by A.
o
a. Prove that A is the union of all the open balls contained in A.
o
b. Prove that A is the largest open subset contained in A.
o
c. Can A be empty if A is not?
11
Problem set 3: topological spaces
Solution p.30
(1) Let {Tα }α∈A be a family of topologies on a non-empty set X.
\
a. Prove that I =
Tα is a topology on X.
α∈A
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
1
(a, b) and (a, b) \
, n ∈ Z+ ;
n
- 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 of
R` × R and as a subspace of R` × R` .
12
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
d(x, y) ≤ ρ(x, y) ≤ 2d(x, y).
2
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 open
in Y ? Which are open in R?
1
A = x , < |x| < 1
2
1
B = x , < |x| ≤ 1
2
1
C = x , ≤ |x| < 1
2
1
D = x , 0 < |x| < 1 and ∈ Z+
x
2. Let X = R` × Ru where R` denotes the topology with basis consisting of all intervals
of the form [a, b) and Ru denotes the topology with basis consisting of all intervals of the
form (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 any
A, 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 m1 + n1 , m, n ∈ Z+ of R.
(3) The boundary of a subset A in a topological space X is defined by
∂A = Ā ∩ X \ A.
(a) Show that Ā = Å 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 = Ū \ U .
˚?
(d) If U is open, is it true that U = Ū
(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 − y 2 ≤ 1}
(f) F = {(x, y) , x 6= 0 and y ≤ x1 }
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 subsets
U such that if n is in U , then any divisor of n belongs to U . Give a necessary
and 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 by
ϕ(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 A0
the sets of accumulation points of A and by ∂A its boundary.
Let f : X −→ Y be a map. Determine the implications between the following
statements:
(i) f is continuous.
(ii) f (A0 ) ⊂ (f (A))0 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 subset
of X and f1 , f2 continuous maps from the closure Ā to Y .
Prove that if f1 and f2 restrict to the same function f : A → Y , then f1 = f2 .
(4) Let {Xα }α∈J be a family of topological spaces and X =
Q
Xα .
α∈J
(a) Give a necessary and sufficient condition for a sequence {un }n∈Z+ to converge
in 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 ρ̄ be the uniform metric on Rω . For x = (xn )n∈Z+ ∈ Rω and 0 < ε < 1, let
Y
P (x, ε) =
(xn − ε, xn + ε).
n∈Z+
(a) Compare P (x, ε) with Bρ̄ (x, ε).
(b) Is P (x, ε) open in the [
uniform topology?
(c) Show that Bρ̄ (x, ε) =
P (x, δ).
δ<ε
(2) We denote by `2 (Z+ ) the set of square-summable real-valued sequences, that is,
(
)
X
`2 (Z+ ) = x = (xn )n∈Z+ ∈ Rω ,
x2n converges .
n≥1
We admit that the formula
!1/2
d(x, y) =
X
(xn − yn )2
n≥1
2
defines a metric on ` (Z+ ).
(a) Compare the metric topology induced by d on `2 (Z+ ) with the restrictions
of the box and uniform topologies from Rω .
(b) Let R∞ denote the subset of `2 (Z+ ) consisting of sequences that have finitely
many 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 a
sequence of continuous functions that converges uniformly to f : X −→ Y .
Let (xn )n≥0 be a sequence in X such that lim xn = x. Prove that
n→∞
lim fn (xn ) = f (x).
n→∞
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 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
1. Show that a topological space is T1 if and only if for any pair of distinct points, each
has 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
Y
P = (0, +∞)ω =
(0, +∞)
n≥1
be the subset of positive sequences.
2. Verify that 0 belongs to P̄ .
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 the
uniform 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 other
source 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 same
connected 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
un − xn if xn = yn
.
vn =
un −xn
if xn 6= yn
yn −xn
(b) Deduce a necessary and sufficient condition for x and y in Rω to lie in the same
connected component for the box topology.
21
Problem 2
Let F be a functor between categories C and C 0 . A functor G : C 0 −→ C is said to be a
left adjoint for F if there is a natural isomorphism
HomC (G(X), Y ) ∼
= HomC 0 (X, F (Y ))
for all objects X ∈ C 0 and Y ∈ C. Similarly, G is called a right adjoint for F if there is a
natural isomorphism
HomC (X, G(Y )) ∼
= HomC 0 (F (X), Y )
0
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 the
trivial 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] −→ U
with 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 Y̊ 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 necessarily compact.
(4) (a) Prove that the Alexandrov compactification of R is homeomorphic to the
unit circle
S1 = {(x, y) ∈ R2 , x2 + y 2 = 1}.
(b) Verify that Z+ ⊂ R is a locally compact Hausdorff space.
(c) Prove that the Alexandrov compactification of Z+ is homeomorphic to
1
, n ∈ Z+ ∪ {0}.
n
23
Final examination
Solution p.58
This exam consists of 6 independent problems. Your may treat them in the order of your
choosing, 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 singletons.
1. Prove that a discrete space is totally disconnected.
2. Does the converse hold?
Problem 3
Let {Xα }α∈J be a family of topological spaces; let Aα ⊂ Xα for each α ∈ J.
Q
1. In α∈J Xα equipped with the product topology, prove that
Y
Y
Aα .
A¯α =
α∈J
α∈J
2. Does the result hold if
Q
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 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).
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, Å = 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 is
injective 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 and
ϕb : 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 = Id{x} , both ϕa and ϕb have a left inverse, but they
have no right inverse as the only candidate would be ψ and ϕa ◦ ψ 6= Id{a,b} .
d. Can a function have more than one right inverse? More than one
left 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
and
µa (w) = a ,
are both left inverses of λ.
µb (w) = b
e. Show that if f has both a left inverse g and a right inverse h, then
f 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 : X m −→ X n .
n−m
z }| {
Consider f (x,1 , . . . , xm ) = (x1 , . . . , xm , x0 , . . . , x0 ) for some x0 ∈ X.
b. Find a bijective map g : X m × X n −→ X m+n .
Consider g ((x1 , . . . , xm ) , (y1 , . . . , yn )) = (x1 , . . . , xm , y1 , . . . , yn )
c. Find an injective map h : X m −→ X ω .
d. Find a bijective map i : X m × X ω −→ X ω .
Consider ‘infinite versions’ of the functions f and g defined above.
e. Find a bijective map j : X ω × X ω −→ X ω .
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 −→ B ω .
Consider k : (1 x` )`∈Z+ , . . . , (n x` )`∈Z+ 7→ (1 x1 , . . . , n x1 , , . . . , 1 x` , . . . , n x` , . . .).
(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 is
infinite, 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 B n ,
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)
ε
and it suffices to choose |x−a| < min{ 2|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 be
continuous 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 of
v.
(3) Let (E, d) be a metric space. Prove that a subset Ω ⊂ E is open if and
only if for every point x ∈ Ω, there exists an open ball containing x and
included in Ω.
The definition seen in class for open sets in a metric space differs only by the fact
that it requires the ball to be centered at the point considered. Therefore, open
sets trivially satisfy the property.
Observe that if a point x is included in a ball B(a, r), the triangle inequality
implies 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 interior
if there exists r > 0 such that any point x in E such that d(a, x) < r is
o
in A. The set A of interior points of A is called the interior of A.
o
a. Prove that A is the union of all the open balls contained in A.
Let Ȧ be the union of all the open balls contained in A and let a be in Ȧ.
By definition of Ȧ and the argument used in (3), there exists a ball B(a, r)
o
o
included in A, so Ȧ ⊂ A. Conversely, let a be in A. By definition, there
exists r > 0 such that B(a, r) ⊂ A so a ∈ Ȧ, hence the result.
o
b. Prove that A is the largest open subset contained in A.
o
First, A is open as the union of open subsets, as proved in a. We argue by
o
contradiction: assume the existence of Ω open such that A
o
Ω ⊂ A. Let
o
x ∈ Ω \ A. Since x ∈
/ A, no ball B(x, r) with r > 0 is contained in A. Since Ω
is open, it must contain such a ball, which contradicts the assumption that
Ω ⊂ A.
o
c. Can A be empty if A is not?
Yes: consider for instance A = Z in E = R, or any strict linear subspace of
Rn and observe that every ball centered at the origin must contain a basis.
29
Problem set 3: topological spaces - Elements of solution
(1) Let {Tα }α∈A be a family of topologies on a non-empty set X.
\
a. Prove that I =
Tα is a topology on X.
α∈A
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 finer
topology than I and consider U ∈ J \ I. Since U ∈
/ I, there exists some
α0 ∈ 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 other
property follows from the observation that if Ba (n) ∩ Bb (m) is either empty
of 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;
- T
with basis elements of the form (a, b) and (a, b)\
21: the K-topology,
,
n
∈
Z
;
+
n
- 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 is
of the form (−∞, x1 ) ∪ (x1 , x2 ) ∪ . . . ∪ (xn , +∞), that is a union of basis elements
for T1 . Therefore, T3 ⊂ T1 and the inclusion is strict since any union of finitely
complemented set is finitely complemented: T3 T1 .
The argument given in Lemma 13.4 for the lower limit topology can be adapted
to show that T1 T4 .
S
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
S 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 . The
inclusion is strict: consider for instance (0, 2] ∈ T4 . No basis element of T2 can
contain 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 union
of basis elements of T3 can have infinite complement as (−∞, 0) ∈ T5 does, so
T5 6⊂ T3 . Conversely, R \ {0} = (−∞, 0) ∪ (0, +∞) is a basis element for T3 that
does 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
T5
∩
are strict:
T3 ⊂ T 1 ⊂ T 2 ⊂ T 4
(4) Let L be a straight line in R2 . Describe the topology L inherits as a
subspace 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 considering
intersections 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 vertical
and 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 is
slanted, it carries the lower limit topology.
- In the case of L ⊂ R` × R, the same line of reasoning shows that L receives
the lower limit topology if it is vertical or has a non-negative slope. If
the 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 induced
topology 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 to
this list, we must also include all of its divisors. For instance, any open set containing 10
will 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+ contains
the divisors of any integer so it is open too.
S
(O2) Let {Uα }α∈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.
p
T
(O3) Let {Ui }1≤i≤p be a family of elements of T and U =
Ui . Let n ∈ U . For every
i=1
1 ≤ i ≤ p, the integer n belongs to Ui so every divisor of n belongs to all the Ui ’s, hence
to 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 form
a 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
d(x, y) ≤ ρ(x, y) ≤ 2d(x, y).
2
Compare the topologies generated by d and ρ.
Denote by Td and Tρ the topologies associated with the metrics d and ρ. We shall prove
that 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 radius
r0 > 0 such that Bd (x, r0 ) ⊂ Bd (a, r). The inequalities satisfied by d and ρ imply that
x ∈ Bρ (x,
r0
) ⊂ Bd (x, r0 ).
2
r0
1
r0
, then d(x, y) <
so y ∈ Bd (x, r0 ).
2
2
2
Lemma 13.3 in [M] then implies that Td ⊂ Tρ . The converse inclusion can be proved by
the same argument, using the other inequality satisfied by ρ and d and we can conclude
that Td = Tρ .
Indeed, if ρ(x, y) <
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 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 .
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 .
n
T
Any element of T1 ∨T2 is the union of subsets of X of the form U =
Ui with Ui ∈ T1 ∪T2
i=1
for each i. Such a U is an intersection of elements of T1 and of T2 , all of which belong
to T assumed finer so U belongs to T . Since T is stable under unions, it follows that
T1 ∨ 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 sets
are open in Y ? Which are open in R?
A = x , 21 < |x| < 1 = (−1, − 12 ) ∪ ( 12 , 1) is open in R as the union of two basis elements.
It is also open in Y by definition.
B = x , 21 < |x| ≤ 1 = [−1, − 12 ) ∪ ( 12 , 1] is not open in R since no open interval containing 1 is included in B. It is open in Y as the intersection of Y with (−2, − 21 ) ∪ ( 12 , 2),
open in R.
C = x , 12 ≤ |x| < 1 = (−1, − 21 ] ∪ [ 12 , 1) is not open in R since no open interval containing 12 is included in C. It is not open in Y for the same reason: no intersection of Y
with an open interval containing 21 is included in C.
D = x , 0 < |x| < 1 and x1 ∈ Z+ = n1 , 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 all
intervals of the form [a, b) and Ru denotes the topology with basis consisting
of all intervals of the form (c, d]. Describe the topology induced on the plane
curve Γ with equation y = ex .
A basis for the subspace topology on Γ is given by the sets [a, b) × (c, d] ∩ Γ. Singletons
are 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
(iv)
equivalent to A ∪ B = B. Therefore, γ(B) = γ(A ∪ B) = γ(A) ∪ γ(B) ⊃ γ(A).
(i)
(O1) The subset X = X \ ∅ = X \ γ(∅) is in T . Moreover, (ii) implies that
X = γ(X) so ∅ = X \ γ(X) is also in T .
(O2) S
Let {Uα }α∈J be a family such that Uα = X \ γ(Aα ) for each α ∈ J, and
U = α∈J Aα . De Morgan’s Laws imply that
\
Aα
X \U =
α∈J
and weTwant 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,
(iii)
γ(X \ U ) ⊂ γ(γ(Aα )) = γ(Aα )
T
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’s
Laws imply that
!
n
n
n
\
\
\
X\
Ui = X \
γ(Ai ) = X \ γ
Ai
i=1
i=1
i=1
where the last equality follows from (iv) by induction. This shows that T is
stable under finite intersections, which concludes the proof that it is a topology
on X.
Let A be a subset of X. Then γ(A) is closed by definition of T and A ⊂ γ(A)
by (ii) so Ā ⊂ γ(A). Conversely, observe that X \ Ā, being open, is of the form
X \ γ(B), that is, Ā = γ(B) for some B ⊂ X. Since A ⊂ Ā, and γ preserves
inclusions, it follows that
(iii)
γ(A) ⊂ γ(Ā) = γ(γ(B)) = γ(B) = Ā,
hence γ(A) = Ā.
35
(2) (a) Show that a topological space X is Hausdorff if and only if the
diagonal ∆ = {(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, there
exist disjoint open sets Ux 3 x and Uy 3 y. By definition of the product
topology, U = Ux × Uy is a neighborhood of (x, y) and by the preliminary
observation, U ∩ ∆ = ∅ so X × X \ ∆ is open hence ∆ is closed.
Conversely, assume that ∆ is closed and let x 6= y in X. Since (x, y) belongs
to (X × X) \ ∆ assumed open, there exists a neighborhood V of (x, y) such
that V ∩ ∆ = ∅. Product of open sets form a basis for the topology of
X × 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, guarantees
that U1 and U2 are disjoint neighborhoods of x and y respectively.
(b) Determine the accumulation points of A = m1 + n1 , m, n ∈ Z+ ⊂ R.
Let A0 denote the
points of A. The fact that limn→∞
n set of accumulation
o
1
n
=
0 implies that
, p ∈ Z+ ∪ {0} ⊂ A0 . Let us prove the converse inclusion.
First, observe that if an interval (a, b) with a > 0 contains infinitely many
elements of the form m1 + n1 , then one of the variables m and n must take
only finitely many values, while the other takes infinitely many values. Now
let x ∈ A0 with x > 0. For any ε > 0, the set Bε = (x − ε, x + ε) ∩ A must
be infinite. Without loss of generality, we can assume that
1
1
Bε =
+ , m ∈ F , n ∈ Im
m n
with F finite and at least one Im infinite, say Im0 . For all n ∈ Im0 , we have
1 1
1
1
− ≤ x −
x −
− < ε.
m0
n
m0 n
1
p
For
n large
enough, the left-hand side can be made arbitrarily close to
1 x − m0 , in particular, we get that 21 x − m10 < ε. If x > 0 is not of
1
m0
for any m0 ∈ Z+ , then there exists a positive minimum value
1
1 for the numbers 2 x − m0 and Bε cannot be infinite for arbitrarily small
values of ε.
the form
36
(3) The boundary of a subset A in a topological space X is defined by
∂A = Ā ∩ X \ A.
(a) Show that Ā = Å t ∂A5.
If x ∈ Å, 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 Å ∩ ∂A = ∅.
The interior and boundary of A are included in Ā by definition so the inclusion Ā ⊃ Å t ∂A is trivial. Conversely, let x ∈ Ā. If x has a neighborhood U
such that U ⊂ A, then x ∈ Å. The alternative is that every neighborhood of
x has non-empty intersection with X \ A, that is x ∈ X \ A so that x ∈ ∂A.
Therefore, Ā ⊂ Å 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, Å ⊆ A ⊆ Ā and A is open and
closed if and only if Å = Ā. By (a), this is equivalent to ∂A = ∅.
(c) Show that U is open if and only if ∂U = Ū \ U .
The result of (a) states that U and Ů are complements in Ū , so Ů = Ū \ ∂U
and U is equal to Ů , that is, U is open if and only if U = Ū \ ∂U , which is
equivalent to the condition ∂U = Ū \ U .
˚?
(d) If U is open, is it true that U = Ū
˚. However, the reverse
If U is open, the inclusion U ⊂ Ū implies that U ⊂ Ū
inclusion may fail: consider for instance U = R \ {0} in R. It is open as the
˚ = R ! U.
union of open intervals and Ū = R so that Ū
(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 − y 2 ≤ 1}
(f) F = {(x, y) , x 6= 0 and y ≤ x1 }
Note that, except for (d), a picture is very helpful to determine the boundary
and interior of the subsets at hand before rigorously justifying the intuition, using
what 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, +∞) which
is open as a product of open sets. Another way to see this is to remark that every
element of R2 \ A is of the form (x, y) with y 6= 0, and for any x ∈ R, the basis
element
|y|
|y|
V = (x − 1, x + 1) × y − , y +
2
2
satisfies (x, y) ∈ V ⊂ R2 \ A.
Moreover, the interior of A is empty: every element of A is of the form (x, 0), any
neighbourhood 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, d2 ) ∈
/ A.
We conclude that Å = ∅ and ∂A = A.
(b) Note that B = (0, +∞) × (−∞, 0) ∪ (0, +∞) is open as a product of open
sets. Another way to see this is to consider (x, y) ∈ B, that is, x > 0 and y 6= 0.
Then
x 3x
|y|
|y|
V =
,
× y − ,y +
2 2
2
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 the
continuous map (x, y) 7→ (ln x, 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. If
xy = 0, then V contains asubset of the form (a, b) × (c, d) with 0 < b and cd 6= 0
so x+b
, y+d
, y+c
or x+b
belongs to V ∩ B, which is therefore not empty. We
2
2
2
2
have proved that R ⊂ B̄. The converse inclusion follows from the same argument
invoked to prove that R2 \ A is open.
Since B is open, it follows from (c) in the previous problem that ∂B = B̄ \ B,
that is ∂B is the union of the vertical axis and the positive horizontal axis.
(c) Since A ∪ B = Ā ∪ B̄, it follows form (a) and (b) that C̄ = R ∪ A consists of
the points (x, y) such that x ≥ 0 or y = 0.
Next, C̊ is the right half-plane (0, +∞) × R: this set is open as the product of
open 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
, y+d
2
2
y+c
2
or x+a
,
belongs
to
V
∩
(R
\
C),
which
is
therefore
not
empty.
2
2
It follows from the result proved in (a) of the previous problem that ∂C = C̄ \ C̊
is the union of the vertical axis and the negative horizontal axis.
(d) Every non-empty open interval of R contains infinitely many rational and
irrational numbers, so every product of intervals contains infinitely many elements
of D and R2 \ D. Therefore, ∂D = R2 and, since ∂D = D̄ \ D̊, it follows
immediately that D̊ = ∅.
38
(e) First, observe that the set Ω = {(x, y) , 0 < x2 − y 2 < 1} is open, for instance
as the inverse image of the open set (0, 1) under the map (x, y) 7→ x2 − y 2 , which
is polynomial, hence continuous.
A similar argument, shows that Γ = {(x, y) , 0 ≤ x2 − y 2 ≤ 1} is closed. Since
Ω ⊂ E ⊂ Γ, we get the chain of inclusions Ω ⊂ E̊ ⊂ Ē ⊂ Γ, hence
∂E = Ē \ E̊ ⊂ Γ \ Ω.
In other words, a boundary point (x, y) of E satisfies either x2 = y 2 or x2 −y 2 = 1.
Conversely, assume that x2 − y 2 = 1. Every neighbourhood of (x, y) contains the
points Pδ = (x + δ, y) for δ ∈ (−δ0 , δ0 ) with δ0 > 0. Since
(x + δ)2 − y 2 = 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 boundary
point of E. One can proceed in the same way to verify that the two lines given
by the equation x2 = y 2 are also included in ∂E, which concludes the proof that
∂E consists exactly of the union of the hyperbola with equation x2 − y 2 = 1 and
the lines with equations y = ±x.
It also follows that E̊ = Ω. We have already obtained the inclusion Ω ⊂ E̊.
Conversely, assume that (x, y) is a point in E not in Ω. Then x2 − y 2 = 1 so
(x, y) belongs to ∂E which is disjoint from E̊. This proves that E̊ ⊂ Ω and the
equality.
(f) No new technique is needed to prove that F̊ is the region located strictly
below the branches of the hyperbola with equation xy = 1, that is,
1
F̊ = (x, y) , x 6= 0 and y <
,
x
and that ∂F is the union of the hyperbola and the vertical axis:
1
.
F̊ = (x, y) , x = 0 or y =
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 are
the subsets U such that if n is in U , then any divisor of n belongs
to 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 of
n. 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, then
f (b) ∈ Un , that is, f (b)|n. A necessary condition for continuity is therefore
that 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 it
contains all the divisors of a. The property of f implies that f (b)|f (a) for
every 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 −→ R
defined 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,
a 3a
a 3a
−1
ϕ ( ,
)=
,
∩Q
2 2
2 2
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 denote
by A0 the sets of accumulation points of A and by ∂A its boundary.
Let f : X −→ Y be a map. Determine the implications between the
following statements.
(i) f is continuous.
(ii) f (A0 ) ⊂ (f (A))0 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 converse
is true: let A be a subset of X and x ∈ Ā = A ∪ A0 .
- If x ∈ A, then f (x) ∈ f (A) ⊂ f (A).
- If x ∈ A0 , then (ii) implies that f (x) ∈ (f (A))0 ⊂ f (A).
Therefore, f (Ā) ⊂ 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 and
x ∈ ∂(f −1 (B)). If x ∈
/ f −1 (∂B), there are two possibilities.
Case 1: f (x) ∈ B̊. Then x ∈ f −1 (B̊), open by continuity of f . Since f −1 (B̊) ⊂
f −1 (B), it follows that x is an interior point of f −1 (B), which is a contradiction.
Case 2: f (x) ∈ Y \ B̄. Then x ∈ f −1 (Y \ B̄), open by (i). In particular, there
is a neighborhood U of x such that U ⊂ f −1 (Y \ B̄). Since x ∈ ∂(f −1 (B)), it
follows that there exists some y in U such that f (y) ∈ B, which contradicts the
assumption that f (x) ∈ Y \ B̄.
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 (B̄) for every B ⊂ Y .
Proof of the lemma: if f is continuous, the inverse image of the closed set B̄ is a
closed 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 that
f −1 (B) 6⊂ f −1 (B̄). Let x be an element of f −1 (B) such that f (x) ∈
/ B̄, hence
−1
−1
−1
−1
f (x) ∈
/ B. Since f (B) = f (B) ∪ ∂f (B), it follows that x ∈ ∂f (B).
The fact that f (x) ∈
/ B̄ implies that f (x) ∈
/ B̄ \ B̊ = ∂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 be
a subset of X and f1 , f2 continuous maps from the closure Ā to Y .
Prove that if f1 and f2 restrict to the same function f : A → Y , then
f1 = f2 .
We argue by contradiction: if f1 6= f2 , there exists x ∈ Ā 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 f1−1 (V1 ) and f2−1 (V2 ) are
neighborhoods of x, and so is U = f1−1 (V1 ) ∩ f2−1 (V2 ). Since x ∈ Ā \ A, the
neighborhood U contains some a in A such that f1 (a) = f (a) = f2 (a). Therefore,
f (a) ∈ V1 ∩ V2 which is assumed empty.
Q
(4) Let {Xα }α∈J be a family of topological spaces and X =
Xα .
α∈J
(a) Give a necessary and sufficient condition for a sequence {un }n∈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
,
lim unα = lα .
n→∞
Conversely, assume that limn→∞ πα (un ) = πα (l) for every α ∈ J and let U be
a neighborhood of l. We may assume that U is an intersection of cylinders,
that is,
U = πα−11 (Uα1 ) ∩ πα−12 (Uα2 ) ∩ . . . ∩ πα−1p (Uαp )
since such elements form a basis for the product topology. With our assumption, 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 ≥ max Ni , so limn→∞ un =
1≤i≤n
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 (?) is
certainly necessary. It is not sufficient, however, as the following example
shows.
Let J = Z+ and Xk = R with the standard topology for each k ∈ Z+ so that
X = Rω as a set. Then, consider the sequence (n u)n∈Z+ defined by
1 if k = n
n
uk =
.
0 if k 6= n
Then limn→∞ n uk = limn→∞ πk (n u) = 0 for every k ∈ Z+ so n u converges to
the zero sequence in the product Q
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 (n u)n∈Z+ , which therefore
cannot converge to zero in the box topology.
42
Problem set 6: metrizable spaces - Elements of solution
(1) Let ρ̄ be the uniform metric on Rω . For x ∈ Rω and 0 < ε < 1, let
Y
P (x, ε) =
(xn − ε, xn + ε).
n∈Z+
(a) Compare P (x, ε) with Bρ̄ (x, ε).
With 0 < ε < 1, the definition of ρ̄ implies that
ω
Bρ̄ (x, ε) = y ∈ R , sup |xn − yn | < ε .
n≥1
On the other hand, P (x, ε) = {y ∈ Rω , |xn − yn | < ε for every n ≥ 1} so
the inclusion Bρ̄ (x, ε) ⊂ P (x, ε) holds by definition of the supremum. This
inclusion is strict: the sequence y defined by
1
yn = x n + ε −
n
belongs to P (x, ε) but ρ̄(x, y) = ε so Bρ̄ (x, ε) P (x, ε).
(b) Is P (x, ε) open in the uniform topology?
No: we prove that P (x, ε) contains no ρ̄-ball centered at the element y
introduced above. For any η > 0, the sequence z defined by zn = yn + η2
satisfies ρ̄(y, z) < η hence belongs to Bρ̄ (y, η).
η
Observe that xn + ε − yn < for n large enough, so
2
η
zn − xn = yn + − xn
2
1 η
= ε− +
n 2
> ε
which means that z ∈
/ P (x, ε), which is therefore not a neighborhood of y,
hence not open.
[
(c) Show that Bρ̄ (x, ε) =
P (x, δ).
δ<ε
Let y ∈ P (x, δ) with δ < ε. Then, ρ̄(x, y) ≤ δ < ε so y ∈ Bρ̄ (x, ε), which
proves that Bρ̄ (x, ε) ⊃ P (x, δ). Since δ was arbitrary, we conclude that
[
Bρ̄ (x, ε) ⊃
P (x, δ).
δ<ε
Conversely, if y ∈ Bρ̄ (x, ε), then y ∈ P (x, δy ) with δy = ρ̄(x, y) < ε, so
[
Bρ̄ (x, ε) ⊂
P (x, δ).
δ<ε
43
(2) We denote by `2 (Z+ ) the set of square-summable real-valued sequences:
(
)
X
`2 (Z+ ) = x = (xn )n∈Z+ ∈ Rω ,
x2n converges ,
n≥1
equipped with the metric
!1/2
d(x, y) =
X
(xn − yn )2
.
n≥1
(a) Compare the metric topology induced by d on `2 (Z+ ) with the
restrictions of the box and uniform topologies from Rω .
The inclusion Tρ̄ ⊂ Td follows from the observation that Bd (x, r) ⊂ Bρ̄ (x, r)
for any x ∈ `2 (Z+ ) and r > 0. Indeed, observe that
X
|xk − yk |2 ≤
|xn − yn |2
n≥1
for any fixed k, so that ρ̄(x, y) ≤ d(x, y) for any x, y ∈ `2 (Z+ )6.
This inclusion is strict, as the following example shows. Denote by 0 the
sequence that is constantly equal to 0. We will prove that Bd (0, 1), open in
Td by definition, is not open in the uniform topology. More
precisely, no ball
2 2
Bρ̄ (0, r) with r > 0 is included in Bd (0, 1): let n0 > r and consider the
sequence
r
if n ≤ n0
2
ξn =
.
0 if n > n0
√ r
/ Bd (0, 1).
Then ρ̄(0, ξ) = 2r < r so ξ ∈ Bρ̄ (0, r) but d(0, ξ) = n0 > 1 so ξ ∈
2
Next, we prove that Td ⊂ Tbox : let x ∈ `2 (Z+ ), r > 0 and consider the box
Y
r
r
B=
(xn − n , xn + n ).
2
2
n≥1
Then x ∈ B ⊂ Bd (x, r) since y ∈ B implies d(x, y)2 ≤
X r2
r2
=
.
4n
3
n≥1
Again, the inclusion is strict: consider the open box
Y 1 1
.
B=
− ,
n n
n≥1
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 convergence
and conclude by the sequential characterization of the closure.
44
2
r
and let η be the sequence defined by
r
if n = n0
2
ηn =
.
0 if n 6= n0
Indeed, consider n0 >
Then x ∈
/ B but d(0, x) =
r
2
We have proved that Tρ̄
Td
< r so x ∈ Bd (0, r). Therefore B ∈
/ Td .
Tbox .
(b) Let R∞ denote the subset of `2 (Z+ ) consisting of sequences that
have finitely many non-zero terms. Determine the closure of R∞
in `2 (Z+ ).
`2 (Z+ )
We will prove that R∞
= `2 (Z+ ), using the sequential characterization
of the closure in a metric space. For any x ∈ `2 (Z+ ), let n x be the sequence
defined by
xk if k ≤ n
n
xk =
.
0 if k > n
P
Then, d(n x, x)2 = k>n x2k . This quantity converges to 0 as n → ∞ because
x is square-summable, so lim n x = x in `2 (Z+ ).
n→∞
(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 xn = x. Prove that
n→∞
lim fn (xn ) = f (x).
n→∞
Let ε > 0 and observe that the triangle inequality implies that
d(fn (xn ), f (x)) ≤ d(fn (xn ), f (xn )) + d(f (xn ), f (x)) .
|
{z
} |
{z
}
An
Bn
The convergence of the sequence is assumed uniform so there exists an integer
ε
NA such that d(fn (ξ), f (ξ)) < for any n ≥ NA and any ξ ∈ X.
2
ε
In particular, An < for n ≥ NA .
2
Moreover, the Uniform Limit Theorem guarantees that f is continuous. Thereε
fore, there exists NB such that Bn < for any n ≥ NB .
2
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 every
element 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), d(x, z)) < r,
| {z }
<r
which implies that d(y, z) < r by (4). It follows that B ⊂ B(y, r). The
reverse 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 the
topology induced by d.
Let B be a closed ball, that is B = Bc (x, r) = {y ∈ X , d(x, y) ≤ r} for
some x ∈ X and r > 0. Let y ∈ B. The open ball B y, 2r is a neighborhood
of y contained in B:
d(x, z) ≤ max(d(x, y) , d(y, z)) ≤ r
| {z } | {z }
(4)
≤r
<r/2
r
2
for any z ∈ B y, , 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 xn = x for some x ∈ X.
n→∞
Since d(xn , x) can be made arbitrarily small, there exists an integer n0 such
that 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-class
Elements of solution
Problem 1
1. Show that a topological space is T1 if and only if for any pair of distinct
points, 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 neighborhood
of y that does not contain x. Similarly, X \ {y} is a neighborhood of x that does not
contain y. Conversely, assume that distinct points have neighborhoods that does not
contain the other and let x ∈ X. Then if y 6= x, there is a neighborhood of y that does
not 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.
Ξ̊ = (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 that
d−1 ((a, b)) is not empty. Let (x, y) ∈ d−1 ((a, b)) and d = d(x, y). Then, by the triangle
inequality,
b−d
b−d
(p, q) ∈ B(x,
) × B(y,
) ⇒ d(p, q) < b.
2
2
The triangle inequality also implies that d(p, q) ≥ d(x, y) − d(x, p) − d(v, y) so
d−a
d−a
) × B(y,
) ⇒ a < d(p, q).
2
2
d−a
It follows that B(x, r) × B(y, r) with r = min b−d
,
is a neighborhood of (x, y)
2
2
contained in d−1 ((a, b)), which is therefore open.
(p, q) ∈ B(x,
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) belongs
to 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 continuous.
Problem 3
We prove that the box topology on Rω is not metrizable.
1. Recall the definition of the box topology on Rω .
Q
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
Y
P = (0, +∞)ω =
(0, +∞)
n≥1
be the subset of positive sequences.
2. Verify that 0 belongs to P̄ .
Q
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 ∈ P̄ .
3. Prove that no sequence (pn )n≥1 ∈ P ω converges to 0 in the box topology.
Let (n u)n≥1 be a sequence of elements of P and consider the open box
Y
B=
(−n un , n un ).
n≥1
n
Then 0 ∈ B, but no u belongs to B, since the nth term of n u lies outside the nth interval
in the product defining B.
4. Conclude.
In a metrizable space, closure points of are limits of sequences. Here, 0 is a closure point
of P that is the limit of no sequence of elements of P . Therefore, the box topology on Rω
is 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 ρ̄(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 converges
for T∞ .
Assume that fn converges uniformly to f and let 0 < ε < 1. Then for n ≥ N 2ε and all
x ∈ X,
ε
min{|f (x) − g(x)|, 1} = |fn (x) − f (x)| < ,
2
so
ε
ρ̄(fn , f ) = sup min{|fn (x) − f (x)|, 1} ≤ < ε,
2
x∈X
which means that fn converges to f in the uniform topology.
Conversely, assume that limn→∞ ρ̄(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 (n u)n≥1 be
such that each (n uk )k≥1 is bounded:
|n uk | ≤ Mn
for all k ≥ 1
n
and assume that limn→∞ u = u, uniformly. Then there exists an integer n0 such that
(?)
∀n ≥ n0 , sup |n uk − uk | < 1.
k≥1
If u were unbounded, it would admit a subsequence uk` such that lim`→∞ |uk` | = +∞.
Since n0 u is bounded by Mn0 , this would imply that
lim |n0 uk` − 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 (n u)n≥1 be a uniformly convergent sequence of elements of R∞ and u = limn→∞ n u.
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,
|n u − uk | = |uk | > η
for some k
since n u has only finitely many non-zero terms. This implies that supk≥1 |n uk − uk | ≥ η
for all n ≥ 1, which contradicts the uniform convergence of (n u)n≥1 .
Conversely, any sequence u in c0 (Z+ ) is the uniform limit of its truncations: let n u be the
sequence defined by
uk if k ≤ n
n
uk =
.
0 if k > n
Then, n u ∈ c0 (Z+ ) and
sup |n uk − uk | = sup |uk | −→ 0
k≥1
k>n
n
so u converges uniformly to u.
50
n→∞
Midterm 2: take-home
Elements of solution
Problem 1
The purpose of this problem is to study the connected components of Rω in various topologies. In what follows, B and U respectively denote the sets of bounded and unbounded
sequences. Note that Rω = B t U .
The sequence whose terms are constantly equal to 0 is denoted by 0. The connected
component 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 of
A-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 component
map sxn ,yn is polynomial hence continuous from [0, 1] to R. Since each sx,y (t) is a realvalued 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 x
is bounded.
We know that B is closed in Rω for the uniform topology. A similar argument
shows that U is also closed, so that they constitute a separation of Rω in this
topology. 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 · sup |xn |
n≥1
for every t ∈ [0, 1]. It follows that it is continuous and that every s0,x (t) is
bounded 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 in
the 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 component
if 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 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.
Let us prove the following:
Lemma. Let (αn )n∈Z+ and (βn )n∈Z+ be fixed sequences of real numbers such
that αn 6= 0 for all n. Then the map
ϕ:
.
Rω
−→
Rω
(un )n∈Z+ 7−→ (αn un + βn )n∈Z+
is a homeomorphism in the box topology.
un
Proof. Since ϕ is bijective with inverse (un )n∈Z+ 7−→
− βn
it suffices
αn
n∈Z+
ω
to prove that every map from
Q R to itself with (non-constant) affine components
is continuous. Let W = n≥1 (sn , tn ) be a basis element for the box topology.
Then,
^
Y sn
tn
−1
− βn ,
− βn
ϕ (W ) =
α
α
n
n
n≥1
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 ϕ
defined by
(
un − xn
if xn = yn
u
−
x
n
n
ϕ(u)n =
en ·
if xn 6= yn
yn − xn
for u ∈ Rω . Then ϕ is a homeomorphism of Rω by the lemma and ϕ(x) = 0 ∈ B
while 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 in
the same connected component for the box topology.
Let x, y ∈ Rω . Since B t U is a separation of Rω in the box topology, any
homeomorphism ϕ of Rω must satisfy ϕ(Cx ) ⊂ B or ϕ(Cx ) ⊂ U . By the previous
question, 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].
ω
Let us prove that this map
Q is continuous between R and R equipped with the
box 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
T only for finitely many values of n, the inverse image of W
under 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 path
connected, hence equal to Cx .
Problem 2
Let F be a functor between categories C and C 0 . A functor G : C 0 −→ C is said
to be a left adjoint for F if there is a natural isomorphism
HomC (G(X), Y ) ∼
= HomC 0 (X, F (Y ))
for all objects X ∈ C 0 and Y ∈ C. Similarly, G is called a right adjoint for F if
there is a natural isomorphism
HomC (X, G(Y )) ∼
= HomC 0 (F (X), Y )
for all objects X ∈ C and Y ∈ C 0 .
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 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 properties 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, it
suffices to check that if f is a morphism in Set, then G(f ) is a morphism in Top and
compatibility with compositions in each categroy. Let X and Y be objects in Set, and
f ∈ Hom(X, Y ) that is, f is a map between sets X and Y . Then G(f ) = f by definition
and 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 ))
with
HomSet (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 element of HomTop ((X, T ), G(Y )). In other words, the natural isomorphism realizing the
adjunction 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. ) and
H(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 path
γ : [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 that
B(x, r) ⊂ U . Then for y ∈ B(x, r), the map γ̃ : [0, 1] −→ U defined by
γ(2t)
if 0 ≤ t ≤ 12
γ̃(t) =
2(1 − t)x + (2t − 1)y if 21 ≤ 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 > 0
such that B(x, r) ⊂ U . Since x is an accumulation point, there exists y 6= x
in B(x, r) ∩ Γa . Then, as in the proof that Γa is open, one can concatenate a
path from a to y and the segment from y to x to get a continuous path in U
that connects a to x. It follows that x ∈ Γa , which means that Γa contains
its 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 closed
in 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 path
connectedness are equivalent for open subsets of R2 .
(2) Let X be a topological space and Y ⊂ X a connected subspace.
(a) Are Y̊ and ∂Y necessarily connected?
The answer is negative in both cases. Let Y1 = L ∪ R ⊂ R2 be the union of
the half plane L = {x ≤ 0} and the half-cone R = {x ≥ 0 , |y| ≤ x}. Then
Y1 is connected because both L and R are, and they intersect at the origin.
On the other hand,
Y˚1 = {x < 0} t {x > 0 , |y| < x}
is 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 suffices
to consider a closed interval of R. Another example is that of a closed washer
in R2 : let Y2 = {1 ≤ x2 + y 2 ≤ 4}. It is homeomorphic to the rectangle
[1, 2]×[0, 2π) via polar coordinates, hence connected. However, its boundary
consists of two disjoint circles, closed in R2 :
∂Y2 = {x2 + y 2 = 1} t {x2 + y 2 = 4}
which 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 Y˚3 = (−∞, 0) and ∂Y3 = [0, +∞] are connected, but Y3 is disconnected
as 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, assume
that K is compact in E and that K can be covered by open balls of radius
1. By compactness, K can be covered by finitely many ball of radius 1, say
N . 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 are
not necessarily compact.
Let X be an infinite set, equipped with the discrete metric. Then X is
closed and bounded for the corresponding metric topology. However, the
cover 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 of
some locally compact spaces. The Alexandrov compactification is defined up to
homeomorphism and it follows from the universal property that homeomorphic
spaces have the same compactification.
(a) Prove that the Alexandrov compactification of R is homeomorphic
to the unit circle S1 = {(x, y) ∈ R2 , x2 + y 2 = 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
2t
1 − t2
r(t) =
,
.
1 + t2 1 + t2
This map is continuous because its components are rational functions with
non-vanishing denominators, and takes values in S1 . Its inverse is
y
(x, y) 7−→
,
x+1
continuous on S1 \ {(−1, 0)} for the same reason.
We have proved that R ∼
= S1 .
= S1 \ {(−1, 0)}, which proves that R̃ ∼
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 theAlexandrov compactification of Z+ is homeomorphic
to n1 , n ∈ Z+ ∪ {0}.
Again, observe that n1 , n ∈ Z+ ∪ {0} is closed and bounded in R, hence
compact. Moreover, the map n 7−→ n1 is a homeomorphism between Z+ and
1
1
f+ ∼
,
n
∈
Z
so
Z
,
n
∈
Z
∪ {0}.
=
+
+
n
n
57
Final examination - Elements of solution
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 .
We know that points can be separated from compact sets in Hausdorff spaces. In other
words, for every x ∈ K1 , there exist U1x neighborhood of x and U2x open containing K2
such that
U1x ∩ U2x = ∅.
The family {U1x , x ∈ K1 } is an open cover of K1 compact so we can extract a finite
subcover {U1x1 . . . , U1xn }. Then
K1 ⊂
n
[
def
U1xi = U1 ,
i=1
and U1 is open as the union of open sets. Moreover, K2 ⊂ U2xi for every i ∈ {1, . . . , n}, so
K2 ⊂
n
\
def
U2xi = U2 ,
i=1
open as the finite intersection of open sets. To conclude, observe that U1 and U2 are
disjoint because U1x ∩ U2x = ∅ 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 are
open and the family {{x} , x ∈ K} is a covering of K. The fact that a finite subcover
can be extracted shows that K must be finite.
The converse holds for non-necessarily discrete topologies so the compact subsets of a
discrete space are exactly the finite sets.
Problem 2
A topological space is said totally disconnected if its only connected subspaces
are singletons.
1. Prove that a discrete space is totally disconnected.
In a discrete space X, singletons are open and closed. Therefore, the connected component
of x ∈ X is {x}. Another way to see this is to observe that any non-trivial partition of a
set 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) ∩ Q
with a < b. Such a set contains infinitely many rationals. In particular, singletons are not
open, 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 elements
q 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 {Xα }α∈J be a family of topological spaces; let Aα ⊂ Xα for each α ∈ J.
Q
1. In α∈J Xα equipped with the product topology, prove that
Y
Y
A¯α =
Aα .
α∈J
α∈J
Q
a neighborhood
Let (xα )α∈J be a closure point of α Aα . And consider, for β ∈ J, Q
Uβ of xβ . Since the projection maps are continuous, πβ−1
(U
)
is
open
in
α Xα , hence a
Q β
neighborhood of (xα )α∈J so it contains a point (yα )α∈J ∈ α Aα . In particular yβ ∈ Uβ ∩Aβ
so xβ ∈ A¯β .
Q
Q
Conversely, let (xα )α∈J ∈ α A¯α and U = α Uα a neighborhood
of (xα )α∈J . Then
Q
every Uα contains a point yα ∈ Uα ∩ Aα so (yα )α∈J ∈ U ∩ Aα , which means that
Q
(xα )α∈J ∈ α Aα .
Q
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 still
holds in the box topology. The other part of the proof also carries over without change.
Problem 4
Is R homeomorphic to R2 ?
Assume
ϕ : R −→ R2 and consider the restricted map
the existence of a homeomorphism
2
ϕ̃ = ϕ R\{0} : R \ {0} −→ R \ {f (0)}. Then ϕ̃ is bijective by construction and continuous
as the restriction of a continuous function. Observe that ϕ̃−1 is continuous for the same
reason, which means that ϕ̃ is a homeomorphism between the disconnected space R \ {0},
and the connected connected space R2 \ {f (0)}, which is impossible since connectedness
is 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
lim d(f (xn ), f (x)) = lim d(xn , x) = 0
n→∞
n→∞
so lim f (xn ) = f (x), which proves that f is sequentially continuous at any x in E metric.
n→∞
Assume from now on that E is compact and f an isometry. We want to prove
that 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 compact
as 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 )).
| {z }
∈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) < 2ε
and d(um , k) < 2ε , 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 a
homeomorphism. 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 is
continuous. 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 map
satisfying, 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 intersection
of two elements of T belongs to T , which is guaranteed by (4), and argue by induction.
Finally, we prove that T is stable under
S arbitrary unions. Let {Aα }α∈J be a family of
subsets of X; we want to prove that α∈J ι(Aα ) = ι(B) for some B ⊂ X. Observe that
(2) implies that
!
[
[
ι(Aα ) ⊃ ι
ι(Aα ) .
α∈J
Moreover, ι(Aα ) ⊂
S
α∈J
α∈J ι(Aα ) for every α ∈ J so the result proved in (1) gives
!
[
(3)
ι(Aα ) = ι (ι(Aα )) ⊂ ι
ι(Aα ) .
α∈J
This holds for every α ∈ J so
S
α∈J ι(Aα ) ⊂ ι
[
ι(Aα ) = ι
α∈J
ι(A
)
, hence
α
α∈J
!
[
ι(Aα ) .
S
α∈J
3. Prove that, in this topology, Å = 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) ⊂ Å.
Conversely, ˚
(A) is open so it must be of the form ι(B) for some B ⊂ X. Since ι(B) =
Å ⊂ A, it follows from (3) and the result of (1) that
Å = ι(B) = ι(ι(B)) ⊂ ι(A)
which concludes the proof.
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