Notes On General Topology
Notes On General Topology
Notes On General Topology
Jesper M. Mller
Matematisk Institut, Universitetsparken 5, DK2100 Kbenhavn
E-mail address: moller@math.ku.dk
URL: http://www.math.ku.dk/~moller
Contents
Chapter 1. Sets and maps
1. Sets, functions and relations
2. The integers and the real numbers
3. Products and coproducts
4. Finite and infinite sets
5. Countable and uncountable sets
6. Well-ordered sets
7. Partially ordered sets, The Maximum Principle and Zorns lemma
5
5
9
11
12
14
16
18
21
21
23
24
25
27
30
35
42
46
52
59
63
63
64
67
69
72
75
77
Bibliography
79
CHAPTER 1
A B = {x | x A or x B}
A (B C) = (A B) (A C)
A (B C) = (A B) (A C)
A (B C) = (A B) (A C)
Here, (A, B) is a pair of spaces and maps (f, g) : (X, Y ) (A, B) between pairs of spaces are
defined to be pairs of maps f : X A, g : Y B. The diagonal, X = (X, X), takes a space
X to the pair (X, X). These people say that the product is right adjoint to the diagonal and the
coproduct is left adjoint to the diagonal.
1.5. Relations. There are many types of relations. We shall here concentrate on equivalence
relations and order relations.
1.6. Definition. A relation R on the set A is a subset R A A.
1.7. Example. We may define a relation D on Z+ by aDb if a divides b. The relation
D Z+ Z+ has the properties that aDa for all a and aDb and bDc = aDc for all a, b, c. We
say that D is reflexive and transitive.
1.5.1. Equivalence relations. Equality is a typical equivalence relation. Here is the general
definition.
1.9. Definition. An equivalence relation on a set A is a relation A A that is
Reflexive: a a for all a A
Symmetric: a b b a for all a, b A
Transitive: a b c a c for all a, b, c A
The equivalence class containing a A is the subset
[a] = {b A | a b}
of all elements of A that are equivalent to a. There is a canonical map [ ] : A A/ onto the set
A/ = {[a] | a A} P(A)
of equivalence classes that takes the element a A to the equivalence class [a] A/ containing
a.
A map f : A B is said to respect the equivalence relation if a1 a2 = f (a1 ) = f (a2 ) for
all a1 , a2 A (f is constant on each equivalence class). The canonical map [ ] : A A/ respects
the equivalence relation and it is the universal example of such a map: Any map f : A B that
respects the equivalence relation factors uniquely through A/ in the sense that there is a unique
map f such that the diagram
/B
AA
AA
}>
}
AA
}
AA
}}
[ ]
}} !f
A/
f
(3) x y |x| = |y| is an equivalence relation in the plane R2 . The equivalence class [x] is a
circle centered at the origin and R2 / is the collection of all circles centered at the origin. The
canonical map R2 R2 / takes a point to the circle on which it lies.
def
(, b] = {x A | x b}
and similarly for other types of intervals, [a, b], (a, b], (, b] etc.
If (a, b) = then a is the immediate predecessor of b, and b the immediate successor of a.
Let (A, <) be an ordered set and B A a subset.
M is a largest element of B if M B and b M for all b B. The element m is a
smallest element of B if m B and m b for all b B. We denote the largest element
(if it exists) by max B and the smallest element (if it exists) by min B.
M is an upper bound for B if M A and b M for all b B. The element m
Tis a lower
bound for B if m A and m Tb for all b B. The set of upper bounds is bB [b, )
and the set of lower bounds is bB (, b].
T
If the set of upper bounds has a smallest element, min bB [b, ), it is called the least
upperTbound for B and denoted sup B. If the set of lower bounds has a largest element,
max bB (, b], it is called the greatest lower bound for B and denoted inf B.
1.15. Definition. An ordered set (A, <) has the least upper bound property if any nonempty
subset of A that has an upper bound has a least upper bound. If also (x, y) 6= for all x < y, then
(A, <) is a linear continuum.
1.16. Example.
(1) R and (0, 1) have the same order type. [0, 1) and (0, 1) have distinct
order types for [0, 1) has a smallest element and (0, 1) doesnt. {1} (0, 1) and [0, 1) have the
same order type as we all can find an explicit order isomorphism between them.
(2) R R has a linear dictionary order. What are the intervals (1 2, 1 3), [1 2, 3 2] and
(1 2, 3 4]? Is R R a linear continuum? Is [0, 1] [0, 1]?
(3) We now consider two subsets of R R. The dictionary order on Z+ [0, 1) has the same
order type as [1, ) so it is a linear continuum. In the dictionary order on [0, 1) Z+ each
element (a, n) has (a, n + 1) as its immediate successor so it is not a linear continuum. Thus
Z+ [0, 1) and [0, 1) Z+ do not have the same order type. (So, in general, (A, <) (B, <)
and (B, <) (A, <) represent different order types. This is no surprise since the dictionary
order is not symmetric in the two variables.)
(4) (R, <) is a linear continuum as we all learn in kindergarten. The sub-ordered set (Z+ , <)
has the least upper bound property but it is not a linear continuum as (1, 2) = .
(5) (1, 1) has the least upper bound property: Let B be any bounded from above subset of
(1, 1) and let M (1, 1) be an upper bound. Then B is also bounded from above in R, of
course, so there is a least upper bound, sup B, in R. Now sup B is the smallest upper bound
so that sup B M < 1. We conclude that sup B lies in (1, 1) and so it is also a least upper
bound in (1, 1). In fact, any convex subset of a linear continuum is a linear continuum.
(6) R {0} does not have the least upper bound property as the subset B = {1, 21 , 13 , . . .}
is bounded from above (by say 100) but the set of upper bounds (0, ) has no smallest element.
2. The integers and the real numbers
We shall assume that the real numbers R exists with all the usual properties: (R, +, ) is a
field, (R, +, , <) is an ordered field, (R, <) is a linear continuum (1.15).
What about Z+ ?
1.17. Definition. A subset A R is inductive if 1 A and a A = a + 1 A.
There are inductive subsets of R, for instance R itself and [1, ).
1.18. Definition. Z+ is the intersection of all inductive subsets of R.
We have that 1 Z+ and Z+ [1, ) because [1, ) is inductive so 1 = min Z+ is the
smallest element of Z+ .
1.19. Theorem. (Induction Principle) Let J be a subset of Z+ such that
1 J and n Z+ : n J = n + 1 J
Then J = Z+ .
Proof. J is inductive so J contains the smallest inductive set, Z+ .
10
Proof. We show the contrapositive. Let J be a proper subset of Z+ . Consider the smallest
element n = min(Z+ J) outside J. Then n 6 J and Sn J (for n is the smallest element not in
J meaning that all elements smaller than n are in J). Thus J does not satisfy the hypothesis of
the theorem.
1.23. Theorem (Archimedean Principle). Z+ has no upper bound in R: For any real number
there is a natural number which is greater.
Proof. We assume the opposite and derive a contradiction. Suppose that Z+ is bounded
from above. Let b = sup Z+ be the least upper bound (R has the least upper bound property).
Since b 1 is not an upper bound (it is smaller than the least upper bound), there is a positive
integer n Z+ such that n > b 1. Then n + 1 is also an integer (Z+ is inductive) and n + 1 > b.
This contradicts that b is an upper bound for Z+ .
1.24. Theorem (Principle of Recursive Definitions). For any set B and any function
: map({Sn | n Z+ }, B) B
there exists a unique function h : Z+ B such that h(n) = (h|Sn ) for all n Z+ .
Proof. See [8, Ex 8.8].
This follows from the Induction Principle, but we shall not go into details. It is usually
considered bad taste to define h in terms of h but the Principle of Recursive Definition is a permit
to do exactly that in certain situations. Here is an example of a recursive definition from computer
programing
fibo := func< n | n le 2 select 1 else $$(n-1) + $$(n-2) >;
of the Fibonacci function. Mathematicians (sometimes) prefer instead to apply the Principle of
Recursive Definitions to the map
(
1
n<2
f
(Sn
Z+ ) =
f (n 2) f (n 1) n > 2
Recursive functions can be computed by Turing machines.
jJ Aj 6
jJ
11
Aj J
jJ
Aj
J
Figure 1. The coproduct
3. Products and coproducts
1.25. Definition. An indexed family of sets consists of a collection A of sets, an index set J,
and a surjective function f : J A.
We often denote the set f (j) by Aj and the whole indexed family by {Aj }jJ . Any collection
A can be made into an indexed family by using the identity map A A as the indexing function.
We define the union, the intersection, the product, and the coproduct of the indexed family as
\
[
Aj = {a | a Aj for all j J},
Aj = {a | a Aj for at least one j J}
jj
jj
Aj = {x map(J,
Aj ) | j J : x(j) Aj }
jJ
Aj =
jJ
{(j, a) J
jJ
Aj | a Aj }
jJ
`
(projection)
j : Aj
jJ
Aj
(injection)
jJ
given by j (x) = x(j) and j (a) = (j, a) for all j J. These maps are used in establishing the
identities
Y
Y
a
Y
map(X,
Aj ) =
map(X, Aj ),
map(
Aj , Y ) =
map(Aj , Y )
jJ
jJ
jJ
jJ
jJ
jJ
(codiagonal)
jJ
A1 An , A1 An , A1 An A1 q q An
S
T
Q
`
for jZ+ Aj , jZ+ Aj , jZ+ Aj , jZ+ Aj , respectively. If also Aj = A for all j we write A
Q
for the product jZ+ A, the set of all functions x : Z+ A, i.e. all sequences (x1 , . . . , xn , . . .) of
elements from A. 2
S
1.26. Example.
(1) S1 S2 Sn = nZ+ Sn = Z+ .
T
S
Q
(2) If the collection A = {A} consists of just one set A then jJ A = A = jJ A, jJ A =
`
map(J, A), and jJ A = J A.
(3) There is a bijection (which one?) between {0, 1} = map(Z+
Q, {0, 1}) and P(Z+ ). More
generally, there is a bijection (which one) between the product jJ {0, 1} = map(J, {0, 1})
and the power set P(J).
2 is the formal set within set theory corresponding to the nave set Z [12, V.1.5]
+
Ah j
Aj j
Aj
Even though we shall not specify our (ZF) axioms for set theory, let us mention just one axiom
which has a kind of contended status since some of its consequences are counter-intuitive.
n
12
jJ
Aj
c: J
jJ
Aj
J
Figure 2. A choice function
Here is a special, but often used, case. Let A be any nonempty set and P 0 (A) = P(A){} the
collection of nonempty subsets of A. Then there exists (1.28.(3)) a choice function c : P 0 (A) A
such that c(B) B for any nonempty B A. (The choice function selects an element in each
nonempty subset of A.)
4. Finite and infinite sets
1.29. Definition. A set A is finite if Sn+1 A for some n Z+ . A set is infinite if it is not
finite.
We write X Y if there is a bijection between the two sets X and Y .
1.30. Lemma. Let n Z+ and let B be a proper subset of Sn+1 .
(1) It is impossible to map B onto Sn+1 .
(2) Sm+1 B for some m < n.
13
1.36. Theorem (Characterization of infinite sets). Let A be a set. The following are equivalent:
(1) A is infinite
14
i>1
Then h is injective (if i < j then h(j) A {h(1), . . . , h(i), . . . , h(j 1)} so h(i) 6= h(j)).
(2) (3): 1.4.(2)
(2) = (4): We view Z+ as a subset of A. Then A = (A Z+ ) Z+ is in bijection with the proper
subset A {1} = (A Z+ ) (Z+ {1}).
(4) = (1): This is 1.30.
f
i>1
using 1.20. Note that C {h(1), . . . , h(i 1)} is nonempty since C is infinite (1.33). We claim
that h is bijective.
h is order preserving: If i < j, then
h(i) = min(C {h(1), . . . , h(i 1)}) < min(C {h(1), . . . , h(i 1), . . . , h(j 1)}) = h(j)
because C {h(1), . . . , h(i 1)} ) C {h(1), . . . , h(i 1), . . . , h(j 1)}.
h is surjective: Let c C. We must find a positive integer m such that c = h(m). Our only hope
is
m = min{n Z+ | h(n) c}
(Note that this has a meaning since the set {n Z+ | h(n) c} is nonempty as we can not inject
the infinite set Z+ into the finite set {1, . . . , c 1} = Sc (1.31). Note also that again we use 1.20.)
By definition of m,
h(m) c and h(n) c n m
The last of these two properties is equivalent to n < m h(n) < c, so c 6 {h(1), . . . , h(m 1)},
or c C {h(1), . . . , h(m 1)}, and therefore
h(m) = min(C {h(1), . . . , h(m 1)}) c
by definition of h. Thus h(m) = c.
f
15
nZ+ {0, 1}
= {0, 1} is uncount-
16
6. Well-ordered sets
We have seen that all nonempty subsets of (Z+ , <) have a smallest element and we have used
this property in quite a few places so there is reason to suspect that this is an important property
in general. This is the reason for the following definition. You may think of well-ordered sets as
some kind of generalized versions of Z+ .
1.45. Definition. A set A with a linear order < is well-ordered if any nonempty subset has
a smallest element.
Any well-ordered set has a smallest element. Any element (but the largest) in a well-ordered set
has an immediate successor, the smallest successor [Ex 10.2]. (And any element (but the smallest)
has an immediate predecessor?) A well-ordered set can not contain an infinite descending chain
x1 > x2 > , in fact, a linearly ordered set is well-ordered if and only if it does not contain a
copy of the negative integers Z [Ex 10.4].
Let (A, <) be a well-ordered set and an element of A. The subset
S (A) = S = (, ) = {a A | a < }
is called the section of A by .
The induction principle and the principle of recursive defintions apply not only to Z+ but to
any well-ordered set.
1.46. Theorem (Principle of Transfinite Induction). [Ex 10.7] (Cf 1.22) Let (A, <) be a wellordered set and J A a subset such that
A : S J = J
Then J = A.
Proof. Formally identical to the proof of 1.22.
1.47. Theorem (Principle of Transfinite Recursive Definitions). Let (A, <) be a well-ordered
set. For any set B and any function
: map({S | A}, B) B
there exists a unique function h : A B such that h() = (h|S ) for all A.
1.48. Proposition (Hereditary properties of well-ordered sets).
(1) A subset of a well-ordered set is well-ordered.
(2) The coproduct of any well-ordered family of well-ordered sets is well-ordered [Ex 10.8].
(3) The product of any finite family of well-ordered sets is well-ordered.
Proof.
(1) Clear.
(2) Let J be a well-ordered set and {Aj }jJ a family of well-ordered sets indexed by J. For
i, j J and x Ai , y Aj , define
def
6. WELL-ORDERED SETS
17
(6) The set S = [1, ] = Z+ q {} is well-ordered (1.48.(2)). It has as its largest element.
The section S = [1, ) = Z+ is countably infinite but any other section is finite. Any finite
subset A of [1, ) has an upper bound because the set of non-upper bounds
[
{x [1, ) | a A : x < a} =
Sa
aA
is finite (1.34.(3)) but [1, ) is infinite. S has the same order type as the interval [1 1, 2 1]
in Z+ Z+ .
Which of these well-ordereds have the same order type [Ex 10.3]? Draw pictures of examples
of well-ordered sets.
We can classify completely all finte well-ordered sets.
1.50. Theorem (Finite order types). [Ex 6.4] Any finite linearly ordered set A of cardinality
n has the order type of (Sn+1 , <); in particular, it is well-ordered and it has a largest element.
Proof. Define h : Sn+1 A recursively by h(1) = min A and
h(i) = min(A {h(1), . . . , h(i 1)},
i>1
Then h is order preserving. In particular, h is injective and hence bijective (by the pidgeon hole
principle (1.32)) since the two sets have the same cadinality.
Can you find an explicit order preserving bijection Sm+1 Sn+1 Sm+n+1 ?
So there is just one order type of a given finite cardinality n. There are many countably infinite
well-ordered sets (1.49). Is there an uncountable well-ordered set? Our examples, R and {0, 1} ,
of uncountable sets are not well-ordered (1.49.(2), 1.49.(5)).
1.51. Theorem (Well-ordering theorem). (Zermelo 1904) Any set can be well-ordered.
We focus on the minimal criminal, the minimal uncountable well-ordered set. (It may help to
look at 1.49.(6) again.)
1.52. Lemma. There exists a well-ordered set S = [0, ] with a smallest element, 0, and a
largest element, , such that:
(1) The section S = [0, ) by is uncountable but any other section, S = [0, ) for < ,
is countable.
(2) Any countable subset of S = [0, ) has an upper bound in S = [0, )
Proof. (Cf 1.49.(6)) Take any uncountable well-ordered set A (1.51). Append a greatest
element to A. Call the result A again. Now A has at least one uncountable section. Let be the
smallest element of A such that the section by this element is uncountable, that is = min{
A | S is uncountable}. Put S = [0, ] where 0 is the smallest element of A. This well-ordered
set satisfies (1) and (2). Let C be a countable subset of S = [0, ). We want to show that it has
an upper bound. We consider the set of elements of S that are not upper bounds, i.e
countable
uncountable
z[}| {
z}|{
{x S | x is not an upper bound for C} = {x S | c C : x < c} =
Sc (
S
cC
This set of not upper bounds is countable for it is a countable union of countable sets (1.41.(4)).
But S is uncountable, so the set of not upper bounds is a proper subset.
See [SupplExI : 8] for an explicit construction of a well-ordered uncountable set. Z+ is the
well-ordered set of all finite (nonzero) order types and S is the well-ordered set of all countable
(nonzero) order types. See [Ex 10.6] for further properties of S .
Recall that the ordered set Z+ [0, 1) is a linear continuum of the same order type as [1, )
R. What happens if we replace Z+ by S [Ex 24.6, 24.12]?
18
19
We shall later use Zorns lemma to prove Tychonoffs theorem (2.148) that a product of
compact spaces is compact. In fact, The Axiom of Choice, Zermelos Well-ordering theorem,
Hausdorffs maximum principle, Zorns lemma, and Tychonoffs theorem are equivalent.
Here are two typical applications. Recall that a basis for a vector space over a field is a maximal
independent subset and that a maximal ideal in a ring is a maximal proper ideal.
1.57. Theorem. [Ex 11.8] Any linearly independent subset of a vector space is contained in a
basis.
Proof. Apply Zorns lemma to the poset of independent subsets of the vector space. Any
linearly ordered set of independent subsets has an upper bound, namely its union.
1.58. Theorem. Any proper ideal of a ring is contained in a maximal ideal.
Proof. Apply Zorns lemma to the poset of proper ideals. Any linearly ordered set of proper
ideals has an upper bound, namely its union.
As a corollary of (1.57) we see that R and R2 are isomorphic as vector spaces over Q.
Other authors prefer to work with partial orders instead of strict partial orders.
1.59. Definition. [Ex 11.2] Let A be a set. A relation on A is said to be a partial order
precisely when it is symmetric (that is a a for all a in A), transitive (that is a b and b c
implies a c), and anti-symmetric (that is a b and b a implies a = b).
> SubgroupLattice(AlternatingGroup(5));
Partially ordered set of subgroup classes
---------------------------------------------[1]
--[2]
[3]
[4]
--[5]
[6]
[7]
--[8]
--[9]
Order 1
Length 1
Maximal Subgroups:
Order 2
Order 3
Order 5
Length 15
Length 10
Length 6
Maximal Subgroups: 1
Maximal Subgroups: 1
Maximal Subgroups: 1
Order 4
Order 6
Order 10
Length 5
Length 10
Length 6
Maximal Subgroups: 2
Maximal Subgroups: 2 3
Maximal Subgroups: 2 4
Order 12
Length 5
Maximal Subgroups: 3 5
Order 60 Length 1
Maximal Subgroups: 6 7 8
Table 1. The poset of subgroups of the alternating group A5
CHAPTER 2
22
(2.7)
Proof. (1) is obvious since TB is the coarsest topology containing B. Item (2) is immediate
from (1). Item (3) is proved in the same way since TB = TS iff TB TS and TB TS iff B TS
and TB S.
2.10. Example.
(1) In a metric space, the set B = {B(x, r) | x X, r > 0} of open balls
is (by definition) a basis for the metric topology Td . The collection of open balls of radius n1 ,
n Z+ , is an equivalent topology basis for Td .
(2) The collection of rectangular regions (a1 , b1 ) (a2 , b2 ) in the plane R2 is a topology basis
equivalent to the standard basis of open balls B(a, r) = {x R2 | |x a| < r}. you can always
put a ball inside a rectangle and a rectangle inside a ball.
(3) Let f : X Y be any map. If T is a topology on Y with basis B or subbasis S , then the
pull-back f 1 (T ) is a topology, the initial topology for f , on X with basis f 1 (B) and subbasis
f 1 (S).
(4) More generally, let X be a set, {Yj } a collection of topological spaces, and fj : X Yj ,
j J, a set of maps. Let Tj be the topology on Yj , Bj a basis and Sj , j J a subbasis. Then
S 1
S
S
fj (Tj ), fj1 (Bj ), fj1 (Sj ) are equivalent subbases on X. The topology they generate
is called the initial topology for the maps fj , j J.
2. ORDER TOPOLOGIES
23
a, b X, a < b.
2.14. Lemma. The collection of all open rays together with all open intervals is a basis for the
order topology T< .
(2.7)
24
Uk Xk open, k J.
Q
The set k1 (Uk ) consists of the points (xj )
Xj with kth coordinate in Uk . Alternatively,
S
k1 (Uk ) consists of all choice functions c : J jJ Uj such that c(k) Uk (Figure 2).
Q
2.17. Definition. The product topology on jJ Xj is the topology with subbasis
[
SQ =
{j1 (U ) | U Xj open}
jj
The product topology is the coarsest topology making all the projection maps j :
j J, continuous.
This becomes particularly simple when we consider finite products.
Xj Xj ,
(a b, +) = {a} (b, +)
{x} Y
x>a
show that the order topology is coarser than the product of the discrete and the order topology. On
the other hand, the sets {x}(c, d) is a basis for = Xd Y< (2.15, 2.19) and {x}(c, d) = (xc, xd)
is open in the order topology since it is an open interval.
The corollary shows that the order topology on X Y does not yield anything new in case the
second factor is unbounded in both directions. As an example where this is not the case we have
already seen Z+ Z+ and Io2 (2.16.(4), 2.16.(7)).
25
Xj (3) is the
jJ
Uj of open set Uj
Xj .
The lemma says that an open subset of an open subset is open and that a closed subset of a
closed subset is closed.
The next theorem says that the subspace and the product space operations commute.
Q
Q
2.26. Theorem. Let Yj Xj , j J. The subspace topology that Yj inherits from Xj is
the product topology of the subspace topologies on Yj .
Q
Proof. The subspace topology on Yj has subbasis
Y
Y
[
[ Y
Yj SQ Xj =
Yj
{k1 (Uk )} =
{ Yj k1 (Uk )}, Uk Xk open,
kj
kJ
Yj has subbasis
[
SQ Yj =
{k1 (Yk Uk )},
Uk Xk open
kJ
[8, Ex 2.2]
k1 (Yk ) k1 (Uk ) = (
Yj ) k1 (Uk ).
26
2.27. Subspaces of linearly ordered spaces. When (X, <) is a linearly ordered set and
Y X a subset we now have two topologies on Y . We can view Y as a subspace of the topological
space X< with the order topology or we can view Y as a sub-ordered set of X and give Y the order
topology. Note that these two topologies are not the same when X = R and Y = [0, 1) {2} R:
In the subspace topology {2} is open in Y but {2} is not open in the order topology because Y
has the order type of [0, 1]. See 2.29 for more examples. The point is that
Any open ray in Y is the intersection of Y with an open ray in X
The intersection of Y with an open ray in X need not be an open ray in Y
However, if Y happens to be convex then open rays in Y are precisely Y intersected with open
rays in X.
2.28. Lemma (Y< Y X< ). Let (X, <) be a linearly ordered set and Y X a subset. The
order topology on Y is coarser than the subspace topology Y in general. If Y is convex, the two
topologies on Y are identical.
Proof. The order topology on Y has subbasis
S< = {Y (, b) | b Y } {Y (a, ) | a Y }
and the subspace topology on Y has subbasis
S = {Y (, b) | b X} {Y (a, ) | a X}
Clearly, S< S , so the order topology on Y is coarser than the subspace topology in general.
If Y is convex and b X Y then b is either a lower or an upper bound for Y as we cannot
have y1 < b < y2 for two points y1 and y2 of Y . If b is a lower bound for Y , then Y (, b) = ,
and if b is an upper bound for Y , then Y (, b) = Y . Therefore also S T< so that in fact
T< = T .
2.29. Example.
(1) S 21 = 1 Z+ {2 1} = [1 1, 2 1] is a convex subset of Z+ Z+ .
The subspace topology (2.16.(4)) is the same as the order topology.
(2) The subset Z+ Z+ is not convex in Z Z so we expect the subspace topology to be strictly
finer than the order topology. Indeed, the subspace topology that Z+ Z+ inherits from the
discrete space Z Z is discrete but the order topology is not discrete (2.16.(4)(5)).
(3) Consider X = R2 and Y = [0, 1]2 with order topologies. Y is not convex so we expect the
subspace topology to be strictly finer than the order topology. Indeed, the subspace topology
on [0, 1]2 , which is [0, 1]d [0, 1], is strictly finer than Io2 (2.16.(7)): The set
1
1
[0, ] [0, 1] = ([0, 1] [0, 1]) (, 2)
2
2
is open in the subspace topology on Y but it is not open in the order topology on Y as any
basis open set (2.14) containing 12 1 also contains points with first coordinate > 12 .
(4) Consider X = R2 and Y = (0, 1)2 with order topologies. Y is not convex so we expect the
subspace topology to be strictly finer than the order topology. But it isnt! The reason is that
(0, 1) does not have a greatest nor a smallest element (2.21).
(5) The subset Q of the linearly ordered set R is not convex but nevertheless the subspace
topology inherited from R is the order topology. Again, the reason seems to be that Q does
not have a greatest nor a smallest element.
27
28
If A Y X, Y Int(A) IntY (A) for Y Int(A) is a relatively open set contained in A. These
two sets are equal if Y is open but they are distinct in general (consider A = [0, 1) [0, 1] = Y ).
2.36. Definition. A subset A X is said to be dense if A = X, or, equivalently, if every
open subset of X contains a point of A.
2.37. Proposition. Let A be a dense and U an open subset of X. Then A U = U .
Proof. The inclusion A U U is general. For the other inclusion, consider a point x U .
Let V be any neighborhood of x. Then V (A U ) = (V U ) A is not empty since V U is a
neighborhood of x and A is dense. But this says that x is in the closure of A U .
2.38. Limit points and isolated points. Let X be a topological space and A a subset of
X.
2.39. Definition (Limit points, isolated points). A point x X is a limit point of A if
U (A {x}) 6= for all neighborhoods U of x. The set of limit points1 of A is denoted A0 . A
point a A is an isolated point if a has a neighborhood that intersects A only in {a}.
Equivalently, x X is a limit point of A iff x A {x}, and a A is an isolated point of A
iff {a} is open in A. These two concepts are almost each others opposite:
x is not a limit point of A x has a neighborhood U such that U (A {x}) = {x}
x is an isolated point of A {x}
2.40. Proposition. Let A be a subset of X and A0 the set of limit points of A. Then AA0 = A
and A A0 = {a A | a is not an isolated point of A} so that
A A0 A is closed
A A0 A has no isolated points
A A0 = A is discrete
A0 = A is closed and discrete
If B A then B 0 A0 .
Proof. It is clear that all limit points of A are in A, so that A A0 A, and that all points
in the closure of A that are not in A are limit points,
A A = {x X A | all neighborhoods of x meet A} A0 A,
so that A = A (A A) A A0 . This shows that A A0 = A. From the above discussion we
have that the set of limit points in A, A A0 = A (A A0 ), is the set of non-isolated points of A.
If A A0 = then all all points of A are isolated so that the subspace A has the discrete topology.
We have A0 = A A0 , A A0 = A is closed and discrete.
2.41. Example.
(1) Q = R = R Q (R Q contains (Q {0}) 2). Cl Int Q = ,
Int Cl Q = R.
1The set of limit points of A is sometimes denotes Ad and called the derived set of A
29
(2) Let Cr R2 be the circle with center (0, r) and radius r > 0. Then
[
[
[
[
C1/n ,
Cn
C1/n =
Cn = (R {0})
nZ+
nZ+
nZ+
nZ+
The first set of decreasing circles (known as the Hawaiian Earring [8, Exmp 1 p 436]) is closed
because it is also the intersection of a collection (which collection?) of closed sets.
(3) The set of limit points of K = { n1 | n Z+ } is {0} in R and R` and in RK .
(4) Let X be a linearly ordered space (2.13). Closed intervals [a, b] are closed because their
complements X [a, b] = (, a) (b, ) are open. Therefore the closure of an interval of the
form [a, b) is either [a, b) or [a, b]. If b has an immediate predecessor b then [a, b) = [a, b ] is
closed so that [a, b) = [a, b). Otherwise, [a, b) = [a, b] because any neighborhood of b contains
an interval of the form (c, b] for some c < b and [a, b) (c, b], which equals [a, b] or (c, b], is not
empty.
2.42. Convergence, the Hausdorff property, and the T1 -axiom. Let xn , n Z+ , be a
sequence of points in X, ie a map Z+ X.
2.43. Definition. The sequence xn converges to the point x X if for any neighborhood U
of x there exists some N such that xn U for all n > N .
If a sequence converges in some topology on X it also converges in any coarser topology but
not necessarily in a finer topology.
2.44. Example. In the Sierpinski space X = {0, 1} (2.2.(3)), the sequence 0, 0, converges
to 0 and to 1. In R with the finite complement topology (2.2.(4)), the sequence 1, 2, 3, converges
to any point. In RK , the sequence n1 does not converge; in R and R` it converges to 0 (and to no
other point). The sequence n1 converges to 0 in R and RK , but does not converge in R` . In the
ordered space [1, ] = Z+ q {} (1.49.(6)), the sequence n converges to (and to no other point).
Can you find a sequence in the ordered space [0, ] (1.52) that converges to ?
2.45. Definition (Separation Axioms). A topological space is T1 -space if points are closed:
For any two distinct points x1 6= x2 in X there exists an open set U such that x1 U and x2 6 U .
A topological space X is a T2 -space (or a Hausdorff space) if there are enough open sets to
separate points: For any two distinct points x1 6= x2 in X there exist disjoint open sets, U1 and
U2 , such that x1 U1 and x2 U2 .
All Hausdorff spaces are T1 . X is T1 iff all finite subsets are closed. Cofinite topologies are T1
by construction and not T2 when the space has infinitely many points. Particular point topologies
are not T1 (on spaces with more than one point).
All linearly ordered [Ex 17.10] or metric spaces are Hausdorff; in particular, R is Hausdorff.
R` and RK are Hausdorff because the topologies are finer than the standard Hausdorff topology
R.
2.46. Theorem. A sequence in a Hausdorff space can not converge to two distinct points.
A property of a topological space is said to be (weakly) hereditary if any (closed) subspace of
a space with the property also has the property. Hausdorffness is hereditary and also passes to
product spaces.
2.47. Theorem (Hereditary properties of Hausdorff spaces). [Ex 17.11, 17.12] Any subset of
a Hausdorff space is Hausdorff. Any product of Hausdorff spaces is Hausdorff.
2.48. Theorem. Suppose that X is T1 . Let A be a subset of and x a point in X. Then
x is a limit point All neighborhoods of x intersect A in infinitely many points
Proof. =: If all neighborhoods of x intersect A in infinitely many points, then, clearly,
they also intersect A in a point that is not x.
=: Assume that x is a limit point and let U be a neighborhood of x. Then U contains a point
a1 of A different from x. Remove this point from U . Since points are closed, U {a1 } is a new
neighborhood of x. This new neighborhood of x contains a point a2 of A different from x and a1 .
In this way we recursively find a whole sequence of distinct points in U A.
30
6. Continuous functions
Let f : X Y be a map between two topological spaces.
2.49. Definition. The map f : X Y is continuous if all open subsets of Y have open preimages in X: V open in X = f 1 (V ) open in Y
If TX is the topology on X and TY is the topology on Y then
(2.50)
where BY is a basis and SY a subbasis for TY . The finer the topology in Y and the coarser the
topology in X are, the more difficult is it for f : X Y to be continuous.
2.51. Theorem. Let f : X Y be a map between two topological spaces. The following are
equivalent:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
f is continuous
The preimage of any open set in Y is open in X: V open in Y = f 1 (V ) open in X
The preimage of any closed set in Y is closed in X: C closed in Y = f 1 (C) closed in X
f 1 (B ) (f 1 (B)) for any B Y
f 1 (B) f 1 (B) for any B Y
f (A) f (A) for any A X
For any point x X and any neighborhood V Y of f (x) there is a neighborhood U X
of x such that f (U ) V .
(4)
(4) = (2): Let B be any open set in Y . Then f 1 (B) = f 1 (B ) (f 1 (B)) f 1 (B) so
f 1 (B) is open since it equals its own interior.
(3) (5): Similar to (2) (4).
A f 1 (f (A)) f 1 (f (A))
(3) = (6):
= A f 1 (f (A)) f (A) f (A).
f 1 (f (A)) is closed
(6)
f (f 1 (B))B
(restriction)
(corestriction)
6. CONTINUOUS FUNCTIONS
31
or
open
open
in X V is
in Y
closed
closed
A X is continuous A X
Y is continuous
The embedding topology is the only topology on X with these two properties.
32
(2.50)
A
X is continuous g 1 (TX ) TA g 1 (f 1 TY ) TA
g
(f g)1 TY ) TA A
X
Y is continuous
by definition of the embedding topology. The identity map of X is a homeomorphism whenever
X is equipped with a topology with these two properties.
2.58. Definition (Embedding). An injective continuous map f : X Y is an embedding if
the topology on X is the embedding topology for f , ie TX = f 1 TY .
This means that the injective continuous map f : X Y is an embedding if and only if for all
U X:
open
open
1
(2.51) U is
in X U = f (V ) for some
V Y
closed
closed
open
open
f (U ) = f (X) V for some
V Y f (U ) is
in f (X)
closed
closed
Alternatively, the injective map f : X Y is an embedding if and only if the bijective map
f (X)|f : X f (X) is a homeomorphism. An embedding is a homeomorphism followed by an
inclusion. The inclusion A , X of a subspace is an embedding. Any open (closed) continuous
injective map is an embedding.
2.59. Example.
(1) The map f (x) = 3x + 1 is a homeomorphism R R.
(2) The identity map R` R is bijective and continuous but not a homeomorphism.
(3) The map [0, 1) S 1 : t 7 (cos(2t), sin(2t)) is continuous and bijective but not a homemorphism. The image of the open set [0, 12 ) is not open in S 1 .
(4) Find an example of an injective continuous map R R2 that is not an embedding.
(5) The obvious bijection [0, 1) {2} [0, 1] is continuous but not a homeomorphism (the
domain has an isolated point, the codomain has no isolated points). There does not exist any
continuous surjection in the other direction.
(6) The spaces [1 1, 2 1] Z+ Z+ and K = { n1 | n Z+ } R are homeomorphic.
(7) The map R R R : t 7 (t, t) is an embedding. For any continuous map f : X Y , the
map X X Y : x 7 (x, f (x)) is an embedding; see (3.25) for a generalization.
(8) Rn embeds into S n via stereographic projection.
(9) R2 embeds in R3S. Does R3Simbed in R2 ? (See notes on algebraic topology for the answer.)
(10) Are the spaces Cn and C1/n of 2.41.(2) homeomorphic?
(11) A knot is in embedding of S 1 in R3 (or S 3 ). Two knots, K0 : S 1 , R3 and K1 : S 1 , R3 ,
are equivalent if there exists a homeomorphism h of R3 such that h(K0 ) = K1 . The fundamental
problem of knot theory [1] is to classify knots up to equivalence.
2.60. Lemma. If f : X Y is a homeomorphism (embedding) then the corestriction of the
restriction f (A)|f |A : A f (A) (B|f |A : A B) is a homeomorphism (embedding) for any subset
A of X (and any subset B of Y containing
Q
Qf (A)). If the maps fj : Xj Yj are homeomorphisms
(embeddings) then the product map fj :
Xj Yj is a homeomorphism (embedding).
Proof. In case of homeomorphisms there is a continuous inverse in both cases. In case of
embeddings, use that an embedding is a homeomorphism followed by an inclusion map.
2.61. Lemma (Composition of embeddings). Let X
/Y
6. CONTINUOUS FUNCTIONS
33
2.62. Maps into products. There is an easy test for when a map into a product space is
continuous.
Q
2.63. Theorem (Characterization of the product topology). Give Yj the product topology.
Then
Q
(1) the projections j : Q
Yj Yj are continuous, and,
(2) for any map f : X jJ Yj into the product space we have
j
f Y
f Y
X
Yj is continuous j J : X
Yj Yj is continuous
jJ
jJ
The product topology is the only topology on the product set with these two properties.
S
Proof. Let TX be the topology on X and Tj the topology on Yj . Then SQ = jJ j1 (Tj )
Q
is a subbasis for the product topology on jJ Yj (2.17). Therefore
f: X
(2.50)
Yj is continuous f 1 (
jJ
j1 (Tj )) TX
jJ
f 1 (j1 (Tj )) TX
jJ
j J : (j f )1 (Tj ) TX
j J : j f is continuous
by definition of continuity (2.50).
We now show that theQproduct topology is the unique topology with these properties. Take two
copies of the product set jJ Xj . Equip one copy with the product topology and the other copy
with some topology that has the two properties of the theorem. Then the identity map between
these two copies is a homeomorphism.
The reason for the great similarity between 2.63 and 2.57 is that in both cases we use an initial
topology.
2.64. Example. Suppose that J and K are sets and that (Xj )jJ and (Yk )kK are indexed
families of topological spaces. Given a map g : J K between index sets and an indexed family of
continuous maps (fj : Yg(j) Xj )jJ . Then there is a unique map between product spaces such
that
Q
Q
/ jJ Xj
kK Yk
g(j)
Yg(j)
fj
/ Xj
(2)
Aj Aj and equality holds if Aj = Xj for all but finitely many j J.
Q
S
Proof. (1) Let (xj ) be a point of Xj . Since BQ = jJ j1 (Uj ) is a subbasis for the product
topology we have:
Y
Y
(xj )
Aj k J : k1 (Uk )
Aj 6= for all neighborhoods Uk of xk
k J : Uk Ak 6= for all neighborhoods Uk of xk
k J : xk Ak
Y
(xj )
Aj
34
Q
Aj
It follows that a product of closed sets is closed. (Whereas a product of open sets need not be
open in the product topology.)
2.66. Maps out of coproducts.
`
2.67. Theorem. Let f : jJ Xj Y be a map out of a coproduct space. Then
a
f:
Xj Y is continuous f j : Xj Y is continuous for all j J
jJ
where j : Xj
jJ
35
(A is open in X)
= U open in X
(f is open)
= f (U ) open in Y
= f (U ) open in f (A)
show that f |A : A f (A) is open.
BQ
Xj Xj is open.
2.71. Lemma (Characterization of open or closed continuous maps). Let f : X Y be a continuous map.
(1) f is open if and only if f 1 (Int B) = Int f 1 (B) for all B Y .
(2) f is closed if and only if f (A) = f (A) for all A X
Proof. See Solution June 04 (Problem 1) and Solution Jan 05 (Problem 1).
0 1 x 0
f (x) = x 0 x 1
1 1x2
is continuous and closed (2.140.(1)). It is not open for f ([1, 1/2)) = {0} is not open.
2.2. Quotient topologies and quotient maps. Quotient maps are continuous surjective
maps that generalize both continuous, open surjective maps and and continuous, closed surjective
maps.
2.73. Definition (Quotient topology). Let X be a topological space, Y a set, and p : X Y
a surjective map. The quotient topology on Y is the collection
{V Y | p1 (V ) is open in X}
of subsets of Y .
36
The quotient topology2 is the finest topology on Y such that p : X Y continuous. There are
as many open sets in Y as possible without destroying continiuty of f : X Y . Quotient topology
and embedding topology (2.56) are dual concepts.
2.74. Lemma (Characterization of the quotient topology). Suppose that Y has the quotient
topology with respect to the map p : X Y . Then
(1) p : X Y is continuous, and,
(2) for any map g : Y Z out of Y
g
Y
Z is continuous X
Y
Z is continuous
The quotient topology is the only topology on Y with these two properties.
Proof. This is because
g
Y
Z is continuous g 1 (TZ ) TY p1 g 1 (TZ ) TX
p
(pg)1 (TZ ) TX X
Y
Z is continuous
by definition of the quotient topology.
If we give Y some topology with these two properties then the identity map between the two
topologies is a homeomorphism.
2.75. Definition (Quotient map). A surjective continuous map p : X Y is a quotient map
if Y has the quotient topology.
Quotient maps and embeddings (2.58) are dual concepts.
A surjective map p : X Y is a quotient map when the quotient topology on Y , TY = {V
Y | p1 (V ) is open in X}, is the finest topology on Y such that p is continuous. This means that
p is a quotient map if and only if
p1 (V ) is open in X V is open in Y
holds for all V Y .
S
Subsets A of X of the form A = p1 (B) = yB p1 (y) are called saturated sets. They
are the subsets
unions of fibres f 1 (y), y Y . The saturation of A X is the union
S that are
1
1
f f (A) = yf (A) f (y) of all fibres that meet A.
2.76. Proposition. For a surjective map p : X Y the following are equivalent:
(1) p : X Y is a quotient map
open
open
1
(2) For all V Y we have: p (V ) is
in X V is
in Y .
closed
closed
open
open
(3) p : X Y is continuous and maps saturated
sets to
open sets
closed
closed
Proof. Condition (1) and condition (2) with the word open are clearly equivalent. Suppose
now that: p1 (V ) is open V is open. Then we get
p1 (C) is closed X p1 (C) is open p1 (Y C) is open
Y C is open C is closed
for all C Y . This shows that the two conditions of (2) are equivalent. The content of (3) is just
a reformulation of (2).
Equivalently, a surjective continuous map p : X Y is quotient if and only if the maps
Saturated open (closed) sets o U p(U ) / Open (closed) sets
in X
in Y
p1 (V )V
are bijections.
2.77. Corollary.
(1) Any
open
closed
continuous surjective map is a quotient map.
2The quotient topology is sometimes called the final topology [2, I.2.4] with respect to the map f
37
open
open
(2) A quotient map f : X Y is
if and only if all
sets A X have
closed
closed
open
saturations f 1 f (A).
closed
(3) A bijective continuous map is a quotient map if and only if it is a homeomorphism.
Proof. Let f : X Y be an open map. Then
f open
f is continuous
f 1 (V ) is open
for all V Y . This shows that f is quotient. Suppose next that f : X Y is quotient. Then
(2.68)
There are quotient maps that are neither open nor closed [8, Ex 22.3] [2, Ex 10 p 135] and
there are (non-identity) quotient maps that are both open and closed (2.82) [2, Ex 3 p 128].
f
/ Y g / Z be continuous
2.78. Corollary (Composition of quotient maps). Let X
maps. Then
f and g are quotient = g f is quotient = g is quotient
Proof. The first assertion is a tautology: Assume that f and g are quotient. Then
(g f )1 (V ) open in X f 1 g 1 (V ) open in X
f is quotient
g 1 (V ) open in Y
g is quotient
V open in Z
f continuous
(g f )1 (V ) is open in X
g f quotient
V is open in Z
2.79. Example.
(1) The projection map 1 : R R R is (2.72.(1)) open, continuous,
and surjective so it is a quotient map. The restriction 1 |H {(0, 0)} is continuous and surjective, even bijective, but it is not a quotient map (2.76) for it is not a homeomorphism: {(0, 0)}
is open and saturated in H {(0, 0)} but 1 ({(0, 0)} = {0} is not open.
Thus the restriction of a quotient map need not be a quotient map in general. On the positive
side we have
2.80. Proposition. The restriction-corestriction of a quotient map p : X Y to an open (or
closed) saturated subspace A X is a quotient map p(A)|p|A : A p(A).
Proof. (Similar to the proof of 2.69.) Let p : X Y be a quotient map and B Y an
open set. (The case where B is closed is similar.) The claim is that B|p|p1 (B) : p1 (B) B is
quotient. For any U B the implications
p1 (U ) open in p1 (B)
= p1 (U ) open in X
= U is open in Y
= U is open in B
(p is quotient)
38
[x]V [x] X is open. We shall often say that X/R is the space obtained by identifying equivalent
points of X.
A continuous map f : X Y respects the equivalence relation R if equivalent points have
identical images, that is if x1 Rx2 = f (x1 ) = f (x2 ). The quotient map X X/R respects the
equivalence relation R and it is the universal example of such a map.
2.81. Theorem (The universal property of quotient spaces). Let R be an equivalence relation
on the space X and let f : X Y be a continuous map.
(1) The map p : X X/R respects the equivalence relation R.
(2) If the continuous map f : X Y respects R then there exists a unique continuous map
f : X/R Y such that
f
/Y
XA
AA
}>
}
AA
}
A
}}f
p AA
}
}
X/R
commutes. (We say that f factors uniquely through X/R.) Conversely, if f factors
through X/R then f respects R.
(3) If f exists then: f is quotient f is quotient
Proof. If f exists then clearly f respects R. Conversely, if f respects R then we can define
f [x] = f (x) and this map is continuous by 2.74 and it is the only possibility. The rest follows from
2.78: f is quotient pf is quotient f is quotient.
The theorem says that there is a bijective correspondence
Continuous maps X Y o
that respect R
f f
gpg
/ Continuous maps
X/R Y
39
commutes. Note that f is bijective. The bijective continuous map f : X/f Y is a homeomorphism if and only if f is quotient (2.77.(3), 2.81). In particular, all quotient maps have (up
to homeomorphism) the form X X/R for some equivalence relation R on X.
(2) Let f : X Y be any surjective continuous map. The induced map f : X/f Y is a continuous bijection but in general not a homeomorphism. Instead, the topology on the quotient
space X/f is finer than the topology on Y . This can sometimes be used to show that X/f is
Hausdorff if Y is Hausdorff.
(3) Let f : X Y be any continuous map. Then f has a canonical decomposition
p
f (X)
Y
X
X/f
where p is a quotient map, f is a continuous bijection, and is an inclusion map.
(4) Let X be a topological space and A1 , A2 , . . . , Ak a finite collection of closed subsets. Consider
the equivalence relation where the equivalence classes are the sets A1 , A2 , . . . , Ak together with
the sets {x} for x 6 A1 A2 Ak . The quotient space X/(A1 , . . . , Ak ) is obtained from X by
identifying each of the sets Ai to the point p(Ai ). The quotient map S
p : X X/(A1 , . . . , Ak ) is
closed because (2.77) closed sets A X have closed saturations A Ai A6= Ai . A continuous
map f : X Y factors through the quotient space X/(A1 , . . . , Ak ) if and only if it sends each of
the sets Ai X to a point in Y (2.81). The restriction p|X(A1 Ak ) : X(A1 Ak )
X/(A1 , . . . , Ak ) {p(A1 ), . . . , p(Ak )} to the complement of A1 Ak is a homeomorphism
(2.69). In case of just one closed subspace A X, the quotient space is denoted X/A.
(5) The standard map f : [0, 1] S 1 that takes t [0, 1] to (cos(2t), sin(2t)) is quotient
because it is continuous and closed. (If you cant see this now, we will prove it later (2.139.(1)).)
The induced map [0, 1]/{0, 1} S 1 is a homeomorphism. More generally, the standard map
Dn /S n1 S n is a homeomorphism where Dn Rn , the unit disc, is the set of vectors of
length 1.
(6) Let R be the equivalence relation zero or not zero on R. The quotient space R/R is
homeomorphic to Sierpinski space {0, 1}.
`
S
(7) There is an obvious continuous surjective map f : Z+ S 1 = Z+ S 1 Cn (2.41.(2))
that takes Z+ {1} to the point common to all the circles. This map is continuous because its
restriction to each of the open sets {n}S 1 is continuous (2.53.(5). However, f is not a quotient
map (2.76) for the image of the closed saturated set consisting of the points n (cos( 2 ), sin( 2 ))
is not closed as it does not contain
all its limit points. The induced bijective continuous
S
map f : Z+ S 1 /Z+ {1} ` Cn is therefore not aShomeomorphism. There is an obvious
continuous surjective map g : Z+ S 1 = Z+ S 1 C1/n (2.41.(2)) that takes Z+ {1}
to the point common to all the circles. This map is continuous because its restriction to each
of the open sets {n} S 1 is continuous (2.53.(5)). However, g is not a quotient map for the
image of the closed saturated set Z+ {1} is not closed as it does not
S contain all its limit
points. The induced bijective continuous map g : Z+ S 1 /Z+ {1} C1/n is therefore not
a homeomorphism either. (Actually
(2.97.(7)), the quotient space Z+ S 1 /Z+ {1}, known as
W
the countable wedge of circles nZ+ S 1 [8, Lemma 71.4], is not homeomorphic to any subspace
of the plane.)
(8) [8, p 451] Let P4g be a regular 4g-gon and with edges labeled a1 , b1 , a1 , b1 , . . . , ag , bg , ag , bg in
counter-clockwise direction. The closed orientable surface Mg of genus g 1 is (homeomorphic
to) the quotient space P4g /R where R is the equivalence relation that makes the identifications
1
1 1
a1 b1 a1
1 b1 ag bg ag bg on the perimeter and no identifications in the interior of the polygon.
See [8, p 452] for the case g = 2. Are any of these surfaces homeomorphic to each other? [8,
Thm 77.5]
40
(9) [8, p 452] Let P2g be a regular 2g-gon with edges labeled a1 , a1 , . . . , ag , ag in counterclockwise direction. The closed non-orientable surface Ng of genus g 1 is the quotient
space P2g /R where R is the equivalence relation that makes the identifications a21 a2g on the
perimeter and no identifications in the interior of the polygon. For g = 1 we get the projective
plane RP 2 and for g = 2 we get the Klein Bottle [8, Ex 74.3].
2.84. Example. (The adjunction space) [8, Ex 35.8] [5, p 93] [10, Chp 1, Exercise B p 56]
f
i ?_
/ Y consisting of a space X and a continuous map f : A Y
Consider the set-up X o
A
defined on the closed subspace A X. Let R be the smallest equivalence relation on X q Y such
that aRf (a) for all a A; the equivalence classes of R are {a} q f (a) for a A, {x} for x X A,
anf f 1 (y) q {y} for y Y . The adjunction space is the quotient space
X f Y = X q Y /R
for the equivalence relation R. Let f : X X f Y be the map X X q Y X f Y and let
i : Y X f Y be the map Y X q Y X f Y . These two continuous maps agree on A in the
sense that f i = i f and the adjunction space is the universal space with this property. For any
other space Z receiving maps X Z Y that agree on A there exists a unique continuous map
X f Y such that the diagram
f
A
_
i
X
/Y
i
/ X f Y
!
#,
Z
commutes. The map i : Y X f Y is closed for closed sets B Y X q Y have closed
saturations f 1 (B) q B. Since i is injective it is an embedding; its image is a closed subspace
of X f Y homeomorphic to Y . The map f |X A : X A X f Y is open for open sets
U X A X q Y have open saturations U q . Since f |X A is injective it is an embedding; its
image is an open subspace of X f Y homeomorphic to X A. The quotient map X q Y X f Y
is closed if the map f is closed for then also closed subsets B X X q Y have closed saturations
B f 1 f (B A) q f (A B). We shall later see [8, Ex 35.8] that X f Y is normal when X and
Y are normal.
Only few topological properties are preserved by quotient maps. The reason is that surjective
open maps and surjective closed maps are quotient maps so that any property invariant under
quotient maps must also be invariant under both open and closed maps.
As we saw in 2.83.(6) the quotient space of a Hausdorff space need not be Hausdorff, not even
T1 . In general, the quotient space X/R is T1 if and only if all equivalence classes are closed sets.
(For instance, if X and Y are T1 then also the adjunction space X f Y is T1 .) The quotient space
X/R is Hausdorff if and only if any two distinct equivalence classes are contained in disjoint open
saturated sets. We record an easy criterion for Hausdorfness even though you may not yet know
the meaning of all the terms.
2.85. Proposition. If X is regular and A X is closed then the quotient space X/A is
Hausdorff.
The product of two quotient maps need not be a quotient map [2, I.5.3] in general but here
is an important case where it actually is the case.
2.86. Theorem. Let p : A B and q : C D be quotient maps. If B and C are locally
compact Hausdorff spaces (2.169) then p q : A C B D is a quotient map.
Proof. Using Lemma 2.87 below we see that the map p q is the composition
AC
p1
/B C
1q
/B D
41
2.87. Lemma (Whitehead Theorem). [5, 3.3.17] Let p : X Y be a quotient map and Z a
locally compact space. Then
p 1: X Z Y Z
is a quotient map.
Proof. Let A X Z. We must show: (p 1)1 (A) is open = A is open. This means
that for any point (x, y) (p 1)1 (A) we must find a saturated neighborhood U of x and a
neighborhood V of y such that U V (p 1)1 (A).
Since (p 1)1 (A) is open in the product topology there is a neighborhood U1 of x and a
neighborhood V of y such that U1 V (p 1)1 (A). Since Y is locally compact Hausdorff we
may assume (2.169) that V is compact and U1 V (p 1)1 (A). Note that also p1 (pU1 ) V
is contained in (p 1)1 (A). The tube lemma 2.142 says that each point of p1 (pU1 ) has a
neighborhood such that the product of this neighborhood with V is contained in the open set
(p 1)1 (A). Let U2 be the union of these neighborhoods. Then p1 (pU1 ) U2 and U2 V
(p 1)1 (A). Continuing inductively we find open sets U1 U2 S
Ui Ui+1 such
that p1 (pUi ) S
Ui+1 and Ui+1S V (p 1)1 (A). The open set U = Ui is saturated
S because
U p1 (pU ) = p1 (pUi ) Ui+1 = U . Thus also U V is saturated and U V Ui V
(p 1)1 (A).
For instance, if p : X Z is a quotient map, then also p 1 : X [0, 1] Z [0, 1] is a quotient map. This fact is important for homotopy theory.
42
8. Metric topologies
If X is a set with a metric d : X X [0, ) the collection {Bd (x, ) | x X, > 0} of balls
Bd (x, ) = {y X | d(x, y) < } is a basis for the metric topology Td induced by d.
2.88. Definition. A metric space is the topological space associated to a metric set. A topological space is metrizable if the topology is induced by some metric on X.
Hausdorff dimension, fractals, or chaos are examples of metric, rather than topological, concepts.
2.89. Theorem (Continuity in the metric world). Let f : X Y be a map between metric
spaces with metrics dX and dY , respectively. The following conditions are equivalent:
(1) f is continuous
(2) x X > 0 > 0y X : dX (x, y) < dY (f (x), f (y)) <
(3) x X > 0 > 0 : f (BX (x, )) BY (f (x), )
Proof. Essentially [8, Ex 18.1].
Td0
Proof. Td Td0 if and only if the identity map (X, Td0 ) (X, Td ) is continuous [8, Ex 18.3]
2.91. Lemma (Standard bounded metric). Let d be a metric on X. Then d0 (x, y) = min{d(x, y), 1}
is a bounded metric on X (called the standard bounded metric corresponding to d) that induce the
same topology on X as d.
Proof. Either use the above proposition or use that for any metric the collection of balls of
radius < 1 is a basis for the metric topology. These bases are the same for the two metrics.
2.92. Theorem (Hereditary properties of metrizable spaces). Any subspace of a metrizable
space is a metrizable. Any countable product of metrizable spaces is metrizable.
Proof. See [8, Ex 21.1] for the first assertion. To prove the second assertion, let Xn , n Z+ ,
be a countable collection of metric spaces. We may assume that each Xn has diameter at most 1
(2.91). Put
1
d((xn ), (yn )) = sup{ dn (xn , yn ) | n Z+ }
n
Q
for points (xn ) and (yn ) of Xn and convince yourself that d is a metric. The idea here is that
1
1
n dn (xn , yn ) n becomes small when n becomes large. For any > 0
d((xn ), (yn )) n N :
1
dn (xn , yn )
n
8. METRIC TOPOLOGIES
43
2.93. The first countability axiom. Which topological spaces are metrizable? To address
this question we need to build up an arsenal of metrizable and non-metrizable spaces and to identify
properties that are common to all metrizable spaces. Here are the first such properties: All metric
spaces are Hausdorff and first countable.
2.94. Definition (Neighborhood basis). A neighborhood basis at x X is a collection of
neighborhoods of x such that any neighborhood of x contains a member of the collection.
2.95. Definition (First countable spaces). Let X be a space and x a point in X. We say that
X has a countable basis at x if there is a countable neighborhood basis at x. X is first countable
if all points of X have a countable neighborhood basis.
All metrizable spaces are first countable since {B(x, 1/n) | n Z+ } is a countable neighborhood basis at x.
2.96. Proposition (Hereditary properties of first countable spaces). [8, Thm 30.2] Any subspace of a first countable space is first countable. Any countable product of first countable spaces
is first countable.
Q
Proof. The first assertionQis immediate. Let Xn be a countable product of first countable
spaces. Let
Q (xn ) be a point of Xn . Let Bn be countable basis at xn Xn . The collection of all
products Un where Un Bn for finitely many n and Un = Xn for all other n is then a countable
(1.41) basis at (xn ).
2.97. Example.
(1) R is first countable because it is a metric space.
(2) R` is first countable. The collection of half-open intervals [a, b) where b > a is rational is a
countable basis of neighborhoods at the point a. Is R` metrizable? (3.3) [8, Ex 30.6]
(3) Is RK first countable? Is it metrizable?
(4) The ordered space S = [0, ] (1.52) is not first countable at but it is first countable at
any other point. Let {Un } be any countable collection of neighborhoods of . Choose an <
such that (an , ] Un and chose a such that an < a < for all n. This is possible because
the countable set {an } [0, ) of left end-points has an upper bound in [0, ) (1.52.(2)). Then
(a, ] is a neighborhood of that does not contain any of the Un because it does not even
contain any of the intervals (an , ]. The section S , however, is first countable: {1} = [1, 2)
is open so S has a finite basis at the first element 1. For any other element, > 1, we can
use the countable collection of neighborhoods of the form (, ] for < (2.31.(5)). Is S
metrizable? (2.163)[8, Exmp 3 p 181, Ex 30.7].
(5) RJ is not first countable at any point when J is uncountable: Let {Un }nZ+ be any countable
collection of neighborhoods of, say, the point (0)jJ of RJ . I claim that there is a neighborhood
that does not contain any of the Un . This is because there is an index j0 J such that
j0 (Un ) = R for all n. Indeed, the set of js for which this is not true
[
{j J | n Z+ : j (Un ) 6= R} =
{j J | j (Un ) 6= R}
nZ+
(R
is a countable union of finite sets, hence countable (1.41.(3)). Then the neighborhood j1
0
{1}) of (0)jJ does not contain any of the neighborhoods Un in the countable collection. (RJ
does not even satisfy the sequence lemma (2.99.(1)) [8, Exmp 2 p 133].)
(6) The closed (2.83.(4)) quotient map R R/Z takes the first countable space R to a space
that is not first countable [5, 1.4.17] at the point corresponding to Z. This can be seen by a
kind of diagonal argument: Let {Un }nZ be any countable collection of open neighborhoods
of Z R. Let U be the open neighborhood of Z such that U (n, n + 1), n Z, equals
Un (n, n + 1) with one `
point deleted.WThen U does not contain any of the Un . `
(7) The quotient map p : nZ+ S 1 nZ+ S 1 is closed (2.83.(4)). The domain nZ+ S 1 =
W
Z+ S 1 R R2 is first countable (2.96) but the image nZ+ S 1 (2.83.(7)) is not: Let
{Un }nZ+ be any collection of saturated neighborhoods of Z+ {1}. The saturated neighborhood U which at level n equals Un with one point deletedW(cf Cantors diagonal argument
(1.43)) does not contain any of the Un . It follows (2.96) that nZ+ S 1 does not embed in R2
44
nor in any other first countable space. For instance, the universal property of quotient spaces
(2.81) gives a factorization
`
(2.98)
nZ+
S1
KKK
KKK
K
p KKK
W%
nZ+
s9
s
s
s
sss
sss f
1
nZ+ S
S1
of the continuous map f such that m (f |{n} S 1 ) is the identity function when m = n and
the constant function when m 6= n. The induced Q
map f is an injective continuous map. It can
not be an embedding
for
the
countable
product
S 1 of circles is first
W 1
Q countable (2.96). The
topology on S is finer than the subspace topology inherited from S 1 .
These examples show that the uncountable product of first countable (even metric) spaces and
the quotient of a first countable space may fail to be first countable (2.97.(5), 2.97.(7)). Some
linearly ordered spaces are first countable, some are not (2.97.(1), 2.97.(4)).
In a first countable (eg metric) space X, the points of the closure of any A X can be
approached by sequences from A in the sense that they are precisely the limit points of convergent
sequences in A. This is not true in general (2.97.(4)).
2.99. Lemma. Let X be a topological space.
(1) (The sequence lemma) Let A X be a subspace and x a point of X. Then
x is the limit of a sequence of points from A = x A
The converse holds if X is first countable.
(2) (Continuous map preserve convergent sequences) Let f : X Y be a map of X into a
space Y . Then
f is continuous = f (xn ) f (x) whenever xn x for any sequence xn in X
The converse holds if X is first countable.
Proof. (1) The direction = is clear. Conversely, suppose that x A. Let Un be a countable
basis at x. We may assume that U1 U2 Un Un+1 as we may replace Un by
U1 . . . Un . For each n choose a point xn A Un . We claim that the sequence (xn ) converges
to x. Let U be any neighborhood of x. Then Un U for some n so that xm Um Un U for
m n.
(2) The direction = is clear. Conversely, suppose that f (xn ) f (x) whenever xn x. We want
to show that f is continuous, ie (2.51.(6)) that f (A) f (A) for any A X. Let x A. Since X
is first countable, there is by (1) a sequence of points an A converging to x. By hypothesis, the
sequence f (an ) f (A) converges to f (x). Thus f (x) f (A) by (1) again.
We say that X satisfies the sequence lemma1 if for any A X and for any x A there is a
sequence of points in A converging to x.
To summarize:
X is metrizable = X is 1st countable = X satisfies the sequence lemma (is Frechet)
Examples show that neither of these arrows reverse.
The largest element is a limit point of [0, ) but it is not the limit of any sequence in [0, )
as any such sequence has an upper bound in [0, ) (1.52.(2)). Thus [0, ] does not satisfy the
sequence lemma. Hnece it is not 1st countable and not metrizable.
1Aka a Frech
et space
8. METRIC TOPOLOGIES
45
2.100. The uniform metric. Let Y be a metric space with a bounded metric d and let J
be a set. We shall discuss uniform convergence which is a metric, not a topological, concept.
Q
2.101. Definition. [8, pp 124, 266] The uniform metric on Y J = jJ Y = map(J, Y ) is the
metric given by d(f, g) = sup{j J | d(f (j), g(j))}.
It is easy to see that this is indeed a metric.
2.102. Theorem. On RJ we have: product topology uniform topology box topology
Proof. Omitted.
46
9. Connected spaces
2.104. Definition. The topological space X is connected if it is not the union X = X0 X1
of two disjoint open non-empty subsets X0 and X1 .
Two subsets A and B of a space X are separated if A B = = A B. This means that the
two sets are disjoint and neither contains a limit point of the other. Two disjoint open (closed)
sets are separated. If C A and D B and A and B are separated, then C and D are separated.
A separation of X consists of two separated non-empty subsets A and B with union X = A B.
2.105. Theorem. The following are equivalent:
(1) X is connected
(2) The only clopen (closed and open) subsets of X are and X
(3) X has no separations
(4) Every continuous map X {0, 1} to the discrete space {0, 1} is constant.
Proof. We show that the negated statements are equivalent.
(1) = (2): Suppose that X = U1 U2 where U1 and U1 are disjoint, open, and non-empty.
Then U1 is an open, closed, non-empty, proper subset of X.
(2) = (3): If C is a closed, open, nonempty, proper subset of X then X = C (X C) is a
separation of X.
(3) = (4): Suppose that X = A B where A and B are separated. Then A A, for A does
not meet B, so A is closed. The map f : X {0, 1} given by f (A) = 0 and f (B) = 1 is continuous
since all closed subsets of the co-domain have closed pre-images.
(4) = (1): Let f : X {0, 1} be a surjective continuous map. Then X = f 1 (0) f 1 (1) is
the union of two disjoint non-empty open subsets.
2.106. Theorem. If X is connected, then f (X) is connected for any continuous map f : X Y .
Proof. We use the equivalence of (1) and (4) in Theorem 2.105. Let f : X Y be a continuous map. If f (X) is not connected, there is a non-constant continuous map f (X) {0, 1} and
hence a non-constant continuous map X {0, 1}. So X is not connected.
2.107. Example.
(1) R{0} = R R+ is not connected for it is the union of two disjoint
open non-empty subsets.
(2) We shall later prove that R is connected (2.118) and that the connected subsets of R are
precisely the intervals, rays, R, and .
(3) R` is not connected for [a, b) is a closed and open subset whenever a < b (2.31.(4)). In fact,
any subset Y of R` containing at least two points a < b is disconnected as Y [a, b) is closed
and open but not equal to or Y . (R` is totally disconnected.)
(4) RK is connected [8, Ex 27.3].
(5) Q is totally disconnected (and not discrete): Let Y be any subspace of Q containing at least
two points a < b. Choose an irrational number t between a and b. Then Y (t, ) = Y [t, )
is an open, closed, non-empty, proper subset of Y .
(6) Particular point topologies (2.2.(3)) are connected.
A subspace of X is said to be connected if it is connected in the subspace topology. A subspace
of a connected space need obviously not be connected. So how can we tell if a subspace is connected?
2.108. Lemma. Let Y X be a subspace. Then
Y is connected Y is not the union of two separated non-empty subsets of X
Proof. Suppose that Y = Y1 Y2 is the union of two subspaces. Observe that
(2.109)
9. CONNECTED SPACES
47
Proof. The subsets Y A and Y B are separated (since the bigger sets A and B are
separated) with union Y . By 2.108, one of them must be empty, Y A = , say, so that Y B.
2.111. Corollary. The closure of any connected subspace is connected. Indeed, if C Y C
and C is connected, then Y is connected.
Proof. Let f : Y {0, 1} be a continuous map. Then f (C) is a single point since C is
connected. Moereover, f (y) = f (C) for all y Y since otherwise we could separate y and C by
disjoint open sets. Thus f is constant.
2.112. Theorem. Let (Yj )jJ be an indexed family of connected subspaces of the space X.
Suppose that
S there is an index j0 J such that Yj and Yj0 are not separated for any j J. Then
the union jJ Yj is connected.
S
Proof. Let f : jJ Yj {0, 1} be a continuous map. The image f (Yj ) of Yj is a single
point for each j J since Yj is connected. In fact, f (Yj ) = f (Yj0 ) as otherwise we could separate
Yj and Yj0 by disjoint open sets. Thus f is constant.
2.113. Corollary. The union of a collection of connected subspaces with a point in common
is connected.
Proof. Apply 2.112 with any of the subspaces as Yj0 .
2.114. Corollary. Suppose that for any two points in X there is a connected subspace containing both of them. Then X is connected.
Proof. Let x0 be some fixed point S
of X. For each pointT x X, let Cx be a connected
subspace containing x0 and x. Then X = Cx is connected as Cx 6= (2.113).
2.115. Theorem. Products of connected spaces are connected.
Proof. We prove first that the product X Y of two connected spaces X and Y is connected.
In fact, for any two points (x1 , y1 ) and (x2 , y2 ) the subspace X {y1 } {x2 } Y contains the two
points and this subspace is connected since (2.113) it is the union of two connected subspaces with
a point in common. Thus X Y is connected by Corollary 2.114.
Next, induction shows that the product of finitely many
Q connected spaces is connected.
Finally [8, Ex 23.10], consider an arbitrary product
Xj of connected spaces Xj , j J.
Choose (1.28) a point xj in each of the spaces Xj (assuming
that
all the spaces of the product are
Q
non-empty). For every finite subset F of J let CF Xj be the product of the subspaces Xj if
j F and {xj } if j 6 F . Since CF is connected
for each finite subset F and these subsets have
S
the point (xj )jJ in common, the union F F CF , where F is the collection of all finite subsets of
Q
S
Q
J, is connected (2.113). This union is not all of Xj but its closure is, so F F CF = jJ Xj
is connected (2.111).
2.116. Connected subspaces of linearly ordered spaces. We determine the connected
subsets of R, or, more generally, of any linear continuum.
Recall that a subset C of a linearly ordered set X is convex if a, b C = [a, b] C.
Connected subspaces are convex. But are convex subspaces connected? Not always: The convex
subset [0, 1] is not connected in Z but it is connected in R.
2.117. Lemma. Suppose that X is a linear continuum (1.15). Then
{connected subsets of X} = {convex subsets of X}
The inclusion holds in any linearly ordered space.
Proof. Suppose first that X is any linearly ordered space and let C X a subspace that is
not convex. Then there exist points a < x < b in X such that a and b are in C and x is outside C.
Since C (, x) (x, +) is contained in the union of two separated subsets and meet both of
them, C is not connected (2.110).
Let now X be a linear continuum and C a convex subset of X. We claim that C is connected.
Pick a fixed point a C and note that C is a union of closed intervals with a as one of their
end-points. Therefore it suffices (2.113) to prove the claim in case C = [a, b] is a closed interval.
48
Suppose that [a, b] = A B is the union of two disjoint relative open nonempty subsets A. We
may assume that b B. The sets A and B are closed and open in [a, b]. As the closed interval
[a, b] is closed in X (2.13), A and B are also closed in X (2.35). The nonempty bounded set A has
a least upper bound, c = sup A. Now, c b since b is an upper bound and c A since A is closed
in X. (The least upper bound of any bounded set always belongs to the closure of the set since
otherwise it wouldnt be the least upper bound.) So c < b (for b 6 A) and A [a, b). But A is
also open in [a, b] and in the subspace [a, b) (which has the subspace topology which is the order
topology (2.28)). This contradicts c A for no element of an open subset A of [a, b) can be an
upper bound for A: For any point d A there exists an open interval (x, y) around d such that
[a, b) (x, y) A and since [d, y) 6= , d is not an upper bound for A. (In short, c A since A is
closed in X and c 6 A since A is open in [a, b).) This is a contradiction and therefore A must be
empty.
The inclusion does not hold for all linearly ordered spaces as for instance Z+ is convex but
not connected (there are gaps).
In particular, the connected subsets of R are precisely the convex subsets which are , R, and
all intervals (bounded or unbounded, closed, open or halfopen).
We shall now identify the linearly ordered spaces that are connected.
2.118. Theorem. [2, Ex 7 p 382, Prop 1 p 336] Let X be a linearly ordered space. Then
X is connected in the order topology X is a linear continuum
The connected subsets of a linear continuum X are: X, , intervals and rays.
Proof. =: [8, Ex 24.4] [5, Problem 6.3.2] [2, Ex 7 p 382]. Suppose that X is a linearly
ordered set that is not a linear continuum. Then there are nonempty, proper, clopen subsets of X:
If (x, y) = for some points x < y then (, x] = (, y) is clopen and 6= , X.
If A X is a nonempty subset bounded
from above
which has no least upper bound
T
S
then the set of upper bounds B = aA [a, ) = bB (b, ) is clopen and 6= , X.
Therefore X is not connected (2.105).
=: Assume X is a linear continuum. From 2.117 we know that the the connected and the convex
subsets of X are the same. In particular, the linear continuum X, certainly convex in itself, is
connected in the order topology. Let C be a nonempty convex subset of X. We look at two cases:
C is neither bounded from above nor below. Let x be any point of X. Since x is neither
a lower nor an upper bound for C there exist a, b C so that a < x < b. Then x C by
convexity. Thus C = X.
C is bounded from above but not from below. Let c = sup C be its least upper bound.
Then C (, c]. Let x < c be any point. Since x is neither a lower nor an upper
bound for C there exist a, b C so that a < x < b. Then x C by convexity. Thus
(, c) C (, c] and C is either (, c) or (, c].
The arguments are similar for the other cases. Recall that X also has the greatest lower bound
property [8, Ex 3.13].
The real line R, the ordered square Io2 (2.16.(7)), the (ordinary) (half) line [1, ) = Z+ [0, 1)
(1.49.(6)), and the long (half) line [0, ) [0, 1) (1.52) [8, Ex 24.6, 24.12] are examples of linear
continua.
2.119. Theorem. [Intermediate Value Theorem] Let f : X Y be a continuous map of a
connected topological space X to a linearly ordered space Y . Then f (X) is convex. If Y is a linear
continuum, f (X) is an interval (bounded or unbounded, closed, half-open, or open)
(2.106)
(2.118)
Proof. X connected = f (X) connected = f (X) convex. For subsets of a linear continuum we know (2.118) that connected = convex = interval.
Any linearly ordered space containing two consecutive points, two points a and b with (a, b) =
is not connected as X = (, b) (a, +) is the union of two disjoint open sets.
Any well-ordered set X containing at least two points is totally disconnected in the order
topology. For if C X contains a < b then a 6 C (a, b] 3 b is closed and open in C since (a, b]
is closed and open in X.
9. CONNECTED SPACES
49
2.120. Path connected spaces. Path connectedness is a stronger property than connectedness.
2.121. Definition. The topological space X is path connected if for any two points x0 and x1
in X there is a continuous map (a path) f : [0, 1] X with f (0) = x0 and f (1) = x1 .
The image under a continuous map of a path connected space is path connected, cf 2.106. Any
product of path connected spaces is path connected [8, Ex 24.8], cf 2.112.
2.122. Example. The punctured euclidian plane Rn {0} is path connected when n 2 since
any two points can be joined by a path of broken lines. Thus also the (n 1)-sphere S n1 , which
is the image of Rn {0} under the continuous map x 7 x/|x|, is path connected for (n 1) 1.
Since the unit interval I is connected (2.118), all paths f (I) are also connected (2.106) so all
path connected spaces are, as unions of paths emanating from one fixed point, connected (2.114).
The converse is not true, not all connected spaces are path connected.
2.123. Example.
(1) (Topologists sine curve) Let S be the graph of the function sin(1/x),
0 < x 1, considered as a subspace of the plane R2 . The (closed) topologists sine curve is
the subspace S = S ({0} [1, 1]) of the plane. It follows from 2.106 and 2.111 that S is
connected. We shall soon see that S is not path connected (2.132.(2)).
(2) The ordered square Io2 (2.16.(7)) is connected since it is a linear continuum [8, Ex 3.15] but
it is not path connected. For suppose that f : [0, 1] Io2 is a path from the smallest element
(the lower left corner) f (0) = 0 0 to the largest element (the upper right corner) f (1) = 1 1.
Then f is surjective for the image contains (2.119) the interval [0 0, 1 1] = Io2 . But this
is impossible since the ordered square Io2 contains uncountably many open disjoint subsets (eg
x(0, 1) = (x0, x1), 0 x 1) but [0, 1] does not contain uncountably many open disjoint
subsets (choose a rational number in each of them).
(3) RK (2.11) is connected but not path-connected space [8, Ex 27.3].
50
(3) The path components are path-connected by their very definition. All path-connected spaces are
connected, so each path-component is contained in precisely one component. Thus the components
are unions of path-components.
If X has only finitely many components the components are closed and open. If X = C1
Cn where the finitely many subspaces Ci are connected and separated then the Ci are the
components of X.
The path components can be closed, open, closed and open, or neither closed nor open. However, for locally connected spaces (2.128) components and path components coincide (2.130).
2.127. Locally connected and locally path connected spaces.
2.128. Definition. The space X is locally (path) connected at the point x if every neighborhood
of x contains a (path) connected neighborhood of x. The space X is locally (path) connected if it
is locally (path) connected at each of its points.
Thus a space is locally (path) connected iff it has a basis of (path) connected subsets.
Consider this table
R
R {0}
S
Q
Connected
YES
NO
YES NO
Locally connected YES
YES
NO NO
before you draw any conclusions about the relationship between connected and locally connected
spaces. Note in particular that a space can be connected (even path connected (2.132)) and not
locally connected.
On the positive side, note that open subsets of locally (path) connected spaces are locally (path)
connected and that quotient spaces of locally connected spaces are locally connected (2.131).
2.129. Proposition. The space X is locally (path) connected if and only if open subsets have
open (path) components (open in X or, what is the same (2.25), open in the open set). In a locally
(path-) connected space the (path-) components are clopen.
Proof. Assume that X is locally connected and let U be an open subset. Consider a component, C, of U . The claim is that C is open. Let x be a point in C. Choose a connected neighborhood
V of x such that V U . Since V is connected and intersects C, V C (2.126). This shows that
C is open. In particular, the components (of the open set X) are open and, since they form a
partition, also closed.
Conversely, assume that open subsets have open components. Let x be a point of X and U
a neighborhood of x. Let C be the component of U containing x. Then C is an open connected
neighborhood of x contained in U .
It follows for instance that the open subsets of R are unions of at most countably many open
intervals. For any open set is the union of its components which are open, since R is locally
connected, and connected, so they are open intervals. Since each of the open intervals contains a
rational number, there are at most countably many components. Any closed subset of R is the
complement to a union of at most countably many open intervals.
2.130. Theorem. In a locally path-connected space the path-components and the components
are the same.
Proof. Suppose that X is locally path-connected. Each path-component P is contained
(2.126) in a unique component, C. Since C is connected and P is clopen (2.129), P = C (2.105).
For instance, the components and the path components of a locally Euclidean space, such as
a manifold, are the same. A locally path-connected space is path-connected if and only if it is
connected.
2.131. Proposition. [8, Ex 25.8] [2, I.11.6] Locally (path) connected spaces have locally (path)
connected quotient spaces.
9. CONNECTED SPACES
51
52
53
2.137
2.138
where
1ik
1ik
1ik jJxi
S
j 1ik Jxi
Here are two small lemmas that are used quite often.
2.144. Lemma (Criterion for noncompactness). If X contains an infinite closed discrete subspace, then X is not compact.
Proof. Any closed subspace of a compact space is compact (2.136). A discrete and compact
space is finite.
54
2.145. Lemma (Intersection of a nested sequence of compact sets). [8, Ex 28.5] [5, 3.10.2] Let
C1 C2 T Cn be a descending sequence of closed nonempty subsets of a compact
space. Then Cn 6= .
T
Proof. If Cn = then Cn = for some n Z+ by 2.134.(2).
2.146. Theorem. [Cf [8, Ex 27.5]] A nonempty compact Hausdorff space without isolated
points (2.39) is uncountable.
Proof. Let xn be a sequence of points in X. It suffices (1.39) to show that {xn | n Z+ } =
6 X.
Since X is Hausdorff and has no isolated points there is a descending sequence
V1 V2 Vn1 Vn
of nonempty open sets such that xn 6 V n for all n. These are constructed recursively. Put V0 = X.
Suppose that Vn1 has been constructed for some n Z+ . Since xn is not isolated, {xn } =
6 Vn1 .
Choose a point yn Vn1 distinct from xn and choose disjoint (separated) open sets Un X,
Vn Vn1 such that xn Un and yn
T Vn . Then Un V n = so xn 6 V n .
By construction, the T
intersection V n contains none of the points {xn } of the sequence and
by compactness (2.145), V n 6= . Thus X contains a point that is not in the sequence.
The Alexandroff compactification (2.166) Z+ = K = K {0}, where K is as in 2.11, is a
countable compact Hausdorff space with isolated points.
Is it true that a connected Hausdorff space is uncountable?
2.147. Example.
What is the suspension of the n-sphere S n ? Define f : [0, 1] S n S n+1 to be the (continuous)
map that takes (t, x), t [0, 1], to the geodesic path from the north pole (0, 1) S n+1 , through
the equatorial point (x, 0) S n S n+1 , to the south pole (0, 1) S n+1 . (Coordinates
in Rn+2 = Rn+1 R.) By the universal property of quotient spaces there is an induced
continuous and bijective map f : SS n S n+1 . Since SS n is compact (as a quotient of a
product of two compact spaces) and S n+1 is Hausdorff (as a subspace of a Hausdorff space),
f is a homeomorphism (2.140). We often write SS n = S n+1 , n 0, where the equality sign
stands for is homeomorphic to.
(2) Every injective continuous map S 1 R2 is an embedding. Can you find an injective
continuous map R1 R2 that is not an embedding?
(3) Alexanders horned sphere [7, Example 2B.2] is a wild embedding of the 2-sphere S 2 in
R3 such that the unbounded component of the complement R3 S 2 (2.132.(5)) contains noncontractible loops. It may be easier instead to consider Alexanders horned disc [3, p 232].
(The horned sphere is obtained by cutting out a disc of the standard sphere and replacing it a
horned disc.)
(4) Let X be a Hausdorff space and
X0 X1 Xn1 Xn X
an ascending sequence of closed subspaces. Assume that the topology on X is coherent with
this filtration in the sense that
A is closed A Xn is closed for all n
holds for all subsets A of X. Then any compact subspace C of X is contained in a finite stage
of the filtration. To see this, choose a point tn C (Xn Xn1 ) for all n Z+ for which
this intersection is nonempty. Let T = {tn } be the set of these points. We want to prove that
T is finite. Certainly, T Xn is finite, hence closed in the Hausdorff space Xn for all n 0.
Therefore T is closed. In fact, any subset of T is closed by the same argument and thus T is
discrete. But any closed discrete subspace of the compact space C is finite (2.144).
Q
2.148. Theorem (Tychonoff theorem). The product jJ Xj of any collection (Xj )jJ of
compact spaces is compact.
55
A1 , . . . , Ak A = A1 Ak A
A A : A0 A 6= = A0 A
To prove (2.149), note that the collection A {A1 Ak } is FIP, hence equals A by maximality.
To prove (2.150), note that the collection A {A0 } is FIP: Let A1 , . . . , Ak A. By (2.149),
A1 Ak A so A1 Ak A 6= by assumption.
Step 3. {j (A) | A A} is FIP for all j J.
j (A1 ) j (Ak ) j (A1 Aj ) 6= for any finite collection of sets A1 , . . . , Ak A.
Since Xj is compact and {j (A) | A A} is an FIP collection of closed subsets of Xj , the
T
T
intersection AA j (A) Xj is nonempty (2.134.(2)). Choose a point xj j (A) for each
j J and put x = (xj ) X.
T
Step 3. x AA A.
The point is that j1 (Uj ) A for any neighborhood Uj of xj . This follows from (2.150) since
j1 (Uj ) A 6= , or equivalently, Uj j (A) 6= , for any A A as xj j (A). Since A has
the FIP it follows that j1
(Uj1 ) j1
(Ujk ) A 6= whenever Uji are neighborhoods of the
1
k
finitely many points xji Xji and A A. Thus x A for all A in the collection A; in other words
T
x AA A.
2.151. Compact subspaces of linearly ordered spaces. We show first that any compact
subspace of a linearly ordered space is contained in a closed interval.
2.152. Lemma. Let X be a linearly ordered space and 6= C X a nonempty compact
subspace. Then C [m, M ] for some elements m, M C.
Proof. The claim is that C has a smallest and a largest
element. The proof is by contradicS
tion. Assume that C has no largest element. Then C cC (, c) as there for any a C is a
c C such that a < c. By compactness
C (, c1 ) (, ck ) (, c)
where c = max{c1 , . . . , ck } is the largest (1.50) of the finitely many elements c1 , . . . , ck C. But
this is a contradiction as c C and c 6 (, c).
We now show that the eg the unit interval [0, 1] is compact in R. Note that the unit interval
[0, 1] is not compact in R { 12 }. (The intersection all the closed subspaces [ n1 + 12 , n1 + 12 ] [0, 1],
n 2, is empty but the intersection of finitely many of them is not empty.) The reason for this
difference is that R, but not R { 21 } (1.16.(6)), has the least upper bound property (1.15).
2.153. Theorem. Let X be a linearly ordered space with the least upper bound property. Then
every closed interval [a, b] in X is compact.
56
Proof. Let [a, b] X be a closed interval and A and open covering of [a, b] (with the subspace
topology which is the order topology (2.28)). We must show that [a, b] is covered by finitely many
of the sets from the collection A. The set
C = {x [a, b] | [a, x] can be covered by finitely many members of A}
is nonempty (a C) and bounded from above (by b). Let c = sup C be the least upper bound of
C. Then a c b. We would like to show that c = b.
Step 1 If x C and x < b then C (x, b] 6= .
Proof of Step 1. Suppose first that x has an immediate successor y > x. We can not have x < b < y
for then y would not be an immediate successor. So x < y b. Clearly [a, y] = [a, x] {y} can be
covered by finitely many members of A. Suppose next that x has no immediate successor. Choose
an open set A A containing x. Since A is open in [a, b] and contains x < b, A contains an
interval of the form [x, d) for some d b. Since d is not an immediate successor of x there is a
point y (x, d). Now [a, y] [a, x] [x, d) [a, x] A can be covered by finitely sets from A.
Step 2 c C.
Proof of Step 2. The claim is that [a, c] can be covered by finitely many members of A. From Step
1 we have that C contains elements > a so the upper bound c is also > a. Choose A A such
that c A. Since A is open in [a, b] and a < c, A contains an interval of the form (d, c] for some d
where a d. Since d is not an uppe bound for C, there are points from C in (d, c]. Let y be such
a point. Now [a, c] = [a, y] (d, c] [a, y] A can be covered by finitely many sets from A.
Step 3 c = b.
Proof of Step 3. By Step 2, c C. But then c = b for Step 1 says that if c < b, then c can not be
an upper bound.
For instance, the ordered square Io2 = [0 0, 1 1], Z+ = [1, ], and S = [1, ] are compact
linearly ordered spaces [8, Ex 10.1]. More importantly, since the linearly ordered space R has the
least upper bound property, we get that all closed intervals in R are compact.
See [8, Ex 27.1] for the converse.
2.154. Corollary. Let X be a linearly ordered space and =
6 C X a nonempty subspace.
Then
C is compact and connected = C is a closed interval
The converse holds when X is a linear continuum.
Proof. Since C is compact, C [m, M ] for some m, M C (2.152) and [m, M ] C because
C is connected, hence convex (2.117). In a a linear continuum every closed interval is compact
(2.153) and connected (2.117).
In particular, the compact and connected subsets of the linear continuum R are precisely the
closed intervals.
2.155. Theorem (HeineBorel). A subset C of the euclidean space Rn is compact if and only
if it is closed and bounded in the euclidean metric.
Proof. Let C Rn be compact. Since C is compact and {(R, R)n | R > 0} an open
covering of Rn , there is an R > 0 such that C [R, R]n . Thus C is bounded. Compact
subspaces Hausdorff spaces are closed (2.138), so C is closed.
Conversely, if C is closed and bounded then C is a closed subset of [R, R]n for some R > 0.
But [R, R]n is compact (2.154, 2.143) and closed subsets of compact spaces are compact (2.136);
in particular, C is compact.
Can you now answer question 2.59.(10)?
2.156. Theorem (Extreme Value theorem). Let f : X Y be a continuous map of a nonempty
topological space X into a linearly ordered space Y .
(1) If X is compact then there exist m, M X such that f (X) [f (m), f (M )].
(2) If X is compact and connected then there exist m, M X such that f (X) = [f (m), f (M )].
Proof. (1) Since the image f (X) is compact (2.137) we can apply (2.152).
(2) Since the image f (X) is compact (2.137) and connected (2.106) we can apply (2.154).
57
2.157. Compactness in metric spaces. Since any open covering of compact space contains
a finite subcovering we see that any open covering of a compact metric space contains a subcovering
of open sets of size bounded from below. This observation is expressed in the Lebesgue lemma.
2.158. Lemma (Lebesgue lemma). Let X be a compact metric space. For any open covering
of X there exists a > 0 (the Lebesgue number of the open covering) such that any subset of X
with diameter < is contained in one of the open sets in the covering.
Proof. Cover X by finitely many open balls, X = B(x1 , 1 ) B(x2 , 2 ) B(xk , k ), such
that each double radius ball B(xi , 2i ) is contained in one of the open sets of the covering. Let
be any positive number smaller than all the i . Then any subset of diameter < is contained in
one of the balls B(xi , 2i ) and therefore in one of the open sets from the covering. (Let A X be
a subset with diam(A) < . Choose any point a in A. Then a is in one of the open balls B(xi , i ).
For any point b A, the distance d(b, xi ) d(b, a) + d(a, xi ) < + i < 2i which means that the
point b is contained in B(xi , 2i ).)
The open covering of R consisting of the open intervals (1, 1) and (x 1/|x|, x + 1/|x|) for
all x R with |x| > 1 does not have a Lebesgue number.
2.159. Definition. A map f : X Y between metric spaces is uniformly continuous if for all
> 0 there is a > 0 such that dY (f (x1 ), f (x2 )) < whenever dX (x1 , x2 ) < :
> 0 > 0x1 , x2 X : dX (x1 , x2 ) < dY (f (x1 ), f (x2 )) <
2.160. Theorem (Uniform continuity theorem). Let f : X Y be a continuous map between
metric spaces. If X is compact, then f is uniformly continuous.
Proof. Given > 0. Let be the Lebesgue number of the open covering {f 1 B(y, 12 ) | y
Y }. Then we have for all x1 , x2 X
dX (x1 , x2 ) < = diam{x1 , x2 } <
1
= y Y : {x1 , x2 } f 1 B(y, )
2
1
= y Y : {f (x1 ), f (x2 )} B(y, )
2
= dY (f (x1 ), f (x2 )) <
2.158. Limit point compactness and sequential compactness.
2.161. Definition. A space X is
(1) limit point compact if any infinite subset of X has a limit point
(2) sequentially compact if any sequence in X has a convergent subsequence.
In other words, a space is limit point compact if it contains no infinite closed discrete subspaces
(2.40).
We shall not say much about these other forms of compactness (see [5, 3.10] for a more
thorough discussion).
2.162. Theorem. (Cf [8, Ex 28.4, Ex 35.3]) For any topological space X we have
X is compact = X is limit point compact
1st countable + T1
X is sequentially compact
58
points from A in Uk+1 (2.48). In particular, there is nk+1 < nk such that xnk+1 is in Uk+1 . The
subsequence xnk converges to x.
X is sequentially compact and metrizable = X compact: This is more complicated (but should
also be well-known from your experience with metric spaces) so we skip the proof.
2.163. Example.
(1) The well-ordered space S of all countable ordinals is limit point
compact and sequentially compact but it is not compact. S is not compact (2.152) for it is a
linearly ordered space with no greatest element [8, Ex 10.6]. On the other hand, any countably
infinite subset of S is contained in a compact subset (1.52.(2), 2.153). Therefore (2.162) any
countably infinite subset, indeed any infinite subset (2.40), has a limit point and any sequence
has a convergent subsequence. (Alternatively, use that S is first countable (2.97.(4)).) It
follows (2.162) that S is not metrizable.
59
cX
||
||
|
| c
|} |
XC
CC
CC
CC
C!
/ X
c
commutes.
Proof. We must verify the following points:
The subspace topology on X is the topology on X: The subspace topology X T is clearly the
original topology on X.
X is dense in X: The intersection of X and some neighborhood of has the form X C where
C is compact. Since X is assumed non-compact, X C is not empty.
X is Hausdorff: Let x1 and x2 be two distinct points in X. If both points are in X then there
are disjoint open sets U1 , U2 X X containing x1 and x2 , respectively. If x1 X and x2 = ,
choose an open set U and a compact set C in X such that x U C. Then U 3 x and X C 3
are disjoint open sets in X.
X is compact: Let {Uj }jJ be any open covering of X. At least one of these open sets contains
. If Uk , k J, then Uk = (X C) {} for some compact set C X. There is a finite set
K J such that {Uj }jK covers C. Then {Uj }jK{k} is a finite open covering of X = X {}.
Let now c : X cX be another compactification X. Define c : cX X by c(x) = x for all
x X and c(cX X) = . We check that c is continuous. For any open set U X X, c1 (U )
is open in X and hence (2.25.(2)) in cX since X is open in cX (2.167). For any compact set
C X, c1 (X C) = cX C is open in cX since the compact set C is closed in the Hausdorff
space cX (2.139.(1)). This shows that c is continuous. By construction, c is surjective and hence
(2.140) a closed quotient map by the Closed Map Lemma (2.140).
60
2.37
ClX (X U ) = X X U = X U
is compact and hence closed in the Hausdorff space Y (2.138). But no part of U can stick outside
X U for since U (X U ) is open
X dense
U (X U ) 6= = (X U ) (X U ) 6=
which is absurd. Thus we must have U X U , in particular, U X. This shows that X is
open.
2.168. Example.
(1) The n-sphere S n = Rn is the Alexandroff compactification of the
locally compact Hausdorff space Rn for there is (2.59.(8)) a homeomorphism of Rn onto the
complement of a point in S n .
(2) The real projective plane is the Alexandroff compactification of the Mobius band. More
generally, real projective n-space RP n = MBn is the Alexandroff compactification of the
n-dimensional M
obius band MBn (2.82) which is a locally compact Hausdorff space (indeed a
manifold (3.40)).
(3) Let X be a linearly ordered space with the least upper bound property and let [a, b) be
a half-open interval in the locally compact space X. Then [a, b) = [a, b]. For instance the
Alexandroff compactifications of the half-open intervals [0, 1) R, Z+ = [1, ) Z+ Z+ ,
and S = [1, ) S are [0, 1], [1, ], [1, ] = S , respectively.
(4) The Alexandroff compactification of a countable union of disjoint copies of the real line
a
[
2.171.(4)
R = (Z+ R) = Z+ S 1 =
C1/n
nZ+
nZ+
is the Hawaiian Earring (2.41.(2)) (which is compact by the Heine-Borel theorem (2.155)).
(5) The Warsaw circle W (which compact by the Heine-Borel theorem (2.155)) (2.132.(2)) is a
compactification of R and the quotient space W/(W R) = R = S 1 .
2.169. Corollary (Characterization of locally compact Hausdorff spaces). In the category of
Hausdorff spaces, the following conditions are equivalent:
(1) X is locally compact
(2) X is homeomorphic to an open subset of a compact space
(3) For any point x X and any neighborhood U of x there is an open set V such that
x V V U and V is compact.
Proof. (1) = (2): The locally compact Hausdorff space X is homeomorphic to the open
subspace X {} of the compact Hausdorff space X (2.166).
(2) = (3): Suppose that Y is a compact Hausdorff space and that X Y is an open subset.
X is Hausdorff since subspaces of Hausdorff spaces are Hausdorff (2.47). Let x be a point of X
and U X a neighborhood of x in X. Then U is open also in Y (2.25) so that the complement
Y U is closed and compact (2.136) in the compact space Y . In the Hausdorff space Y we can
separate (2.139.(2)) the disjoint closed subspaces {x} and Y U by disjoint open sets, V 3 x and
W Y U or Y W U . As V is disjoint from W it is contained in the closed subset Y W
and then also V Y W U . Here, V is compact as a closed subspace of the compact space Y
(2.136). Note (2.35) that the closure of V in X equals the closure of V in Y because V , the closure
of V in Y , is contained in U and X.
(3) = (1): Let x be a point of X. By property (3) with U = X there is an open set V such that
x V V X and V is compact. Thus X is locally compact at x.
61
2.170. Corollary.
(1) Closed subsets of locally compact spaces are locally compact.
(2) Open or closed subsets of locally compact Hausdorff spaces are locally compact Hausdorff.
Proof. (1) Let X be a locally compact space and A X a closed subspace. We use the
definition (2.164) directly to show that A is locally compact. Let a A. There are subsets
a U C X such that U is open and C is compact. Then A U A C where A U is a
neighborhood in A and C A is compact as a closed subset of the compact set C (2.136).
(2) If X is locally compact Hausdorff and A X open, it is immediate from 2.169.(2) that A is
locally compact Hausdorff.
An arbitrary subspace of a locally compact Hausdorff space need not be locally compact
(2.171.(1)). The product of finitely many locally compact spaces is locally compact but an arbitrary
product of locally compact spaces need not be locally compact [8, Ex 29.2]; for instance Z
+ is not
locally compact. The image of a locally compact space under an open continuous [8, 29.3] or a
perfect map [8, Ex 31.7] [5, 3.7.21] is locally compact but the image under a general continuous
map of a locally compact space need not be locally compact; indeed, the quotient of a locally
compact space need not be locally compact (2.171.(2)) [5, 3.3.16].
S
2.171. Example.
(1) nZ+ Cn R2 (2.41.(2)) is not locally compact at the origin: Any
neighborhood of 0 0 contains a countably infinite closed discrete subspace so it can not be
contained in any compact subspace (2.144).
(2) The quotient space R/Z (2.97.(6)) is not locally compact at the point corresponding to Z:
Any neighborhood of this point contains an infinite closed discrete subspace so it can not be
contained in any compact subspace
` (2.144).
Q 1
(3) InWdiagram (2.98), the space
S 1 is locally compact but not compact,
S is compact,
and S 1 is not locally compact (at the one point common to all the circles).
(4) (Wedge sums and smash products) A pointed space is a topological space together with one
of its points, called the base point. The wedge sum of two pointed disjoint spaces (X, x0 ) and
(Y, y0 ) is the quotient space
X Y = (X q Y )/({x0 } {y0 })
obtained from the disjoint union of X and Y (2.22) by identifying the two base points. Let
f : X Y X Y be the continuous map given by f (x) = (x, y0 ), x X, and f (y) = (x0 , y),
y Y . The image f (X Y ) = (X {y0 }) ({x0 } Y ) = 21 ({y0 }) 11 ({x0 }) is closed
when X and Y are T1 spaces. In fact, f is a closed map for if C X is closed then f (C) =
{(x, y0 ) | x C} = 11 (C) 21 ({y0 }) is also closed. Thus f is a quotient map onto its image
and its factorization f as in the commutative diagram (2.81.(2))
f
/ X Y
X YI
II
u:
II
uu
u
II
u
II
uu f
$
uu
X Y
is an embedding. We can therefore identify X Y and X {y0 } {x0 } Y . The smash product
is defined to be
X Y = (X Y )/(X Y )
the quotient of the product space X Y by the closed subspace X Y ( = ).
If A X and B Y are closed subspaces, the universal property of quotient maps
(2.81.(2)) produces a continuous bijection g such that the diagram
gX gY
gX/AY /B
/ X/A Y /B
/ X/A Y /B
X Y R
5
RRR
k
RRR
kk k
k
k
k
RRR
RRR
kkkk g
R)
kkkk
(X Y )/(X B A Y )
commutes. However, g may not be a homeomorphism since the product gX gY of the two
closed quotient maps gX : X X/A and gY : Y Y /B may not be a quotient map. If X/A
62
and Y are locally compact Hausdorff spaces ([Ex 31.7] may be useful here) then gX gY is
quotient [Ex 29.11] so that g is a homeomorphism (2.78, 2.81.(3), 2.77.(3)) and
(2.172)
(X Y )/(X B A Y ) = X/A Y /B
in this case.
(5) The torus S 1 S 1 is a compactification of R R with remainder S 1 S 1 .
(6) (The Alexandroff compactification of a product space) If X and Y are locally compact
Hausdorff spaces the map X Y X Y X Y is an embedding. This follows from
(2.80, 2.77.(3)) when we note that the first map embeds X Y into an open saturated subset
of X Y (2.60). The universal property of the Alexandroff compactification (2.166) implies
that
(2.173)
(X Y ) = X Y
(2.173)
Y
Zp = {(an ) | n : an an+1 mod pn }
Z/pn Z
n=1
i
is closed and therefore also compact (2.136). Indeed, the sets {(an )
n=1 | ai aj mod p } are
closed (and open) whenever 0 < i < j because the projection maps are continuous (2.63). The
diagonal ring homomorphism
Z Zp
Z/pn Z,
(a) = (a mod pn )
n=1
n=1
embeds the ring of integers into Zp . By the definition of the product topology (2.17), Z is
dense in Zp in that any neighborhood of a point in Zp contains a point from Z. Thus Zp = Z
is a compact topological ring containing the integers as a dense subring and the map : Z Zp
is a compactification (2.165) of the discrete space Z. As a ring, Zp is quite different from Z. For
1
2
instance,
P n the integer 1 p is invertible in Zp with inverse (1 p) = (1, 1 + p, 1 + p + p , . . .) =
n=0 p . Actually, any x = (x1 , x2 , . . .) Zp with nonzero reduction mod p is invertible
because any xn Z/pn Z with nonzero reduction mod p is invertible [9, Prop 2, p 12].
Some mathematicians prefer to include Hausdorffness in the definition of (local) compactness.
What we call a (locally) compact Hausdorff space they simply call a (locally) compact space; what
we call a (locally) compact space they call a (locally) quasicompact space.
CHAPTER 3
+3 X is Lindel
of
X is 1st countable
If X is metrizable, the three conditions of the top line equivalent.
Proof. Suppose first that X is second countable and let B be a countable basis for the
topology.
X has a countable dense subset: Pick a point bB B in each basis set B B. Then {bB | B B}
is countable (1.41.(2)) and dense.
X is Lindel
of: Let U be an open covering of X. For each basis set B B which is contained in
some open set from the collection U, pick any UB U such that B UB . Then the at most
countable collection {UB } of these open sets from U is an open covering: Let x be any point in
X. Since x is contained in a member U of U and every open set is a union of basis sets, we have
x B U for some basis set B B. But then x B UB .
Any metric space with a countable dense subset is 2nd countable: Let X be a metric space with a
countable dense subset A X. Then the collection {B(a, r) | a A, r Q+ } of balls centered at
points in A and with a rational radius is a countable (1.41.(4)) basis for the topology: It suffices
1Or to be separable
63
64
to show that for any open ball B(x, ) in X and any y B(x, ) there are a A and r Q+
such that y B(a, r) B(x, ). Let r be a positive rational number such that 2r + d(x, y) <
and let a A B(y, r). Then y B(a, r), of course, and B(a, r) B(x, ) for if d(a, z) < r then
d(x, z) d(x, y) + d(y, z) d(x, y) + d(y, a) + d(a, z) < d(x, y) + 2r < .
f space. For each positive
Any metric Lindel
of space is 2nd countable: Let X be a metric LindeloS
rational
number
r,
let
A
be
a
countable
subset
of
X
such
that
X
=
r
aAr B(a, r). Then A =
S
A
is
a
dense
countable
(1.41.(3))
subset:
For
any
open
ball
B(x,
)
and any positive rational
r
rQ+
r < there is an a Ar such that x B(a, r). Then a B(x, r) B(x, ).
3.3. Example. The ordered square Io2 is compact (2.153) and therefore Lindelof but it is not
second countable since it contains uncountably many disjoint open sets (x 0, x 1), x I. Thus
Io2 is not metrizable [8, Ex 30.6].
2. Separation Axioms
3.4. Definition. A space X is called a
T1 -space if points {x} are closed in X
T2 -space or a Hausdorff space if for any pair of distinct points x, y X there exist disjoint
open sets U, V X such that x U and y V
T3 -space or a regular space if points are closed and for any point x X and any closed
set B X not containing x there exist disjoint open sets U, V X such that x U and
BV
T4 -space or a normal space if points are closed and for every par of disjoint closed sets
A, B X there exist disjoint open sets U, V X such that A U and B V .
We have the following sequence of implications
X is normal = X is regular = X is Hausdorff = X is T1
where none of the arrows reverse (3.9, 3.10, [8, Ex 22.6]).
3.5. Lemma. Let X be a T1 -space. Then
(1) X is regular For every point x X and every neighborhood U of x there exists an
open set V such that x V V U .
(2) X is normal For every closed set A and every neighborhood U of A there is an
open set V such that A V V U .
Proof. Let B = X U .
3.6. Theorem. (Cf 2.47) Any subspace of a regular space is regular. Any product of regular
spaces is regular.
Proof. Let X be a regular space and Y X a subset. Then Y is Hausdorff (2.47)). Consider
a point y Y and a and a closed set B X such that y 6 Y B. Then y 6 B and since Y is
regular there exist disjoint open sets U and V such that y U and B V . The relatively open
sets U Y andQV Y are disjoint and they contain y and B Y , respectively.
Let X =
Xj be the Cartesian product of regular space Xj , j J. Then X is Hausdorff
Q
(2.47)). We use 3.5.(1) to show that X is regular. Let x = (xj ) be a point in X and U = Uj
a basis neighborhood of x. Put Vj = Xj whenever Uj = Xj . Otherwise, choose Vj such that
Q
xj Vj V j Uj . Then V = Vj is a neighborhood of x in the product topology and (2.65)
Q
Q
V = V j Uj = U . Thus X is regular.
3.7. Theorem. [8, Ex 32.1] A closed subspace of a normal space is normal.
Proof. Quite similar to the proof (3.6) that a subspace of a regular space is regular.
An arbitraty subspace of a normal space need not be normal (3.32) and the product of two
normal spaces need not be normal ((3.9),[8, Example 2 p 203], [5, 2.3.36]).
3.8. Example (Sorgenfreys half-open interval topology). The half-open intervals [a, b) form
a basis for the space R` (2.11, 2.31.(4), 2.97.(2), 2.107.(3)).
2. SEPARATION AXIOMS
65
R` is 1st countable: At the point x the collection of open sets of the form [x, b) where
b > x is rational, is a countable local basis at x.
R` is not 2nd countable: Let B be any basis for the topology. For each point x choose
a member Bx of B such that x Bx [x, x + 1). The map R B : x 7 Bx is injective
for if Bx = By then x = inf Bx = inf By = y.
R` has a countable dense subset: Q is dense since any (basis) open set in R` contains
rational points.
R` is Lindel
of: It suffices (!) to show thatSany open covering by basis open sets contains
a countable subcovering, ie that if R = jJ [aj , bj ) is covered by a collection of right
half-open intervals [aj , bj ) then R is actually already covered by countably many of these
intervals. (Note that this is true had the intervals been open as R is Lindelof.) Write
[
[
R=
(aj , bj ) R
(aj , bj )
jJ
jJ
as the (disjoint) union of the corresponding open intervals and the complement of this
union. The first set can be covered by countably many of the intervals (aj , bj ) for any
subset of R (with the standard topology) is 2nd countable and hence Lindelof (3.2). Also
the second set can be covered by countably many of the intervals (aj , bj ) simply because
it is countable: The second set
[
[
[
R
(aj , bj ) =
[aj , bj )
(aj , bj ) = {ak | j J : ak 6 (aj , bj )}
jJ
jJ
jJ
consists of some of the left end-points of the intervals. The open intervals (ak , bk ) are
disjoint for ak in this set. But there is only room for countably many open disjoint
intervals in R (choose a rational point in each of them) so there are at most countaly
many left end-dpoints ak in the second set.
R` is normal: Let A, B R` be disjoint closed sets.SFor each point a A RS B there
is an xa R such that [a, xa ) R B. Let UA = aA [a, xa ). Define UB = bB [b, xb )
similarly. Then UA and UB are open sets containing A and B, respectively. If UA UB 6=
then [a, xa ) [b, xb ) 6= for some a A, b B. Say a < b; then b [a, xa ) R B
which is a contradiction So UA and UB are disjoint. (R` is even completely normal [8,
Ex 32.6] by this argument.)
Since R` has a dense countable subset and is not second countable (3.8) it is not metrizable
(3.2).
3.9. Example (Sorgenfreys half-open square topology). The half-open rectangles [a, b)[c, d)
form a basis (2.19) for the product topology R` R` . The anti-diagonal L = {(x, x) | x R} is
a closed (clear!) discrete (L [x, ) [x, ) = {(x, x)}) subspace of the same cardinality as
R. Q Q is a countable dense subspace.
R` R` is not Lindel
of: A Lindelof space can not contain an uncountable closed discrete
subspace (cf 2.144, [8, Ex 30.9]).
R` R` is not normal: A normal space with a countable dense subset can not contain
a closed discrete subspace of the same cardinality as R [5, 2.1.10]. (Let X be a space
with a countable dense subset. Since any continuous map of X into a Hausdorff space is
determined by its values on a dense subspaceQ[8, Ex 18.13], the set of continuous maps
X R has at most the cardinality of RZ = Z R. Let X be any normal space and L a
closed discrete subspace. The Tietze extension theorem (3.15) says that any map L R
extends to continuous map X R. Thus the set of continuous maps X RQhas at least
the cardinality of RL . If L and R have Q
the same cardinality, RL = RR = R R which
is greater (1.42) than the cardinality of Z R.) See [8, Ex 31.9] for a concrete example
of two disjoint closed subspaces that can not be separated by open sets.
R` R` is (completely) regular: since it is the product (3.30) of two (compeletely) regular (3.28) (even normal (3.8)) spaces.
Example 3.8 shows that the arrows of Theorem 3.2 do not reverse. Example 3.9 shows that
the product of two normal spaces need not be normal, that the product of two Lindelof spaces
need not be Lindel
of, and provides an example of a (completely) regular space that is not normal.
66
3.10. Example (A Hausdorff space that is not regular). The open intervals plus the sets
(a, b) K where K = { n1 | n Z+ } form a basis for the topology RK (2.11, 2.123.(3)). This is
a Hausdorff topology on R since it is finer than the standard topology. But RK is not regular.
The set K is closed and 0 6 K. Suppose that U 3 0 and V K are disjoint open sets. We may
choose U to be a basis open set. Then U must be of the form U = (a, b) K for the other basis
sets containing 0 intersect K. Let k be a point in (a, b) K (which is nonempty). Since k is in
the open set V , there is a basis open set containing k and contained in V ; it must be of the form
(c, d). But U V ((a, b) K) (c, d) and ((a, b) K) (c, d) 6= for cardinality reasons. This
is a contradiction
3. NORMAL SPACES
67
3. Normal spaces
Normal spaces can be characterized in two ways.
3.11. Corollary. Let X be a T1 -space. Then the following conditions are equivalent:
(1) X is normal: For any pair of disjoint closed sets A, B X there exist disjoint open sets
U, V X such that A U and B V .
(2) (Urysohns characterization of normality) For any pair of disjoint closed sets A, B X
there exists a continuous function f : X [0, 1] such that f (A) = 0 and f (B) = 1.
(3) (Tietzes characterization of normality) For any closed subset A of X and any continuous
function f : A [0, 1] there exists a continuous function F : X [0, 1] such that F |A =
f.
The proof relies on Urysohns lemma (3.12) and Tietzes extension theorem (3.15).
3.12. Theorem (Urysohn lemma). Let X be a normal space and let A and B be disjoint closed
subsets of X. Then there exists a continuous function (a Urysohn function) f : X [0, 1] such
that f (a) = 0 for a A and f (b) = 1 for b B.
Proof. We shall recursively define open sets Ur X for all r Q [0, 1] such that (make a
drawing!)
(3.13)
r < s = A U0 Ur U r Us U s U1 X B
(X U r )
r>b
the sets
f 1 ([0, a)) =
[
r<a
are open.
Ur
(X U r )
r>b
0 x 0
f (x) = x 0 < x < 1
1 1x
is the Urysohn function with f (A) = 0 and f (B) = 1.
68
3.15. Theorem (Tietze extension theorem). Every continuous map from a closed subspace A
of a normal space X into (0, 1), [0, 1) or [0, 1] can be extended to X.
3.16. Lemma. Let X be a normal space and A X a closed subspace. For r > 0 and any
continuous map f0 : A [r, r] there exists a continuous map g : X R such that
2
1
(3.17)
x X : |g(x)| r and a A : |f0 (a) g(a)| r
3
3
Proof. Since the sets f01 ([r, 31 r]) A and f01 ([ 13 r, r]) A are closed and disjoint in A
they are also closed and disjoint in X. Choose (3.12) a Urysohn function g : X [ 13 r, 13 r] such
that g(x) = 13 r on f01 ([r, 13 r]) and g(x) = 13 r on f01 ([ 13 r, r]). From
1
1 1
1
f01 ([r, r]) = f01 ([r, r]) f01 ([ r, r]) f01 ([ r, r])
3
3
3
3
|
{z
} |
{z
} |
{z
}
g= 13 r
|g| 13 r
g= 13 r
Proof of 3.15. We shall first prove the theorem for functions from A to [0, 1] or rather to
the homeomorphic interval [1, 1] which is more convenient for reasons of notation.
Given a continuous function f : A [1, 1]. We have just seen (3.16) that there exists a
continuous real function g1 : X R such that
1
2
x X : |g1 (x)|
and a A : |f (a) g1 (a)|
3
3
Now Lemma 3.16 applied to the function f g1 on A says that there exists a continuous real
function g2 : X R such that
2
2
12
and a A : |f (a) (g1 (a) + g2 (a))|
x X : |g2 (x)|
33
3
Proceeding this way we recursively (1.24) define a sequence g1 , g2 , of continuous real functions
on X such that
n1
n
n
X
1 2
2
(3.18)
x X : |gn (x)|
and a A : |f (a)
gi (a)|
3 3
3
i=1
P
By the
first inequality in (3.18), the series n=1 gn (x) converges uniformly and the sum function
P
F = n=1 gn is continuous by the uniform limit theorem (2.103). By the first inequality in (3.18),
n1
1X 2
1
|F (x)|
= 3 = 1,
3 n=1 3
3
and by the second inequality in (3.18), F (a) = f (a) for all a A.
Assume now that f : A (1, 1) maps A into the open interval between 1 and 1. We know
that we can extend f to a continuous function F1 : X [1, 1] into the closed interval between
1 and 1. We want to modify F1 so that it does not take the values 1 and 1 and stays the
same on A. The closed sets F11 ({1}) and A are disjoint. There exists (3.12) a Urysohn function
U : X [0, 1] such that U = 0 on F11 ({1}) and U = 1 on A. Then F = U F1 is an extension
of f that maps X into (1, 1).
A similar procedure applies to functions f : A [1, 1) into the half-open interval.
Proof of Corollary 3.11. The Urysohn lemma (3.12) says that (1) = (2) and the converse is clear. The Tietze extension theorem (3.15) says that (1) = (3). Only eg (3) = (2)
remains. Assume that X is a T1 -space with property (3). Let A, B be two disjoint closed subsets.
The function f : A B [0, 1] given by f (A) = 0 and f (B) = 1 is continuous (2.53.(5)). Let
f : X [0, 1] be a continuous extension of f . Then f is a Urysohn function for A and B.
Many familiar classes of topological spaces are normal.
3.19. Theorem. Compact Hausdorff spaces are normal.
Proof. See 2.139.(2).
69
d(x, A)
d(x, A) + d(x, B)
is a Urysohn function with f (A) = {0} and f (B) = {1}. This shows that X is normal (3.11).
3.21. Theorem. [5, Problem 1.7.4]. Linearly ordered spaces are normal.
Proof. We shall only prove the special case that every well-ordered space is normal. The
half-open intervals (a, b], a < b, are (closed and) open (2.31.(5)). Let A and B be two disjoint
closed subsets and let a0 denote the smallest element of X. Suppose that neither A nor B contain
a0 . For any point a A there exists a point xa < a such that (xa , a] is disjoint from B. Similarly,
for any point b B there exists a point xb < b such that (xb , b] is disjoint from A. The proof now
proceeds as the proof (3.8) for normality of R` . Suppose next that a0 A B, say a0 A. The
one-point set {a0 } = [a0 , a+
0 ) is open and closed (as X is Hausdorff). By the above, we can find
disjoint open sets U , V such that A {a0 } U and B V . Then A U {a0 } and B V {a0 }
where the open sets U {a0 } and V {a0 } are disjoint.
In particular, the well-ordered spaces S and S are normal (the latter space is even compact
Hausdorff).
4. Second countable regular spaces and the Urysohn metrization theorem
We investigate closer the class of regular spaces. We start with an embedding theorem that is
used in other contexts as well.
3.22. An embedding theorem. We discuss embeddings into product spaces.
3.23. Definition. A set {fj : X Yj | j J} of continuous functions is said to separate
points and closed sets if for any point x X and any closed subset C X we have
x 6 C = j J : fj (x) 6 fj (C)
3.24. Lemma. Let f : X Y be a map that separates points and closed sets. Assume also that
f is injective (eg that X is T1 ). Then f is an embedding.
Proof. If X is T1 , points are closed so that f separates points, ie f is injective. For any point
x X and any closed set C X we have that
f (x) f (C)
f separates
x C = f (x) f (C)
so we get that f (X) f (C) = f (C). But this equality says that f (C) is closed in f (X). Hence
the bijective continuous map f : X f (X) is closed, so it is a homeomorphism.
3.25. Theorem (Diagonal embedding theorem). Let {fj : X Yj | J} be a family of
continuous functions that separates points and closed sets. Assume that X
Q is T1 or that at least one
of the functions fj is injective. Then the diagonal map f = (fj ) : X jJ Yj is an embedding.
Proof. Also the map f separates points and closed sets because
Y
Y
2.65
f (x) f (C) f (x)
fj (C) f (x)
fj (C)
j J : fj (x) fj (C)
(fj ) separates
for all points x X and all closed subsets C X. The theorem now follows from 3.24.
xC
In particular, if one of the functions fj is injective and separates points and closed sets then
f = (fj ) is an embedding. For instance, the graph X X Y : x (x, g(x)) is an embedding
for any continuous map g : X Y .
70
3.26. A universal second countable regular space. We say that a space X is universal
for some property if X has this property and any space that has this property embeds into X.
3.27. Theorem (Urysohn metrization theorem). The following conditions are equivalent for
a second countable space X:
(1) X is regular
(2) X is normal
(3) X is homeomorphic to a subspace of [0, 1]
(4) X is metrizable
The Hilbert cube [0, 1] is a universal second countable metrizable (or normal or regular) space.
Proof. (1) = (2): Let X be a regular space with a countable basis B. We claim that X is
normal. Let A and B be disjoint closed sets in X. By regularity (3.5.(1)), each point a A X B
has a basis neighborhood Ua B such that a Ua U a X B. Let U1 , U2 , . . . be the elements
in the image of the map A B : a 7 Ua . Then
Un
and U n B = for n = 1, 2 . . .
n=1
Vn
n=1
S
n=1
and V n A = for n = 1, 2 . . .
Un and
U10
U20
= U1 V 1
V10 = V1 U 1
= U2 (V 1 V 2 )
V20 = V2 (U 1 U 2 )
n=1
..
.
..
.
Un0 = Un (V1 V n )
Vn0 = Vn (U1 U n )
Even though we have removed part of Un we have removed no points from A from Un , and we
have removed no pints from B from Vn . Therefore the open sets Un0 still cover A and the open sets
Vn0 still cover B:
[
[
A
Un0 and B
Vn0
n=1
n=1
[
n=1
Un0
[
n=1
Vn0 =
[
mn
[
0
0
Um
Vn0
Um
Vn0 =
m>n
0
0
because Um
Um does not intersect Vn0 X Um if m n and Um
X Vn does not intersect
Vn0 Vn if m > n. (Make a drawing of U1 , U2 and V1 , V2 .)
(2) = (3): Let X be a normal space with a countable basis B. We show that there is a countable
set {fU V } of continuous functions X [0, 1] that separate points and closed sets. Namely, for each
pair U, V of basis open sets U, V B such that U V , choose a Urysohn function fU V : X [0, 1]
(3.12) such that fU V (U ) = 0 and fU V (X V ) = 1. Then {fU V } separates points and closed sets:
For any closed subset C and any point x 6 C, or x X C, there are (3.5) basis open sets U , V
such that x U U V X C (choose V first). Then fU V (x) = 0 and fU V (C) = 1 so that
fU V (x) 6 fU V (C). According to the Diagonal embedding theorem (3.25) there is an embedding
X [0, 1] with the fU V as coordinate functions.
(3) = (4): [0, 1] is metrizable and any subspace of a metrizable space is metrizable. (2.92).
(4) = (1): Any metrizable space is regular, even normal (3.20).
The point is that in a second countable regular hence normal space there is a countable set of
(Urysohn) functions that separate points and closed sets. Therefore such a space embeds in the
Hilbert cube.
71
72
kC(Y )
5. COMPLETELY REGULAR SPACES AND THE STONECECH
COMPACTIFICATION
73
/Y
/ kC(Y ) I
I
f
LLL
r
LLL
rrr
r
L
r
f k =kf LL
LL rrrr k
% yr
I
Q
jC(X)
commutes: The lower triangle commutes by the definition of the map f (2.64) and then the
upper square commutes because k f X = kf X Q
= kf = k Y f . Put X = (X) and
define : X X to be the corestriction of : X jC(X) I to X. Then X is a compact
Q
Hausdorff space (it is a closed subspace of the compact (2.148) Hausdorff space I) and is, by
design, a continuous map with a dense image X in X. This construction is natural: For any
continuous map f : X Y , the induced map f (3.34) takes X into Y for
(3.35)
f (X) = f (X X) f X X = Y f (X) Y Y = Y
and thus we obtain from (3.35) a new commutative diagram of continuous maps
(3.36)
X
/Y
Y
/ Y
/Y
XB
BB
|>
|
BB
||
BB
|| f
B!
|
|
X
f
74
an embedding. The image of this embedding is closed (2.140.(1)) and dense. Thus the embedding
is bijective so it is a homeomorphism.
(1) and (2): If Y is compact Hausdorff then by item (4) Y is a homeomorphism in diagram (3.36)
and hence f = 1
Y f is a possibility in (3.38). It is the only possibility [8, Ex 18.13], for the
image of X is dense in X.
(3) Let X X be a map to a compact Hausdorff space satisfying the above universal property.
/ X that, by uniqueness, are inverse to each other.
Then there exist continuous maps X o
(All universal constructions are essentially unique.)
If X is completely regular then X X is a compactification (3.37.(4)) and it is called the
StoneCech
compactification of X. Conversely, if X has a compactification then X is homeomorphic to a subspace of a compact space and therefore (3.29) completely regular. We state these
observations in
3.39. Lemma. X has a compactification (2.165) if and only if X is completely regular.
The StoneCech
compactification X of a completely regular space X is the maximal compactification in the sense that for any other compactification X cX there is a closed quotient
map X cX map such that
||
||
|
|
|} |
XB
BB
BB
BB
B
/ cX
commutes. The map X cX is closed by 2.140 and surjective because its image is closed and
dense.
The Alexandroff compactification X of a noncompact locally compact Hausdorff space X is
the minimal compactification in the sense that X = cX/(cX X) for any other compactification
X cX.
Indeed we saw in 2.166 that there is a closed quotient map cX X, taking cX X to ,
such that
cX
||
||
|
|
|} |
XC
CC
CC
CC
C!
/ X
6. MANIFOLDS
75
6. Manifolds
3.40. Definition. A manifold is a locally euclidean second countable Hausdorff space.
Manifolds are locally compact and locally path connected since they are locally euclidean. All
locally compact Hausdorff spaces are (completely) regular. Thus manifolds are second countable
regular spaces and hence they are normal, metrizable, Lindelofand they have countable dense
subsets.
The line with two zeroes is an example of a locally euclidean space that is not Hausdorff. The
long line [8, Ex 24.12] is a locally euclidean space that is not 2nd countable.
The nsphere S n and the real projective nspace RP n (2.82) are examples of manifolds. S n
lies embedded in Rn+1 from birth but what about RP n ? Does RP n embed in some euclidean
space?
3.41. Theorem (Embeddings of compact manifolds). Any compact manifold embeds in RN
for some N .
Proof. We may assume that M is connected. Let n be the dimension of M (we assume that
this is well-defined). Let B = {x Rn | |x| 1} be the unit ball in Rn . Cover M by finitely many
closed sets B1 , . . . , Bk such that each Bi is homeomorphic to B and such that the interiors of the
Bi cover M . Let fi : M S n be the continuous map obtained by collapsing the complement to
the interior of Bi to a point. The map f = (f1 , . . . , fk ) : M S n S n is injective: If x lies
in the interior of Bi then fi (x) 6= fi (y) for all y 6= x. Since M is compact and (S n )k Hausdorff, f is
an embedding. Finally, each sphere S n embeds into Rn+1 so that (S n )k embeds into (Rn+1 )k .
In particular RP 2 (2.82) embeds into some euclidean space. Which one?
There is up to homeomorphism just one compact manifold of dimension 1, the circle. Compact
manifolds of dimension 2 are described by a number, the genus, and orientability. In particular,
any simply connected 2-dimensional compact manifold is homeomorphic to S 2 . We do not know
(December 2003) if there are any simply connected compact 3-dimensional manifolds besides S 3
(Poincare conjecture). Classification of 4-dimensional compact manifolds is logically impossible.
Any manifold admits a partition of unity. We shall prove the existence in case of compact
manifolds.
3.42. Theorem (Partition of unity). Let X = U1 Uk be a finite open covering of the
normal space X. The there exist continuous functions i : X [0, 1], 1 i k, such that
(1) {i > 0} Ui
Pk
(2)
i=1 i (x) = 1 for all x X.
Proof. We show first that there is an open covering {Vi } of X such that V i Ui (we shrink
the sets of the covering). Since the closed set X (U2 Uk ) is contained in the open set U1
there is an open set V1 such that
X (U2 Uk ) V1 V 1 U1
by normality (3.5). Now V1 U2 Uk is an open covering of X. Apply this procedure once
again to find an open set V2 such that V 2 U2 and V1 V2 U3 Uk is still an open covering
of X. After finitely many steps we have an open covering {Vi } such that V i Ui for all i.
Do this one more time to obtain an open covering {Wi } such that Wi W i Vi V i Ui for
all i. Now choose a Urysohn function (3.12) i : X [0, 1] such that i (W i ) = 1 and i (X Vi ) =
P
0. Then {x | i (x) > 0} V i Ui and (x) =
i (x) > 1 for any x X since x Wi for some
i. Hence i = i is a well-defined continuous function on M taking values in the unit interval such
Pk
Pk
Pk
(x)
1
i (x)
that i=1 i (x) = i=1 (x)
= (x)
i=1 i (x) = (x) = 1.
Ex 32.7
77
Ex 23.5
well-ordered topology
path connected
p. 155
connected
KS
Thm 24.1
linear continuum
+3 order topology
SK
Thm 32.4
+3 completely normal
+3 locally compact
SK
T1
Hausdorff
regular
+3 completely regular ks
top grp
Ex 33.10
Ex 22.7
Thm 33.1
+3 normal
KS
totally
disconnected
manifold
J
JJJJ
JJJJ
JJJJ
JJJJ
JJJJ
JJJJ
JJJJ locally connected
JJJJ
KS
JJJJ
JJJJ
JJJJ
J(
locally path connected
Ex 36.1
+metrizable
compact
KS
Thm 28.1
compact Hausdorff
+3 Lindelof ks
+metrizable
limit point
KS compact
+3 Sequence lma
Thm 30.1
Thm 34.1
metrizable
KS
1st countable
KS
Thm 30.3
+metrizable
+metrizable
sk
2nd countable
Thm 30.3
countable dense
subset
KS
+1st countable+T1
sequentially
KS compact
CHAPTER 4
Bibliography
[1] Colin C. Adams, The knot book, W. H. Freeman and Company, New York, 1994, An elementary introduction
to the mathematical theory of knots. MR 94m:57007
[2] Nicolas Bourbaki, General topology. Chapters 14, Elements of Mathematics, Springer-Verlag, Berlin, 1998,
Translated from the French, Reprint of the 1989 English translation. MR 2000h:54001a
[3] Glen E. Bredon, Topology and geometry, Graduate Texts in Mathematics, vol. 139, Springer-Verlag, New York,
1993. MR 94d:55001
[4] John P. Burgess, Book review, Notre Dame J. Formal Logic 44 (2003), no. 3, 227251. MR MR675916
(84a:03007)
[5] Ryszard Engelking, General topology, second ed., Sigma Series in Pure Mathematics, vol. 6, Heldermann Verlag,
Berlin, 1989, Translated from the Polish by the author. MR 91c:54001
[6] Ryszard Engelking and Karol Sieklucki, Topology: a geometric approach, Sigma Series in Pure Mathematics,
vol. 4, Heldermann Verlag, Berlin, 1992, Translated from the Polish original by Adam Ostaszewski. MR
94d:54001
[7] Allen Hatcher, Algebraic topology, Cambridge University Press, Cambridge, 2002. MR 2002k:55001
[8] James R. Munkres, Topology. Second edition, Prentice-Hall Inc., Englewood Cliffs, N.J., 2000. MR 57 #4063
[9] J.-P. Serre, A course in arithmetic, Springer-Verlag, New York, 1973, Translated from the French, Graduate
Texts in Mathematics, No. 7. MR MR0344216 (49 #8956)
[10] Edwin H. Spanier, Algebraic topology, Springer-Verlag, New York, 1981, Corrected reprint. MR 83i:55001
79