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INTRODUCTION TO TOPOLOGY
- Basic concepts
Topology is the area of mathematics which investigates continuity and related concepts. Important fundamental notions soon to come are for example open and closed sets, continuity, homeomorphism.
- In preparation – January 24, ALEX K URONYA¨
- Basic concepts Contents
- Constructing topologies
- 2.1. Subspace topology
- 2.2. Local properties
- 2.3. Product topology
- 2.4. Gluing topologies
- 2.5. Proper maps
- Connectedness
- Separation axioms and the Hausdorff property
- 4.1. More on the Hausdorff property
- Compactness and its relatives
- 5.1. Local compactness and paracompactness
- 5.2. Compactness in metric spaces
- Quotient spaces
- 6.1. Quotient topology
- 6.2. Fibre products and amalgamated sums
- 6.3. Group actions on topological spaces
- Homotopy
- The fundamental group and some applications
- 8.1. Definition and basic properties
- 8.2. Applications
- Covering spaces
- Classification of covering spaces
- References
2 ALEX K URONYA¨
Originally coming from questions in analysis and differential geometry, by now topology permeates mostly every field of math including algebra, combinatorics, logic, and plays a fundamental role in algebraic/arithmetic geometry as we know it today.
Definition 1.1. A topological space is an ordered pair (X, τ ), where X is a set, τ a collection of subsets of X satisfying the following properties
(1) ∅, X ∈ τ , (2) U, V ∈ τ implies U ∩ V , (3) {Uα | α ∈ I} implies ∪α∈I Uα ∈ τ.
The collection τ is called a topology on X, the pair (X, τ ) a topological space. The elements of τ are called open sets. A subset F ⊆ X is called closed, if its complement X − F is open. Although the official notation for a topological space includes the topology τ , this is often suppressed when the topology is clear from the context.
Remark 1.2. A quick induction shows that any finite intersection U 1 ∩ · · · ∩ Uk of open sets is open. It is important to point out that it is in general not true that an arbitrary (infinite) union of open sets would be open, and it is often difficult to decide whether it is so.
Remark 1.3. Being open and closed are not mutually exclusive. In fact, subsets that are both open and closed often exist, and play a special role.
The collection of closed subsets in a topological space determines the topology uniquely, just as the totality of open sets does. Hence, to give a topology on a set, it is enough to provide a collection of subsets satisfying the properties in the exercise below.
Exercise 1.4. Prove the following basic properties of closed sets. If (X, τ ) is a topological space, then
(1) ∅, X are closed sets, (2) if F, G ⊆ X are closed, then so is F ∪ G, (3) if {Fα | α ∈ I} is a collection of closed subsets, then ∩α∈I Fα is closed as well.
Example 1.5 (Discrete topological space). Let X be an arbitrary set, τ def = 2X^ , that is, we declare every subset of X to be open. One checks quickly that (X, τ ) is indeed a topological space.
Example 1.6 (Trivial topology). Considering the other extreme, the pair (X, {∅, X}) is also a topological space, with ∅ and X being the only open subsets.
The previous two examples are easy to understand, however not that important in practice. The primordial example of a very important topological space coming from analysis is the real line. In fact R^1 and its higher-dimensional analogues are the
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(1) For every x ∈ X we have d(x, x) = 0; if d(x, y) = 0 for x, y ∈ X then x = y. (2) (Symmetry) For every x, y ∈ X we have d(x, y) = d(y, x). (3) (Triangle inequality) If x, y, z ∈ X are arbitrary elements, then d(x, y) ≤ d(x, z) + d(z, y).
Exercise 1.12. Show that (Rn, d) with d(x, y) = |x − y| is metric space.
Definition 1.13. Let (X, d) be a metric space. The open ball in X with center x ∈ X and radius > 0 is
B(x, ) def = {y ∈ X | d(y, x) < }.
The topology τd induced by d consists of arbitrary unions of open balls in X.
Remark 1.14. A subset U ⊆ (X, d) is open if and only if for every y ∈ U there exists > 0 (depending on y) such that B(y, ) ⊆ U.
Most of the examples of topological spaces we meet in everyday life are induced by metrics (such topological spaces are called metrizable); however, as we will see, not all topologies arise from metrics.
Example 1.15. Let (X, τ ) be a discrete topological space. Consider the function
d(x, y) =
0 if x = y 1 if x 6 = y
One can see quickly that (X, d) is a metric space, and the topology induced by d is exactly τ.
Exercise 1.16. Let (X, d) be a metric space, and define
d 1 (x, y) def =
d(x, y) if d(x, y) ≤ 1 1 if d(x, y) > 1.
Show that (X, d′) is also a metric space, moreover d, d 1 induce the same topology on X (we could replace 1 by any positive real number).
Example 1.17 (Finite complement topology). Let X be an arbitrary set. The finite complement topology on X has ∅ and all subsets with a finite complement as open sets. Alternatively, the closed subsets with respect to the finite complement topology are X and all finite subsets.
For the next example, we will quickly review what a partially ordered set is. Let X be an arbitrary set, ≤ a relation on X (i.e. ≤⊆ X × X). The relation ≤ is called a partial order, and (X, ≤) a partially ordered set if ≤ is reflexive, antisymmetric, and transitive; that is
(1) (Reflexivity) for every x ∈ X we have x ≤ x (2) (Antisymmetry) x ≤ y and y ≤ x implies x = y
INTRODUCTION TO TOPOLOGY 5
(3) (Transitivity) x ≤ y and y ≤ z implies x ≤ z.
Example 1.18 (Order topology). Let (X, ≤) be a partially ordered set. For an element a ∈ X consider the one-sided intervals {b ∈ X | a < b} and {b ∈ X | b < a}. The order topology τ consists of all finite unions of such.
We turn to a marvellous application of topology to elementary number theory. The example is taken from [Aigner–Ziegler, p.5.].
Example 1.19 (There are infinitely many primes in Z). We will give an almost purely topological proof of the fact that there are infinitely many prime numbers in Z. However, the topology we will use is not the one coming from R, but something a lot more esoteric, coming from doubly infinite arithmetic progressions. Let a, b ∈ Z, b 6 = 0. Denote
Na,b def = {a + bn | n ∈ Z}.
A non-empty subset U ⊆ Z will be designated to be open, if it is a union of sets of the form Na,b. Necessarily, we declare the empty set to be open as well. We need to check that the collection of subsets arising this way indeed forms a topology. Obviously, Z is open, and an arbitrary union of open sets is open as well. We are left with showing that the intersection of two open sets U 1 , U 2 ⊆ Z is open as well. To this end, let a ∈ U 1 ∩ U 2 be arbitrary, let a ∈ Na,b 1 ⊆ U 1 and a ∈ Na,b 2 ⊆ U 2 be arithmetic progressions containing a. Then a ∈ Na,b 1 b 2 ⊆ Na,b 1 ∩ Na,b 2 ⊆ U 1 ∩ U 2 , hence U 1 ∩ U 2 is again a union of two-way arithmetic progressions. An immediate consequence of the definition is that a non-empty open subset must be infinite. Another quick spinoff is the fact that the sets Na,b are not only open, but closed as well:
Na,b = Z \
b⋃− 1
i=
Na+i,b.
Since any integer c ∈ Z \ {− 1 , 1 } has at least one prime divisor p, every such c is contained in some N 0 ,p. Therefore
Z \ {− 1 , 1 } =
p prime
N 0 ,p.
Suppose there are only finitely many primes in Z. Then the right-hand side of the above equality is a finite union of closed subsets, hence itself closed in Z. This would imply that {− 1 , 1 } ⊆ Z is open, which is impossible, as it is finite.
Let us now recall how continuity is defined in calculus. A function f : R → R is called continuous if for every x ∈ R and every > 0 there exists δ > 0 such that
|x − x′| < δ whenever |f (x) − f (x′)| <
INTRODUCTION TO TOPOLOGY 7
Exercise 1.22. Prove the following statements.
(1) Constant functions (ie. f : X → Y with f (X) consisting of exactly one element) are continuous. (2) The identity function f : (X, τ ) → (X, τ ), x 7 → x is continuous for every topological space X. (3) Let X, Y, Z be topological spaces. If the functions f : X → Y and g : Y → Z are continuous, then so is the composition g◦f : X → Z given by (g◦f )(x) def = g(f (x)).
Exercise 1.23. Show that every function from a discrete topological space is con- tinuous. Analogously, verify that every function to a trivial topological space is continuous.
Interestingly enough, our definition of continuity is ’global’ in the sense that no reference is made to individual points of the spaces X and Y. In fact, as opposed to the usual definition of continuity of functions on the real line, it is somewhat more delicate — and is less important in general — to define continuity at a given point of a topological space. In order to find the right notion first we need to pin down what it means to be ’close to a given point’.
Definition 1.24. Let (X, τ ) be a topological space, x ∈ X an arbitrary point. A subset N ⊆ X is a neighbourhood of x if there exists an open set U ⊆ x for which x ∈ U ⊆ N.
Remark 1.25. The terminology in the literature is ambiguous; it is often required that neighbourhoods be open. We call such a neighbourhood an open neighbourhood.
Remark 1.26. The intersection of two neighbourhoods of a given point is also a neighbourhood. If U ⊆ X is an open set, then it is a neighbourhood of any of its points. In particular, X is a neighbourhood of every x ∈ X.
Remark 1.27. In a metric space X, a subset N ⊆ X is a neighbourhood of a point x ∈ X if and only if N contains an open ball centered at x.
Definition 1.28. Let (X, τ ) be a topological space, x ∈ X. A collection Bx ⊆ P (X) of subsets all containing x is called a neighbourhood basis of x if
(1) every element of Bx is a neighbourhood of x; (2) every neighbourhood of x contains an element of Bx as a subset.
Example 1.29. Let X = R^1 be the real line with the Euclidean topology, x = 0. Then (^) {(
−
n
n
| n ∈ N
and (^) {[
−
n
n
]
| n ∈ N
8 ALEX K URONYA¨
are both neighbourhood bases of x. To put it in a more general context, let (X, d) be a metric space with the induced topology, x ∈ X arbitrary. Then the collection { B(x,
n
) | n ∈ N
forms again a neighbourhood basis of x ∈ X.
Example 1.30 (Non-examples). Consider again the case X = R^1 , x = 0. The collections of subsets below are not neighbourhood bases of x: {[ 0 ,
n
| n ∈ N
, {(− 1 , n) | n ∈ N}.
Beside their inherent usefulness neighbourhoods and neighbourhood bases serve the purpose letting us define the continuity of a function at a point.
Definition 1.31 (Continuity of a function at a point). Let f : X → Y be a function between topological spaces, x ∈ X. We say that f is continuous at the point x, if for every neighbourhood N of f (x) in Y there exists a neighbourhood M of x in X such that f (M ) ⊆ N.
Remark 1.32. It is enough to require the condition in the definition for the elements of a neighbourhood basis of f (x). To put it more clearly, let Bf (x) be a neighbour- hood basis of f (x) in Y. Then f as above is continuous at x if and only if for every N ∈ Bf (x) there exists a neighbourhood M of x in X with f (M ) ⊆ N.
Remark 1.33. Observe that for an arbitrary map of sets f : X → Y and a subset A ⊆ Y we have f (f −^1 (A)) = A ∩ f (X) ,
hence f (f −^1 (A)) ⊆ A. Therefore, if f : X → Y is a function between topological spaces, f is continuous at a point x if and only if for every neighbourhood N of f (x) in Y , f −^1 (N ) is a neighbourhood of x ∈ X. Combining this observation with Remark 1.32, f is continuous at x precisely if for any neighbourhood basis Bf (x) of f (x) in Y , the collection { f −^1 (N ) | N ∈ Bf (x)
is a neighbourhood basis of x.
The next result is our first theorem; note that as opposed to calculus, it is no longer the definition of continuity, but rather something we need to prove.
Theorem 1.34. Let f : X → Y be a function between topological spaces. Then f is continuous if and only if it is continuous at x for every x ∈ X
Proof. Assume first that f : X → Y is continuous, that is, the inverse image of every open set in Y under f is open in X. Fix a point x ∈ X; we will show that f is continuous at x. Let N be a neighbourhood of f (x) ∈ Y ; this means that
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In general it is not easy to show that two topological spaces are homeomorphic to each other; however, it can be equally difficult to prove that two topological spaces are not homeomorphic. We will see various methods both simple and hard that help us with such questions. For now, let us get back to our investigation of open sets. As there can be many more open sets than we can easily handle in a random topological space, it is often very useful to come up with a small selection of open subsets that determine the whole topology.
Definition 1.39. Let (X, τ ) be a topological space, B ⊆ P (X). The collection B is called a basis for the topology τ , if the open sets in X are precisely the unions of sets in B. A collection S ⊆ P (X) is called a subbasis for the topology τ if the set B(S) consisting of finite intersections of elements of S forms a basis for τ.
As the exercise below shows, if X is an arbitrary set, then any collection of subsets S ⊆ P (X) is a subbasis for some topology on X.
Exercise 1.40. Let S ⊆ P (X) be an arbitrary set of subsets of X; define τ as the collection of arbitrary unions of finite intersections of elements of S. Prove that τ is a topology on X. Also, show that if τ ′^ is a topology on X such that S ⊆ τ ′, then τ ⊆ τ ′.
The topology defined above is called the topology generated by S. It is the smallest (with respect to inclusion of subsets of P (X)) topology where the elements of S are open. Note that the topology generated by a collection S might contain many more open sets than the elements of S (in fact often it contains many more than one would expect).
Example 1.41. Let X = { 1 , 2 , 3 } be a set with just three elements. We will consider various sets of subsets, and calculate the corresponding generated topologies. First take S = {{ 1 }}. Then the set of finite intersections is B(S) = {X, { 1 }}, hence the topology generated by S (that is, the collection of arbitrary unions of elements of B(S)) is {∅, X, { 1 }}. Next, choose S = {{ 1 , 2 } , { 2 , 3 }}. Then the set of finite intersections is B(S) = {X, { 1 , 2 } , { 2 , 3 } , { 2 }}, hence the generated topology is {∅, X, { 1 , 2 } , { 2 , 3 } , { 2 }}.
Exercise 1.42. Show that S = {{x} | x ∈ X} generates the discrete topology on an arbitrary set X.
Exercise 1.43. How many pairwise non-homeomorphic topologies are there on the set X = { 1 , 2 , 3 }?
Example 1.44. Let X = R^1 , and S =
p q ,^
r s
| p, q, r, s ∈ Z, q, s 6 = 0
. Prove that
S generates the Euclidean topology on R^1.
INTRODUCTION TO TOPOLOGY 11
There are several ways to measure how ’large’ a topological space is. Here is a pair of notions based on the cardinality of sets.
Definition 1.45. A topological space (X, τ ) is called first countable, if every point x ∈ X has a countable neighbourhood basis. The topological space (X, τ ) is second countable, if τ has a countable basis.
Exercise 1.46. Is a discrete topological space first countable? Second countable?
Exercise 1.47. Does second countability imply first countability?
Example 1.48. Euclidean spaces are second countable. The following collection gives a countable basis { B(x,
m
) | x ∈ Qn, m ∈ N
Example 1.49. As evidenced by the collection { B(x,
m
) | m ∈ N
every metric space is first countable. However, not every metric space is second countable. As an example we can take any uncountable set (X = R for instance) with the discrete topology. We have seen earlier that the discrete topology is induced by a metric. In this topology every singleton set {x} is open, hence they need to belong to any basis for the topology; however there are uncountably many such sets.
The next two notions we will only need later; this is nevertheless a good place to introduce them.
Definition 1.50. Let f : X → Y be a function between topological spaces; f is called open if for every open set U ⊆ X the image f (U ) ⊆ Y is open as well.
We can define closed functions in a completely analogous fashion.
Remark 1.51. Note that being open or closed is not the same as being continuous. It is true however that every homeomorphism is open and closed at the same time.
Later we will see examples that show that an open map is not necessarily closed, and vice versa.
Exercise 1.52. Come up with a definition for convergence and Cauchy sequences in metric spaces.
Exercise 1.53. (i) Show that the functions s, p : R^2 −→ R given by
s(x, y) = x + y p(x, y) = xy
INTRODUCTION TO TOPOLOGY 13
If α =
x y
∈ Q×, then we set
ordp(α) = ordp(
x y
def = ordp(x) − ordp(y).
Note that the ordp(α) does not depend on the choice of x and y.
Exercise 1.62. Show that for every x, y ∈ Q
(1) ordp(xy) = ordp(x) + ordp(y) (2) ordp(x + y) ≥ min{ordp(x), ordp(y)} with equality if ordp(x) 6 = ordp(y).
Compute the p-adic order of 5 , 100 , 24 , − 481 , −^1228 for p = 2, 3 , 5.
Definition 1.63. With notation as so far, let α, β ∈ Q. Then we set
dp(α, β) def =
0 if α = β , 1 pordp(α−β)^ otherwise.
This is called the p-adic distance of α and β.
Exercise 1.64. Prove that (Q, dp) is a metric space, which is in addition non- archimedean, that is, for every x, y, z ∈ Q one has
dp(x, y) ≤ max{dp(x, z), dp(z, y)}.
Conclude that in (Q, dp) every triangle is isosceles.
Exercise 1.65. Let f : X → Y be a continuous map. If X is first/second countable, then so is f (X).
Exercise 1.66. Let f : X → Y be a function between topological spaces. Show that f is continuous if and only if f (A) ⊆ f (A)
for every subset A ⊆ X.
Exercise 1.67. If f : X → Y is a continuous surjective open map, then F ⊆ Y is closed exactly if f −^1 (F ) ⊆ X is closed.
- Constructing topologies
2.1. Subspace topology. In this section we will start manufacturing new topolo- gies out of old ones. There are various ways to do this, first we discuss topologies induced on subsets.
Definition 2.1. Let X be a topological space, A ⊆ X an arbitrary subset. The relative or subspace topology on A is the collection of intersections with open sets in X. In other words, a subset U ⊆ A is open in the subspace topology if and only if there exists an open subset V ⊂ X such that U = V ∩ A.
14 ALEX K URONYA¨
Notation 2.2. To facilitate discussion and formalize the above definition, we intro- duce some notation. Let (X, τ ) be a topological space, A ⊆ X a subset. We denote the subspace topology on A by
τA^ def = {A ∩ V | V ∈ τ }.
Remark 2.3. Here is another way of thinking about the subspace topology. Let (X, τ ) be a topological space, A ⊆ X a subset, i : A ↪→ X the inclusion function. Then τA is the smallest topology which makes i continuous.
The following is a related notion, which will play an important role in later de- velopments.
Definition 2.4. A pair is an ordered pair (X, A), with X a topological space, and A ⊆ X an arbitrary subset equipped with the subspace topology. A continuous map of pairs f : (X, A) → (Y, B) is a continuous map f : X → Y for which f (A) ⊆ B.
Example 2.5. Let [0, 1] ⊆ R. Open subsets in R are unions of open intervals. Therefore elements of τA are arbitrary unions of sets of the form [0, 1] ∩ (a, b) with a, b ∈ R. For example [0, 12 ) = [0, 1] ∩ (^12 , 2) is open in [0, 1].
Remark 2.6. A subset U ⊆ A ⊆ X which is open in the subset topology in A will typically not be open in X.
Exercise 2.7. Prove the following statements.
(1) If f : X → Y is a map, A ⊆ X a subspace, then f |A : A −→ Y given by f |A(x) = f (x) whenever x ∈ A, is a continuous function. (2) Let X, Y be topological space, Y 1 ⊆ Y 2 ⊆ Y subspaces, f : X → Y 2 a continuous function. Then f as a function X → Y is also continuous. If f (X) ⊆ Y 1 , then f as a function from X to Y 1 is continuous as well.
Proposition 2.8. Let (X, τ ), (Y, τ ′) be topological spaces, A, B ⊆ X closed subsets such that X = A ∪ B. Assume we are given continuous functions f : (A, τA) → Y , g : (B, τB ) → Y such that f |A∩B = g|A∩B.
Then there exists a unique continuous function h : X → Y for which
h|A = f and h|B = g.
Proof. Set
h(x) def =
f (x) if x ∈ A g(x) if x ∈ B.
This gives a well-defined function h : X → Y by assumption.
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Remark 2.11. Even if A ⊆ X seems naively relatively large (think Q ⊆ R for example) it can happen that int A = ∅. In a similar vein, although Q ⊆ R is in a way smaller, we have Q = R.
Exercise 2.12. Prove that A X = X − intX (X − A).
Proposition 2.13. With notation as above, let Y ⊆ X, B a basis for the topology τ. Then
BY def = {B ∩ Y | B ∈ B}
is a basis of the subspace topology τY. In a similar fashion, if x ∈ Y ⊆ X is an arbitrary point, Bx ⊆ τ a neighbourhood
basis of x, then (Bx)Y def = {N ∩ Y | N ∈ Bx} is a neighbourhood basis of x in τY.
Proof. Left as an exercise.
Definition 2.14. Let again be (X, τ ) a topological space, A ⊆ X a subset. The boundary of A denoted by ∂A is defined as
∂A def = A ∩ X − A.
Exercise 2.15. Show that ∂A = A − intX A.
Definition 2.16. A subset A ⊆ X of a topological space is called dense, if A = X. A subset A ⊆ X is callled nowhere dense, if int A = ∅.
As immediate examples observe that Q ⊆ R is dense, while Z ⊆ R is nowhere dense. Another simple situation is of course the discrete topology: in this case no subset different from X is dense, and no non-empty subset is nowhere dense.
Exercise 2.17. Is there an uncountable nowhere dense set in [0, 1]?
Definition 2.18 (Limit point). Let (X, τ ) be a topological space, A ⊆ X arbitrary. A point x ∈ X is a limit point or accumulation point or cluster point of A, if x ∈ A − {x}. A point x ∈ A is an isolated point of A if there exists an open (in X) neighbour- hood U of x for which U ∩ A = ∅.
A limit point x of A may or may not lie in A.
Remark 2.19. Note that a point x is a limit point of A if and only if every neigh- bourhood of x intersects A in a point different from x.
Exercise 2.20. List all limit points of the following sets: (0, 1] ⊆ R,
n |^ n^ ∈^ N
R,(0, 1) ∪ { 3 } ⊆ R, Q ⊆ R.
It is intuitively plausible that there is a close relation between the closure of a subset and its limit points.
INTRODUCTION TO TOPOLOGY 17
Proposition 2.21. Let (X, τ ) be a topological space, A ⊆ X a subset, denote A′ the set of limit points of A. Then
A = A ∪ A′^.
Proof. First we prove that A ∪ A′^ ⊆ A. The containment A ⊆ A is definitional. Let x ∈ A′. Then every neighbourhood of x intersects A, hence x ∈ A. For the other direction, let x ∈ A − A. As x ∈ A, every open neighbourhood intersects A, as x 6 ∈ A, the intersection point must be a point of A other than x. Therefore x ∈ A′.
Corollary 2.22. A subset A ⊆ X is closed if and only if A contains all of its limit points.
A fundamental and closely related notion is the convergence of sequences. Since in general it behaves rather erraticly and certainly not according to our intuition trained in Euclidean spaces, it is rarely discussed in this generality, in spite of the fact that there is nothing complicated about it.
Definition 2.23. Let (X, τ ) be a topological space, (xn) a sequence of points in X, x ∈ X arbitrary. We say that the sequence xn converges to x if for every neighbourhood B of x there exists a natural number MB such that xn ∈ B whenever n ≥ MB. This fact is denoted by xn → x.
The limit point of a subset of a topological space and the limit of a convergent sequence are different (although admittedly closely related) notions, and one should exercise caution not to confuse them. It is routine to check using the neighbourhood basis of a point consisting of open balls that the above definition is equivalent to the usual one in a metric space. It is very important to point out that in a general topological space the limit of a convergent sequence is not unique. One reason for this is that if U ⊆ X is a minimal open set (i.e. it contains no other non-empty open sets) and x ∈ U is the limit of a sequence (xn), then so is any other element y ∈ U. In extreme cases a sequence may converge to all points of the given topological space X (this happens for example in a trivial topological space, where every sequence of points converges to every point). Worse, continuity can no longer be characterized with the help of convergent sequences.
Example 2.24 (Non-uniqueness of limits of sequences). Let X = { 1 , 2 , 3 }, τ = ∅, { 1 , 2 } , X. It is easy to see using the definition that all sequences in X converge to 3, while sequences with eventually only 1’s and 2’s in them converge to 1,2, and
Exercise 2.25. Let X be a second countable topological space, A ⊆ X an uncount- able set. Verify that uncountably many points of A are limit points of A.
Exercise 2.26. Prove that a subset U ⊆ X is open if and only if ∂U = U \ U.
INTRODUCTION TO TOPOLOGY 19
This means that there exists a closed subset F ′^ ⊆ X with F ′^ ⊇ A ∩ U such that x 6 ∈ F ′. As x belongs to the left-hand side of (1), x ∈ U , hence x ∈ U − (F ∩ U ).
Consider now the subset G def = (X − U ) ∪ (U ∩ F ′) ⊆ X. Being the complement of the open set U − U ∩ F ′^ in X, G is closed, moreover x 6 ∈ G. However, since A = (A ∩ U ) ∪ (A − U ) ⊆ (F ′^ ∩ U ) ∪ (X − U ) = G, we have found a term on the left-hand side of (1) which does not contain x, a contradiction.
Exercise 2.32. Find examples of open subsets in R that are not the interiors of their closures.
Taking this route a bit further, we arrive at the following result, which motivates the local study of topologies. A collection {Uα | α ∈ I} of open subsets of X whose union is X is called an open cover of X.
Proposition 2.33. Let X be a topological space, A ⊆ X an arbitrary subspace, (Uα | α ∈ I) an open cover of X. Then A is closed in X if and only if A ∩ Uα is open in Uα for every α ∈ I.
Proof. The set A is closed in X if and only if X − A is open in A. Since all the Uα’s are open in X, X − A is open in X if and only if Uα ∩ (X − A) is open in Uα for every α ∈ I.
We say that being closed is a local property, as it can be tested on some (any) open cover of X. Note that it is very important that we take a cover of X, and not one of A.
Definition 2.34. A subset A ⊆ X is called locally closed in X, if every a ∈ A has an open neighbourhood Ua ∈ X such that A ∩ Ua is closed in Ua.
Example 2.35. The subset (− 1 , 1) × { 0 } ⊆ R^2 is locally closed.
Note that every open set is locally closed in any topological space by definition. Luckily it turns out that the structure of locally closed subsets is actually simpler than one might guess.
Proposition 2.36. A subspace A ⊆ X of a topological space is locally closed if and only if A = F ∩ U , where F ⊆ X is closed, and U ⊆ X is open.
Proof. Assume first that A has the shape A = F ∩ U , with F closed, and U open in
X. Then we can verify that A is locally closed immediately by taking Ua def = U for all a ∈ A in the definition of local closedness. Conversely, let A be locally closed, and (Ua | a ∈ A) an collection of suitable open
subsets of X. Set U def = ∪a∈AUa. Then A ⊆ U , and the fact that being closed is a local property does the trick.
Proposition 2.37. Let f : X → Y be a function between topological spaces, U ⊆ τX a collection, such that the union of some of its elements equals X. Then f is continuous if and only if f |Uα : Uα → Y is continuous for every Uα ∈ U.
20 ALEX K URONYA¨
Proof. To come.
2.3. Product topology. We are looking for a way to put a topology on the Carte- sian product X × Y of two sets that is in a way ’natural’. The product of the two sets does not come alone, but with two projection functions πX : X × Y → X, πX (x, y) = x, and πY : X × Y → Y , πY (x, y) = y.
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We will require that both projection functions prX and prY be continuous. As we will see, this defines a unique topology on the Cartesian product, the minimal (containing the least number of open sets) for which the continuity of the projections holds. Let us have a look, which open sets are needed. Let U ⊆ X be an open set. Then π− X 1 (U ) = U × Y ⊆ X × Y
has to be open, and analogously for open subsets V ⊆ Y. Moreover, since
π− X^1 (U ) ∩ π− Y 1 (V ) = (U × Y ) ∩ (X × V ) = U × V ,
all subsets of the form U × V are open when U ⊆ X and V ⊆ Y are open. It is quickly checked that the intersection of such sets is again of the same form, and hence form the basis of a topology.
Definition 2.38. Let X, Y be topological spaces. The product topology on the set X ×Y consists of arbitrary unions of subsets of the form U ×V , with U ⊆ X,V ⊆ Y open.
It is important to point out that not all subsets of the product topology are products of open sets. The same definition holds for the product of finitely many topological spaces. All our results for two topological spaces will hold for arbitrary finite products. However, we will mostly content ourselves with the case of two spaces for ease of notation.
Remark 2.39. We can define analogously the product of infinitely many topologi- cal spaces. If {Xα | α ∈ I} is a collection of topological spaces, then the product topology on ×α∈I Xα is given by the basis of open sets
×α∈I Uα
where Uα ⊆ Xα are open for all α ∈ I, and Uα = Xα for all but finitely many α’s.
Lemma 2.40. Let (X, τX ) and (Y, τY ) be topological spaces, U and V bases for τX and τY , respectively. Then the collection of sets
W def = {S × T | S ∈ U, T ∈ V}
