*Room 3.13 in Herschel Building

MAS3706: Topology Dr. Zinaida Lykova School of Mathematics, Statistics and Physics Newcastle University *Room 3.13 in Herschel Building

These lectures concern metric and topological spaces and continuous maps between topological spaces. We shall define these terms and study their properties. This material is a part of point set topology, also called general topology. You can consult the following books. 1. W. A. Sutherland, Introduction to Metric and Topological Spaces, Clarendon Press, Oxford, 1975. 2. J. R. Munkres, Topology, Prentice Hall, New York, 2000. Topology is used in applications, for example, in many parts of Pure Mathematics, in differential equations, in dynamical systems, in mechanics, in physics and in mathematical economics. The web page: http://www.mas.ncl.ac.uk/ nzal/teaching.html 2

Teaching Information - MAS3706 Venue: Tuesday 11.00 12.00, PERB.G.05 Thursday 16.00 17.00, HERB.G.LT2 Thursday 17.00 18.00, HERB.G.LT2 Drop-in Sessions: Thursday 17.00 18.00, HERB.G.LT2 14.02.2019, 7.03.2019, 21.03.2019, 9.05.2019 Office hours: Tuesday 12.00 13.00, HERB.3 Room 3.13 Thursday 13.00 14.00, HERB.3 Room 3.13 1. Assessment of course work (10%). Assessment: Assignments: Assignments are to be handed in by 10am on Friday: 15.02.2019, 8.03.2019, 22.03.2019, 10.05.2019. 2. 2 hours examination at end of Semester (90%). Past Exam Papers: Exam Papers Online : https://crypt.ncl.ac.uk/exam.papers/ Module code: MAS3706 and MAS3209 Title: Topology 3

MAS3706 Lectures, Tutorials and Drop-in Sessions with dates: Week 1, 29th January Tuesday 11.00 12.00, Lecture, PERB.G.05 31st January Thursday 16.00 17.00, Lecture, HERB.G.LT2 31st January Thursday 17.00 18.00, Lecture, HERB.G.LT2 Week 2, 5th February Tuesday 11.00 12.00, Lecture, PERB.G.05 7th February Thursday 16.00 17.00, Lecture, HERB.G.LT2 7th February Thursday 17.00 18.00, Tutorial 1, HERB.G.LT2 Week 3, 12th February Tuesday 11.00 12.00, Lecture, PERB.G.05 14th February Thursday 16.00 17.00, Lecture, HERB.G.LT2 14th February Thursday 17.00 18.00, Drop-in, HERB.G.LT2 Week 4, 19th February Tuesday 11.00 12.00, Lecture, PERB.G.05 21th February Thursday 16.00 17.00, Lecture, HERB.G.LT2 22th February Thursday 17.00 18.00, Lecture, HERB.G.LT2 Week 5, 26th February Tuesday 11.00 12.00, Lecture, PERB.G.05 28th February Thursday 16.00 17.00, Lecture, HERB.G.LT2 28th February Thursday 17.00 18.00, Tutorial 2, HERB.G.LT2 Week 6, 5th March Tuesday 11.00 12.00, Lecture, PERB.G.05 7th March Thursday 16.00 17.00, Lecture, HERB.G.LT2 7th March Thursday 17.00 18.00, Drop-in, HERB.G.LT2 Week 7, 12th March Tuesday 11.00 12.00, Lecture, PERB.G.05 14th March Thursday 16.00 17.00, Lecture, HERB.G.LT2 14th March Thursday 17.00 18.00, Tutorial 3, HERB.G.LT2 Week 8, 19th March Tuesday 11.00 12.00, Lecture, PERB.G.05 21st March Thursday 16.00 17.00, Lecture, HERB.G.LT2 21st March Thursday 17.00 18.00, Drop-in, HERB.G.LT2 Week 9, 26th March Tuesday 11.00 12.00, Lecture, PERB.G.05 28th March Thursday 16.00 17.00, Lecture, HERB.G.LT2 28th March Thursday 17.00 18.00, Lecture, HERB.G.LT2 Week 10, 30th April Tuesday 11.00 12.00, Lecture, PERB.G.05 2nd May Thursday 16.00 17.00, Lecture, HERB.G.LT2 2nd May Thursday 17.00 18.00, Tutorial 4, HERB.G.LT2 Week 11, 7th May Tuesday 11.00 12.00, Lecture, PERB.G.05 9th May Thursday 16.00 17.00, Revision Lecture, HERB.G.LT2 9th May Thursday 17.00 18.00, Drop-in, HERB.G.LT2 Week 12 - Revision week, 14th May Tuesday 11.00 12.00, Revision Lecture, PERB.G.05 16th May Thursday 16.00 17.00, Revision Lecture, HERB.G.LT2 16th May Thursday 17.00 18.00, Drop-in, HERB.G.LT2 4

Contents 1. Open Subsets of R and Continuous Functions Page 6 2. Metric Spaces Page 13 3. Topological Spaces Page 27 4. Hausdorff Spaces and Limits Page 35 5. Closed Sets Page 39 6. Separation axioms Page 43 7. Basic Topological Concepts Page 45 8. Compactness Page 51 9. Continuity Page 56 10. Metric Spaces Again Page 60 11. Completeness in Metric Spaces Page 62 12. Connectedness Page 68 13. Picard s Theorem Page 70 5

1 Open Subsets of R and Continuous Functions Definition 1.1. A subset U of R is open if around each point p of U there is an open interval (p h, p + h) wholly contained in U: (p h, p + h) U. Remark 1.2. You should think of an open set as containing none of the points at the edge of the set. Example 1.3. Every open interval (a, b) is an open set in R. Proof. If p (a, b), let h be the distance from p to the nearest endpoint: h = min(b p, p a). Then (p h, p + h) (a, b). Example 1.4. R is open (obviously). Example 1.5. The empty set is vacuously open. Example 1.6. The closed interval [a, b] is not open in R. Proof. If you take p to be a then every open interval (p h, p + h) contains points outside [a, b]. Example 1.7. A singleton is not open in R. Proof. Every singleton is a closed interval since {a} = [a, a] so no singleton is open. 6

The classic ɛ δ definition of continuity is the following. Definition 1.8. Let S be an open subset of R. A function f : S R is continuous on S for each a S and each ɛ > 0 there is a δ > 0 such that x a < δ = f(x) f(a) < ɛ. Remark 1.9. Speaking very loosely, f is continuous at a if f(x) is close to f(a) when x is close to a. Working with this definition is not easy: you spend a lot of time juggling inequalities. Fortunately, there is an equivalent definition of continuity, couched in the language of sets rather than numbers, which is much easier to work with. More importantly, the use of set-theoretic language allows us to port this definition to many other spaces, including those whose points are not numbers. The study of such spaces is called topology. Definition 1.10. Let f : A B be a function. If T is a subset of B then the inverse image or pre-image of T under f is the subset of A consisting of all the elements of A that f maps into T : f 1 (T ) = {a A : f(a) T }. In other words, a f 1 (T ) f(a) T. Example 1.11. If f : R R is defined by f(x) = x 2 then f 1 ([0, 9]) = [ 3, 3], f 1 ([ 5, 4]) = (the empty set) f 1 (R) = R. Note that the use of the notation f 1 does not imply that the function f possesses an inverse. The inverse image notation can be used with any function whatsoever. 7

Theorem 1.12. Let S be an open subset of R. A function f : S R is continuous the inverse image f 1 (U) under f of each open set U is open. That is, f is continuous for every open subset U in R, f 1 (U) is open. Proof. [= ] Suppose f is continuous and let U be open. It is not necessary that each point of U be the image of a point of S. Indeed, f 1 (U) may very well be empty. Recall that the empty set is open. If a f 1 (U) then f(a) U. But U is open, so (f(a) ɛ, f(a) + ɛ) U for some ɛ > 0. By continuity, there exists δ > 0 such that Thus Therefore f 1 (U) is open. x a < δ = f(x) f(a) < ɛ. (a δ, a + δ) f 1 ((f(a) ɛ, f(a) + ɛ)) f 1 (U). [ =] Now suppose that the inverse image of each open set is open. Let a S and let ɛ > 0. We have to prove that there is δ > 0 such that x a < δ = f(x) f(a) < ɛ. The interval (f(a) ɛ, f(a) + ɛ) is open. Therefore its inverse image f 1 ((f(a) ɛ, f(a) + ɛ)) is open, and contains a, so, by the definition of open set, it contains an interval (a δ, a + δ) about a for some δ > 0. Thus x a < δ = x (a δ, a + δ) = x f 1 ((f(a) ɛ, f(a) + ɛ)) f(x) (f(a) ɛ, f(a) + ɛ) f(x) f(a) < ɛ. Therefore f is continuous at a. 8

Remark 1.13. If f is not continuous one can find an open subset U in the codomain of f such that the inverse image f 1 (U) of U under f is not open. Example 1.14. If f : R R is defined by f(x) = 3 for x 1 and f(x) = x 2 for x > 1, then, for U = ( 4, 0), the inverse image of U under f is not open. f 1 (U) = (, 1] Thus, instead of using the ɛ δ definition, we could define a continuous function to be one for which the inverse image of each open set is open. We have claimed that this definition is easier to work with. Here is a sample. The composition of two functions f : A B and g : B C is the function (g f) : A C defined by (g f)(a) = g(f(a)), a A. Theorem 1.15. Let and be continuous. Then so is f : R R g : R R g f : R R. Proof. U R open = g 1 (U) open, by continuity of g, = f 1 (g 1 (U)) open, by continuity of f, i.e., (g f) 1 (U) is open. 9

Definition 1.16. Let S be an open subset of R. A function f : S R is called open if the image of every open subset V of S is open in R. Remark 1.17. Note that not every continuous map is open. For example, a constant map f(x) = 7 is not an open map. Remark 1.18. If we are to generalise the open set definition of continuity to spaces other than R, we need to isolate the factors that make it work in R. It turns out that, rather than any specific properties of individual open sets, it is the way that open sets can be combined that is important. These are summarised in the following theorem. We use the notation (U λ ) λ Λ to denote a family of sets which is indexed by the members of a set Λ. In many cases, Λ will be N, the set of positive integers, and then the sets in the family can be listed as U 1, U 2, U 3,.... The notation allows Λ to be uncountable, for we shall sometimes need to consider uncountable families. If (U λ ) λ Λ is a family of subsets of a set X, then their union is the subset of X U λ = λ Λ {x X : x U λ for at least one λ Λ} and their intersection is the subset of X U λ = {x X : x U λ for every λ Λ} λ Λ If Λ = N then we write n=1 U n and n=1 U n. 10

Theorem 1.19. Let τ be the collection of all open subsets of R. Then 1. τ contains R and the empty set ; 2. τ is closed under arbitrary unions: U λ τ for all λ Λ = U λ τ; λ Λ 3. τ is closed under finite intersections: U 1, U 2,..., U n τ = U 1 U 2... U n τ. Proof. 1. R is open, since if p R then for every h > 0. The empty set is open. (p h, p + h) R 2. Let U λ be open for all λ Λ and let p λ Λ U λ. Then p U λ0 for some λ 0 Λ. By assumption, U λ0 is open and, for some h > 0, we have (p h, p + h) U λ0 λ Λ U λ. Therefore λ Λ U λ is open. 3. Let U 1, U 2,..., U n be open and let p U 1 U 2... U n. Then, for each i: 1 i n, there exists h i > 0 such that (p h i, p + h i ) U i. Let Then for all 1 i n, so h = min(h 1, h 2,..., h n ) > 0. (p h, p + h) (p h i, p + h i ) U i Thus U 1 U 2... U n is open. (p h, p + h) U 1 U 2... U n. 11

Note the restriction in 3 to finite intersections of open sets. This is because the intersection of an infinite family of open sets need not be open. Example 1.20. The intersection of the infinite family of open intervals ( 1 n, 1 ) (n = 1, 2, 3,...) n is the singleton {0}, and no singleton is open. Example 1.21. If a, b R, then (, a) = (a n, a) n=1 and (b, ) = (b, b + n) are unions of open sets, and so are open. n=1 12

2 Metric Spaces Definition 2.1. A metric on a set X is a function d : X X R, (x, y) d(x, y), which assigns a distance d(x, y) between each pair of points x, y X, satisfying the following three axioms: [M1] For all x, y X, d(x, y) > 0 if x y and d(x, x) = 0; [M2] d(x, y) = d(y, x) for all x, y X. [M3] d(x, z) d(x, y) + d(y, z) for all x, y, z X. The latter is called the triangle inequality. Remark 2.2. [M1] says that each point is distant 0 from itself, but distinct points are at positive distance apart. [M2] says that the distance from x to y is the same as the distance from y to x. [M3] gets its name from the fact that the length of any side of a triangle is less than the sum of the lengths of the other two sides. It says that a journey from x to z doesn t get any shorter if you have to take a detour via y, but may possible get longer. The axioms [M1-[M3] have been chosen so as to give metrics just enough properties to define a topology. Definition 2.3. X. A metric space is a pair (X, d), where X is a set and d is a metric on In practice, we usually refer to the metric space X. Examples of Metric Spaces In most cases, verification of [M1] and [M2] is trivial, and we need only check [M3]. Example 2.4. The real line R with the usual metric d(x, y) = x y is a metric space. 13

Proof. Let us verify [M1]-[M3]. [M1] For all x, y R, d(x, y) = x y > 0 if x y and d(x, x) = x x = 0. [M2] For all x, y R, [M3] For all x, y, z R, d(x, y) = x y = y x = d(y, x). d(x, z) = x z = (x y) + (y z) x y + y z = d(x, y) + d(y, z). Therefore d is a metric and (R, d) is a metric space. Example 2.5. The complex plane C with the usual metric d(z 1, z 2 ) = z 1 z 2 is a metric space. The verification of [M1]-[M3] is the same as in the previous example. Example 2.6. The plane R 2 = {a = (a 1, a 2 ) : a 1, a 2 R} with the taxicab metric is a metric space. d 1 ((a 1, a 2 ), (b 1, b 2 )) = a 1 b 1 + a 2 b 2 Proof. Let us verify [M1]-[M3]. [M1] For all a = (a 1, a 2 ), b = (b 1, b 2 ) R 2, d 1 (a, b) = a 1 b 1 + a 2 b 2 > 0 if a b and d 1 (a, a) = d 1 ((a 1, a 2 ), (a 1, a 2 )) 14

= a 1 a 1 + a 2 a 2 = 0. [M2] For all a = (a 1, a 2 ), b = (b 1, b 2 ) R 2, d 1 (a, b) = d 1 ((a 1, a 2 ), (b 1, b 2 )) = a 1 b 1 + a 2 b 2 = b 1 a 1 + b 2 a 2 = d 1 ((b 1, b 2 ), (a 1, a 2 )) = d 1 (b, a). [M3] For all a = (a 1, a 2 ), b = (b 1, b 2 ), c = (c 1, c 2 ) R 2, observe that d 1 ((a 1, a 2 ), (b 1, b 2 )) = a 1 b 1 + a 2 b 2 = (a 1 c 1 ) + (c 1 b 1 ) + (a 2 c 2 ) + (c 2 b 2 a 1 c 1 + c 1 b 1 + a 2 c 2 + c 2 b 1 = d 1 ((a 1, a 2 ), (c 1, c 2 )) + d 1 ((c 1, c 2 ), (b 1, b 2 )). Therefore d 1 is a metric and (R 2, d 1 ) is a metric space. Example 2.7. Euclidean n-space with the euclidean metric R n = {a = (a 1, a 2,..., a n ) : a 1,..., a n R} d 2 ((a 1, a 2,..., a n ), (b 1, b 2,..., b n )) is a metric space. = (a 1 b 1 ) 2 + + (a n b n ) 2 Example 2.8. Let X be the set of points on the surface of a sphere, and let the distance between two points be the length of the shorter of the two great circle arcs joining the two points. Then (X, d) is a metric space. This is the metric used in measuring distances on the Earth s surface, e.g., when we are told that Sydney is 10,840 miles from London 1. Example 2.9. The complex plane C with the French Railway metric 2 : 1 The distance between Sydney and London is less than 8,000 miles if we use the metric in the previous example. 2 Folklore suggests that the shortest rail journey between any two French towns is via Paris. 15

d f (z 1, z 2 ) = { 0 z 1 = z 2 z 1 + z 2 z 1 z 2 is a metric space. Example 2.10. We can define at least one metric, the discrete metric, on any set X: that is, distinct points are distance 1 apart. d(x, x) = 0 and d(x, y) = 1 if x y, Proof. Verification of [M1] and [M2] is trivial, and we need only check [M3]: for all x, y, z X, d(x, z) d(x, y) + d(y, z). If the LHS is 0, then there is nothing to prove. Otherwise LHS = 1 so x z. But then y x or y z; in either case the RHS is at least 1, and so LHS RHS. Example 2.11. Let C[0, 1] be the set of all continuous complex-valued functions defined on the closed interval [0, 1], and let and d (f, g) = sup{ f(x) g(x) : 0 x 1}, d 1 (f, g) = 1 Then d and d 1 are metrics on C[0, 1] and (C[0, 1], d ) and (C[0, 1], d 1 ) are metric spaces. 0 f(x) g(x) dx. Proof. We shall prove that d 1 is a metric. Let us verify [M1]-[M3]. [M1] For all f, g C[0, 1], by properties of integrals, d 1 (f, g) = 1 f(x) g(x) dx > 0 if f g 0 and d 1 (f, f) = 1 f(x) f(x) dx = 1 0dx = 0. 0 0 16

[M2] For all f, g C[0, 1], d 1 (f, g) = = 1 0 1 0 f(x) g(x) dx g(x) f(x) dx = d 1 (g, f). [M3] For all f, g, h C[0, 1] and all x [0, 1], we have f(x) g(x) f(x) h(x) + h(x) g(x). Thus, for all f, g, h C[0, 1], (by properties of integrals) = 1 d 1 (f, g) = 1 0 1 0 f(x) g(x) dx f(x) h(x) + h(x) g(x) dx f(x) h(x) dx + 1 0 0 h(x) g(x) dx (by the linearity of integrals) = d 1 (f, h) + d 1 (h, g). Therefore d 1 is a metric and (C[0, 1], d 1 ) is a metric space. Open Balls in Metric Spaces Our next task is to show how a metric gives rise to a topology. Definition 2.12. Let (X, d) be a metric space, let a X, and let r R, r > 0. The open ball with centre a and radius r is the set B(a, r) = {x X : d(a, x) < r} consisting of all points of X whose distance from a is less than r. 17

The motivating example here comes from Euclidean 3-space, where an open ball is the interior of a sphere. The following examples show that open balls in other metric spaces can look quite different. Example 2.13. Consider the metric space (R, d) from Example 2.4. An open ball B(a, r) in the real line R is an open interval (a r, a + r). B(a, r) = {x R : d(a, x) < r} = {x R : x a < r} Example 2.14. Consider the metric space (C, d) from Example 2.5. For a C and r > 0, the open ball B(a, r) in the complex plane C B(a, r) = {z C : d(a, z) < r} = {z C : z a < r} is the disc consisting of all points interior to the circle with centre a and radius r. 18

Example 2.15. Consider the metric space (R 3, d 2 ) from Example 2.7. For a = (a 1, a 2, a 3 ) R 3 and r > 0, the open ball in Euclidean 3-space really does look like a ball. B(a, r) = {x R 3 : d 2 (a, x) < r} = {x = (x 1, x 2, x 3 ) R 3 : (x1 a 1 ) 2 + (x 2 a 2 ) 2 + (x 3 a 3 ) 2 < r} Example 2.16. Consider the metric space (R 2, d 1 ) with the taxicab metric d 1 from Example 2.6. For a = (a 1, a 2 ) R 2 and r > 0, the open ball B(a, r) = {b R 2 : d 1 (a, b) < r} = {b = (b 1, b 2 ) R 2 : b 1 a 1 + b 2 a 2 < r}. For 0 = (0, 0) R 2 and r > 0, the open ball B(0, r) = {(x, y) R 2 : d 1 ((0, 0), (x, y)) < r} = {(x, y) R 2 : x 0 + y 0 < r} consists of all points interior to the square enclosed by the lines x + y = r, x y = r, y x = r, y x = r. 19

Example 2.17. Consider the complex plane C with the French Railway metric d f from Example 2.9. For a C and r > 0, the open ball B(a, r) = {z C : d f (a, z) < r}. Recall that d f (a, z) = { 0 if z = a a + z if z a. Thus the open ball B(a, r) = {a} if r < a, and B(a, r) = {a} {z C : z < r a } if r a. 20

Example 2.18. Consider the metric space (X, d) with the discrete metric d from Example 2.10. For a X and r > 0, the open ball B(a, r) = {z X : d(a, z) < r}. Recall that Therefore d(x, x) = 0 and d(x, y) = 1 if x y. B(a, r) = { {a} if r 1 X if r > 1. Remark 2.19. Open balls need not be spherical! Open Sets in Metric Spaces We shall now use open balls to define a topology on any metric space. The method is the one we used to define the usual topology on the real line. Definition 2.20. A subset U of a metric space (X, d) is open if and only if for each u U there is a positive real number r such that B(u, r) U. Thus an open set in a metric space is one which contains an open ball about each of its points. In the case of R with the usual metric, this is just the definition of open set given in 1.1. In fact, the following theorem, and its proof, are modelled on Theorem 1.19. Theorem 2.21. Let (X, d) be a metric space and let τ be the collection of all open subsets of X. Then 1. τ contains X and the empty set ; 2. τ is closed under arbitrary unions: U λ τ for all λ Λ = λ Λ U λ τ; 3. τ is closed under finite intersections: U 1, U 2,..., U n τ = U 1 U 2... U n τ. 21

The open sets in a metric space (X, d) form a topology which is called the metric topology. Proof. 1. X is obviously open, and is open by the vacuous truth principle. 2. Let U λ be open for all λ Λ and let x λ Λ U λ. Then x U λ0 for some λ 0 Λ. By assumption, U λ0 is open and, for some r > 0, we have B(x, r) U λ0. Thus B(x, r) U λ0 λ Λ U λ. Therefore λ Λ U λ is open. 3. Let U 1, U 2,..., U n be open and let x U 1 U 2... U n. By assumption, for each i: 1 i n, U i is open. Therefore, there exists r i > 0 such that B(x, r i ) U i. Let Then Hence Therefore U 1 U 2... U n is open. r = min(r 1, r 2,..., r n ) > 0. B(x, r) B(x, r 1 ) U 1, B(x, r) B(x, r 2 ) U 2,... B(x, r) B(x, r n ) U n. B(x, r) U 1 U 2... U n. 22

In the case of R with the usual metric, B(u, r) = (u r, u + r), so the usual metric induces the usual topology. Open Balls Are Open Sets We can now justify the name open ball. Theorem 2.22. Every open ball B(a, r) in a metric space (X, d) is an open set. Proof. Let u B(a, r) and let r = r d(a, u) > 0. For all x B(u, r ), d(x, u) < r = d(x, a) d(x, u) + d(u, a) < r + d(u, a) = r. Then B(u, r ) B(a, r), so B(a, r) is open. Corollary 2.23. A subset of a metric space is open it is a union of open balls. Proof. = If U is open then for each u U there is an open ball B(u, r u ) U. It follows that U = B(u, r u ), u U a union of open balls. = Any union of open balls is open by Theorem 2.21(2). Example 2.24. Consider the metric space (C n, d 1 ), where d 1 ( v, w) = n v j w j, j=1 23

for v = (v 1,..., v n ), w = (w 1,..., w n ) C n. Give the definition of an open ball in the metric space (C n, d 1 ). For a C n and r > 0, the open ball B( a, r) = { v C n : d 1 ( a, v) < r} = { v = (v 1,..., v n ) C n : n v j a j < r}. j=1 Are the following subsets of the metric space (C n, d 1 ) open? (i) where i 2 = 1; S 1 = { v C n : n v j (1 + ji) < 5}, j=1 (ii) Justify your answer. S 2 = 25 p=1 { v C n : n v j 1 < 2p}. j=1 Proof. (i) The set S 1 = { v C n : is the open ball B( a, 5) with centre n v j (1 + ji) < 5} j=1 a = (1 + i, 1 + 2i,..., 1 + ni) and radius r = 5 in the metric space (C n, d 1 ) and therefore, by Theorem 2.22, it is open. (ii) The set S 2 = 25 p=1 { v C n : n v j 1 < 2p} j=1 is the union of open balls B( b, 2p), p = 1, 2,..., 25, with centre b = (1, 1,..., 1) and radius r = 2p in (C n, d 1 ) and therefore, by Theorem 2.21, it is open. 24

Example 2.25. Consider the metric space (C[0, 1], d ), where C[0, 1] is the set of all continuous complex-valued functions defined on the closed interval [0, 1], and d (f, g) = sup{ f(x) g(x) : 0 x 1}. Are the following subsets of the metric space (C[0, 1], d ) open? and (i) S 1 = {f C[0, 1] : sup f(t) exp(t) < 10} 0 t 1 (ii) S 2 = Justify your answer. {f C[0, 1] : sup f(t) t 2 < 1 p }. p=1 0 t 1 Proof. (i) The set S 1 = {f C[0, 1] : sup f(t) exp(t) < 10} 0 t 1 is the open ball B(g, 10) with centre g(t) = exp(t), t [0, 1], and radius r = 10 in the metric space (C[0, 1], d ) and therefore, by Theorem 2.22, it is open. (ii) Note that for all p N g {f C[0, 1] : sup f(t) t 2 < 1 p } p=1 g {f C[0, 1] : 0 t 1 sup g(t) t 2 < 1 0 t 1 p sup f(t) t 2 < 1 0 t 1 p } sup g(t) t 2 = 0 0 t 1 for all p N 25

g(t) = t 2, t [0, 1]. Therefore S 2 = {g} is a singleton and is not open in the metric space (C[0, 1], d ), because every open ball B(g, r) in (C[0, 1], d ) contains also all continuous functions h such that h(t) = g(t) + p, where p C : p < r. 26

3 Topological Spaces Definition 3.1. such that Let X be a set. Then a topology on X is a family τ of subsets of X [T1] τ includes X and ; [T2] τ is closed under arbitrary unions: U λ τ for all λ Λ = U λ τ; λ Λ [T3] τ is closed under finite intersections: U 1, U 2,..., U n τ = U 1 U 2... U n τ. Remark 3.2. A topology on X is a collection of subsets of X which contains X and and is closed under arbitrary unions and finite intersections. Here the phrase arbitrary unions allows the index set Λ in [T2] to be any set whatsoever, no matter how large. By contrast [T3] allows only intersections of finitely many sets U 1, U 2,..., U n. Definition 3.3. A topological space is a pair (X, τ), where X is a set and τ is a topology on X. It is customary to refer to the topological space X when it is obvious which topology τ is involved. The elements of X are usually called points. The sets which make up the topology τ are called the τ-open sets, or, where no confusion arises, simply the open sets 3. Note that [T3] has the alternative formulation: [T3 ] : U 1, U 2 τ = U 1 U 2 τ. since closure under finite intersections follows by repetition. This alternative version is easier to use when we wish to verify that a collection of sets is a topology. 3 In the early days of topology, there were several competing sets of topology axioms (all more or less equivalent). The eventual triumph of [T1]-[T3] and the adoption of open set as the fundamental concept were largely due to the influence of the mythical French mathematician Nicolas Bourbaki. 27

Examples of Topological Spaces Example 3.4. Let X = R. It is proved in Theorem 1.19 that the open subsets of R, as defined in 1.1, form a topology τ d on R. This topology is called the usual topology on R. From now on, unless stated otherwise, R will always be assumed to have the usual topology. Example 3.5. Consider a metric space (X, d). It is proved in Theorem 2.21 that the open subsets of X, as defined in 2.20, form a topology τ d on X, called the metric topology on (X, d). Example 3.6. Let X be any set, and let τ be the collection of all subsets of X: Clearly τ satisfies [T1]-[T3]: [T1] τ includes X and. τ = {S X}. [T2] τ is closed under arbitrary unions, since any union of subsets of X is again is a subset of X. [T3] τ is closed under finite intersections, since any intersection of subsets of X is again is a subset of X. Therefore τ is a topology. We call this the discrete topology on X. In the discrete topology, every set is an open set. Example 3.7. Let X be any set, and let τ = {X, }. Then it is easy to see that τ is a topology on X. [T1] τ includes X and. 28

[T2] τ is closed under arbitrary unions, since X = X τ. [T3] τ is closed under finite intersections, since X = τ. It is called the indiscrete topology. Example 3.8. Let X be any nonempty set, so X has at least one member p. Let τ p consist of the empty set plus all subsets of X which have p as a member: τ p = { } {S X : p S}. Then τ p is a topology on X. It is called a particular point topology. Example 3.9. Another topology on R. Let τ s.i. consist of R,, and all the semi-infinite open intervals (a, ), a R. Then τ s.i. is a topology on R. Proof. [T1]: τ s.i. includes R and. [T2]: Let (a λ, ), λ Λ, be a family of members of τ s.i.. Then (i) if the a λ, λ Λ are bounded below, their union is (inf a λ, ) τ s.i. ; (ii) if they are not bounded below, their union is R τ s.i.. [T3] n (a i, ) = (max(a 1, a 2,..., a n ), ) τ s.i.. i=1 Remark 3.10. Note that every semi-infinite open interval (a, ) is an open subset of the topological space R with the usual topology, so τ s.i. τ d. On the other hand, the finite open interval (a, b) is not in τ s.i.. Thus two topological spaces (R, τ d ) and (R, τ s.i. ) are different. 29

Equivalent Metrics Definition 3.11. Two metrics d 1 and d 2 on a set X are topologically equivalent if they generate the same topology on X, that is, a subset U of X is d 1 -open if and only if it is d 2 -open. Remark 3.12. This implies that open sets are the same in both cases, as are the closed sets, the compact sets, the connected sets and the continuous functions which we will consider later. Theorem 3.13. Suppose that there are strictly positive real numbers K 1 and K 2 such that, for every x, y X, d 1 (x, y) K 2 d 2 (x, y) and d 2 (x, y) K 1 d 1 (x, y). Then the metrics d 1 and d 2 are topologically equivalent. Proof. Suppose that U X is open in the topology induced by d 1. We claim that U is open in the topology induced by d 2. Let u U. By assumption, there exists ɛ > 0 such that Then, for all x X such that d 2 (x, u) < Thus, for r = ɛ K 2, B 1 (u, ɛ) = {x X : d 1 (u, x) < ɛ} U. ɛ K 2, we have d 1 (x, u) K 2 d 2 (x, u) < K 2 ɛ K 2 = ɛ. B 2 (u, r) = {x X : d 2 (u, x) < r} B 1 (u, ɛ) U. 30

Therefore U is open with respect to d 2. The same proof works for the following statement: if U X is open in the topology induced by d 2, then U is open in the topology induced by d 1. Therefore d 1 and d 2 generate the same topology on X. Example 3.14. Consider the following metrics on R n : d 1 (x, y) = Σ n i=1 x i y i, and d 2 (x, y) = Σ n i=1 x i y i 2 d (x, y) = sup{ x i y i : 1 i n} for x = (x 1, x 2,..., x n ) and y = (y 1, y 2,..., y n ) from R n. We shall show that d 1, d 2 and d are topologically equivalent. Proof. 1. For all x = (x 1, x 2,..., x n ) and y = (y 1, y 2,..., y n ) R n, we have and (i) d (x, y) = max{ x i y i : 1 i n} Σ n i=1 x i y i = d 1 (x, y) (ii) d 1 (x, y) = Σ n i=1 x i y i Σ n i=1 max{ x i y i : 1 i n} = Σ n i=1d (x, y) = n d (x, y). Thus, by Theorem 3.13, the metrics d 1 and d are topologically equivalent. 2. For all x = (x 1, x 2,..., x n ) and y = (y 1, y 2,..., y n ) R n, we have (i) d (x, y) = max{ x i y i : 1 i n} and Σ n i=1 x i y i 2 = d 2 (x, y) (ii) d 2 (x, y) = Σ n i=1 x i y i 2 31

Σ n i=1 max{ x i y i 2 : 1 i n} = n d (x, y). Thus, by Theorem 3.13, the metrics d 2 and d are topologically equivalent. 3. By Part 1, for all x, y R n, d 1 (x, y) n d (x, y) and, by Part 2, d (x, y) d 2 (x, y). Thus, for all x, y R n, d 1 (x, y) n d 2 (x, y). By Part 2, for all x, y R n, d 2 (x, y) n d (x, y) and, by Part 1, d (x, y) d 1 (x, y). Thus, for all x, y R n, d 2 (x, y) n d 1 (x, y). Therefore, by Theorem 3.13, the metrics d 2 and d 1 are topologically equivalent. It follows that all three metrics generate the same topology. Example 3.15. Let C[0, 1] be the set of all continuous complex-valued functions defined on the closed interval [0, 1], and let d (f, g) = sup{ f(x) g(x) : 0 x 1}, and d 1 (f, g) = 1 0 f(x) g(x) dx be metrics on C[0, 1]. Then d and d 1 are NOT topologically equivalent. 32

Proof. 1. We shall show that, for every f, g C[0, 1], By definition of d, for all x [0, 1], Hence by integration theory, d 1 (f, g) d (f, g). f(x) g(x) sup f(y) g(y) = d (f, g). y [0,1] d 1 (f, g) = 1 0 1 0 f(x) g(x) dx d (f, g)dx = d (f, g). This implies that, for every f C[0, 1] and r > 0, B (f, r) = {g C[0, 1] : d (f, g) < r} {g C[0, 1] : d 1 (f, g) < r} = B 1 (f, r). Thus if U C[0, 1] is open in the topology induced by d 1, then U is open in the topology induced by d. 2. Let ˇ0 denote the constant function with value 0 for all t [0, 1]. We shall show that d -open ball B (ˇ0, 1) is not open in the topology induced by d 1 (i.e., not d 1 -open). For any ɛ > 0, there exists a continuous function { nt + 1 if t [0, 1 n g ɛ (t) = ] 0 if t [ 1, 1], n where 1 2n < ɛ. Then it is easy to see that d 1 (ˇ0, g ɛ ) = 1 0 0 g ɛ (x) dx and = 1 n 0 ( nt + 1)dt = 1 2n < ɛ 33

d (ˇ0, g ɛ ) = sup 0 g ɛ (t) t [0,1] sup nt + 1 = 1. t [0,1] Thus, for every ɛ > 0, there is g ɛ B 1 (ˇ0, ɛ), but g ɛ does not belong to B (ˇ0, 1). Therefore B (ˇ0, 1) is not d 1 -open. It means that d and d 1 are not topologically equivalent. 34

4 Hausdorff Spaces and Limits Definition 4.1. A topological space (X, τ) is Hausdorff if for each pair of distinct points x 1 and x 2 in X there are disjoint open sets U 1 and U 2 such that x 1 U 1 and x 2 U 2. Example 4.2. The topological space R with the usual topology τ d is Hausdorff. Proof. Let x 1 and x 2 be distinct points in R, let r = x 1 x 2 /4 and let U 1 = (x 1 r, x 1 + r) and U 2 = (x 2 r, x 2 + r). We know that U 1 and U 2 are open in the usual topology τ d. It is clear that x 1 U 1 and x 2 U 2. For u U 1 U 2, we have 4r = x 1 x 2 x 1 u + u x 2 < r + r = 2r, a contradiction. Hence U 1 U 2 =. Example 4.3. The topological space X with the discrete topology τ is Hausdorff. Proof. Let x 1 and x 2 be distinct points in X. Recall that the discrete topology τ is the collection of all subsets of X; see Example 3.6. Let U 1 = {x 1 } and U 2 = {x 2 }. By definition of the discrete topology, U 1 and U 2 are in τ, so open. It is clear that x 1 U 1, x 2 U 2 and U 1 U 2 =. Example 4.4. The topological space X with the indiscrete topology τ is not Hausdorff provided X contains more than one point. Proof. Let x 1 and x 2 be distinct points in X. Recall that the indiscrete topology τ = {X, }; see Example 3.7. Every open set U 1 τ such that x 1 U 1 has to be X and therefore x 2 U 1. Thus there are no disjoint open sets U 1 and U 2 in the indiscrete topology such that x 1 U 1 and x 2 U 2. 35

Theorem 4.5. Every metric space (X, d) is Hausdorff. For a proof see Solutions 2, Qu.11. Limits in Metric and Topological Spaces Recall the definition of limit from the basic analysis course. Definition 4.6. We say a sequence (t n ) n=1 of real numbers converges to t R as n if for every ε > 0 there is N N such that for all n N, t n t < ε. Definition 4.7. Let (X, d) be a metric space. We say a sequence (x n ) n=1, x n X, converges to x X as n if for every ε > 0 there is N N such that, for all n N, d(x n, x) < ε, that is, x n B(x, ε). We write with respect to the metric d. lim x n = x n Remark 4.8. It is easy to see that a sequence (x n ) n=1, x n X, converges to x X as n if and only if the sequence of non-negative real numbers t n = d(x n, x) 0 as n. Example 4.9. Consider the metric space (C, d), where d(z 1, z 2 ) = z 1 z 2. Show that the sequence (z n ) n=1, where z n = (1 1 ) + i C, converges to 1 as n. n 2 n 36

Proof. Note that d(z n, 1) = z n 1 = (1 1 n ) + i 2 1 = 1 n n + i 2 n { ( = 1 ) 2 ( ) } 2 1/2 1 + n 2 n = { 1 n + 1 } 1/2 0 2 2 2n as n. Thus lim n z n = z with respect to the metric d. Theorem 4.10. limit. In a metric space (X, d), every convergent sequence has a unique Proof. Suppose that (x n ) n=1, x n X, converges to x X and also to y X as n. If x y, we may choose ε > 0, say ε = 1 d(x, y), 2 so that the open balls B(x, ε) and B(y, ε) are disjoint. Since x n is supposed to belong to both of those open balls for sufficiently large n, we get a contradiction. Thus x = y. Definition 4.11. Let (X, τ) be a topological space. We say a sequence (x n ) n=1, x n X, converges to x X as n if for every open set U containing x, there is N N such that x n U for all n N. We write lim n x n = x with respect to the topology τ. Example 4.12. Let (X, τ) be the topological space with the indiscrete topology τ. Then any sequence (x n ) n=1, x n X, converges to every point x X as n. 37

Proof. Recall that the indiscrete topology τ = {X, }; see Example 3.7. Every open set U τ such that x U has to be X and therefore x n U for all n 1. Thus lim n x n = x with respect to the indiscrete topology. Recall that the topological space (X, τ) with the indiscrete topology is not Hausdorff if X has more than one point. Theorem 4.13. limit. In a Hausdorff space (X, τ), every convergent sequence has a unique Proof. Suppose that (x n ) n=1, x n X, converges to x X and also to y X as n with respect to the topology τ, If x y, the Hausdorff condition implies that there are disjoint open sets U 1 and U 2 such that x U 1 and y U 2. Since x n is supposed to belong to both of those open sets for sufficiently large n, we get a contradiction. Thus x = y. 38

5 Closed Sets Definition 5.1. A subset F of a topological space (X, τ) is called closed if its complement F c is open. Recall that, for S X, the complement of S in X is X \ S = {x X : x S}. In practice, we often work with subsets of a fixed set X and then we write S c instead of X \ S. We write S cc instead of (S c ) c ; it is obvious that S cc = S. Complementation obeys de Morgan s Laws: and more generally (A B) c = A c B c, (A B) c = A c B c, ( λ Λ U λ)c = λ Λ U c λ, ( λ Λ U λ)c = λ Λ U c λ. Example 5.2. Every closed interval [a, b] of R is a closed set in the usual topology. Its complement [a, b] c is the union (, a) (b, ) of two open sets and so is open. Example 5.3. Every open interval (a, b) of R is not closed in the usual topology. Its complement (a, b) c is the union (, a] [b, ) which is not an open subset of R. Proposition 5.4. Every singleton {y} of a metric space (X, d) is closed in the metric topology. Proof. It is clear that the complement of {y} is the set S = X \ {y} = {x X : d(x, y) > 0}. We have to show that, for each u S, there is an open ball B(u, r ) wholly contained in S. Let r = d(u, y) > 0. It is obviuos that y does not belong B(u, r ). Therefore B(u, r ) X \ {y}, so S is open. Hence {y} is closed. 39

Definition 5.5. Let (X, d) be a metric space, let a X, and let r R, r > 0. The closed ball with centre a and radius r is the set B(a, r) = {x X : d(a, x) r} consisting of all points of X whose distance from a is less than or equal to r. Proposition 5.6. The closed ball B(a, r) of a metric space (X, d) is closed in the metric topology. Proof. It is clear that the complement of B(a, r) is the set which is open; see Solution 2, Qu.4. {x X : d(x, a) > r}, Example 5.7. Consider the metric space (C[0, 1], d ), where C[0, 1] is the set of all continuous complex-valued functions defined on the closed interval [0, 1], and d (f, g) = sup{ f(x) g(x) : 0 x 1}. Are the following subsets of the metric space (C[0, 1], d ) closed? and (i) S 1 = {f C[0, 1] : sup f(t) exp(t) 10} 0 t 1 (ii) S 2 = Justify your answer. {f C[0, 1] : sup f(t) t 2 < 1 p }. p=1 0 t 1 40

Proof. (i) The set S 1 = {f C[0, 1] : sup f(t) exp(t) 10} 0 t 1 is the closed ball B(g, 10) with centre g(t) = exp(t), t [0, 1], and radius r = 10 in the metric space (C[0, 1], d ) and therefore, by Proposition 5.6, it is closed. (ii) We proved in Example 2.25 that {f C[0, 1] : sup f(t) t 2 < 1 p } = {g}, p=1 0 t 1 where g(t) = t 2, t [0, 1]. By Proposition 5.4, the singleton {g} in the metric space (C[0, 1], d ) is closed. Every statement involving open sets can be translated into a corresponding statement involving closed sets, simply by taking complements. or Warning: In topology, a set can be (i) open and closed, e.g., every point in X with the discrete topology; (ii) open and not closed, e.g., an open inverval (a, b) in R with the usual topology; (iii) closed and not open, e.g., a closed inverval [a, b] in R with the usual topology; (iv) neither open nor closed, e.g., an inverval (a, b] in R with the usual topology. In most spaces, most sets are neither open nor closed. Theorem 5.8. Then Let κ be the family of all closed subsets in a topological space (X, τ). [T1 ] κ includes and X; [T2 ] κ is closed under arbitrary intersections: if each F λ, λ Λ, is closed, then F λ is closed; λ Λ [T3 ] κ is closed under finite unions: if all F 1, F 2,..., F n are closed, then F 1 F 2... F n is closed. 41

Proof. [T1 ] is closed since its complement X is open. X is closed since its complement is open. [T2 ] Each Fλ c is open, so, by [T2], the union λ Λ F λ c is open, and hence its complement ( Fλ) c c = Fλ cc = F λ λ Λ λ Λ λ Λ is closed. [T3 ] F c 1, F c 2..., F c n are open, so, by [T3], the finite intersection is open, and hence its complement F c 1 F c 2... F c n (F1 c F2 c... Fn) c c = F1 cc F2 cc... Fn cc is closed. = F 1 F 2... F n 42

6 Separation axioms In metric spaces (X, d), the metric d measures the separation of two distinct points, and, in general, it is a very useful tool. Since most topological spaces are not metrisable, this tool is not available to us. However we can generalise useful properties of metrics, in the form of separation axioms. Definition 6.1. Let (X, τ) be a topological space. Then (X, d) is (0) T 0 if for each pair of points x y X, there is an open set containing one of the points and not the other. (1) T 1 if for each pair of points x y X, there are open sets about each of the points not containing the other point. (2) T 2 (also called Hausdorff ) if for each pair of points x y X, there are open sets U x and V y such that U V =. (Reg.) Regular if for each given a closed set A and a point x / A, there are open sets U A and V x such that U V =. (3) A space which is regular and T 1 is called T 3. Remark 6.2. In this context we make an abuse of terminology. The word axiom is used here in the meaning of requirement contained in a definition (which can be fulfilled or not, depending on the cases). Proposition 6.3. The following holds T 3 = T 2 = T 1 = T 0. Proof. The proof of the implications is left as an exercise. Proposition 6.4. A topological space X is T 1 if and only if the points are closed. Proof. ( ) Suppose that X is T 1. For any y x, there is an open set U y containing y and not x. Then X \{x} = y x U y, which being the union of open sets is open. Hence its complement {x} is closed. ( ) If points are closed and x, y are distinct points in X, then X \ {x} is an open set containing y and not x, while X \ {y} is an open set containing x and not y. Thus X is T 1. Lemma 6.5. There exists a topological space that satisfies none of the separation axioms. 43

Proof. Let X be a set with more than two elements. Then (X, τ ind ) with the indiscrete topology τ ind = {X, } is such a space. Prove that (X, τ ind ) is not T 0. Lemma 6.6. There exists a topological space that is T 0 but it is not T 1. Proof. Let X be a set that contains at least two points. Fix p X and let the particular point topology on p. That is τ p = { } {S X : p S}. Then (X, τ p ) is a T 0 space. But it is not T 1 since p cannot be separated by any other point x X (every open set containing x X must contain p as well). Lemma 6.7. There exists a space that is T 1 but it is not T 2. Proof. Let X be an infinite space. Let the co-fnite topology on X defined by τ = { } {S X : X \ S is finite}. Then (X, τ) is a T 1 space. Indeed for x, y X let U = X \ {y} and V = X \ {x}. Then U and V are open, with x U and y V, and y / U and x / V. However (X, τ) is not T 2, since it does not contain disjoint open sets. Indeed let A, B X be open and suppose that A B =. Then, by de Morgan s Laws, X = X \ (A B) = (X \ A) (X \ B). However X \ A and X \ B are finite, hence we would have that X is finite which is a contradiction. Example 6.8. Consider the space X = {a, b, c}. Define the topology τ = {, {a}, {b, c}, X}. Then (X, τ) is regular. Hint: Consider all possible combinations for the closed subsets and the points which are not contained there. These closed subsets are open as well. Lemma 6.9. There exists a regular space that is not T 2. Proof. Consider the space X = {a, b, c}. Define the topology τ = {, {a}, {b, c}, X}. We have shown above that (X, τ) is regular. By Proposition 6.4, (X, τ) is not T 1 since {b} is not a closed set; hence it is not even T 2. 44

7 Basic Topological Concepts Interior of a Set Definition 7.1. Let A be a subset of a topological space (X, τ). Then a point p is an interior point of A if there is an open set U such that p U and U A. Remark 7.2. Let A be an open set. Then all points p of A are interior points of A. Definition 7.3. The subset A of A consisting of all the interior points of A is called interior of A. Example 7.4. Consider R with the usual topology and the closed interval A = [a, b] R. Then every point r of the open interval (a, b) is an interior point of [a, b]. One can see that r (a, b) and the open subset (a, b) A, but a and b are not interior points. Therefore [a, b] = (a, b). Example 7.5. Consider R with the usual topology and the subset Z R. Every point of Z is NOT an interior point of Z, since every nonempty open interval contains points which are not in Z. Therefore Z =. Proposition 7.6. Let (X, τ) be a topological space and let A X. Then A is the union of all open subsets of A, and, hence, it is the largest open set. Proof. By definition, p is an interior point of A it lies in an open subset of A. The = part tells us that A is contained in the union of the open subsets U λ, λ Λ, of A: A λ Λ U λ. 45

The = part tells us that the union of all the open subsets U λ, λ Λ, of A is contained in A : U λ A. λ Λ By the definition of a topology T 2, any union of open subsets is open. Thus A is the largest open subset of A. Definition 7.7. A neighbourhood of x in a topological space (X, τ) means an open set containing x. Closure of a Set Definition 7.8. Let A be a subset of a topological space (X, τ). Then a point p is a closure point of A if every open set containing p meets A. Definition 7.9. Let A be a subset of a topological space (X, τ). The set of all closure points of A is called the closure of A, and is denoted by A. Example 7.10. Consider R with the usual topology and A = (0, 1). The point 0 is a closure point of (0, 1) since every open set containing 0 contains an open interval ( ɛ, ɛ) and hence a point of (0, 1). The point 1 is also a closure point of (0, 1) since every open set containing 1 contains an open interval (1 ɛ, 1 + ɛ) and hence a point of (0, 1). The closure of A = (0, 1) is [0, 1]. Example 7.11. Consider R with the usual topology and A = Q R the set of rational numbers. Then, every point of R is a closure point for Q, since every nonempty open interval contains points which are in Q. Thus the closure of Q is R. Theorem 7.12. Let A be a subset of a topological space (X, τ). Then 46

(i) (A ) c = A c ; (ii) (A) c = (A c ), that is, the complement of the interior is the closure of the complement, and the complement of the closure is the interior of the complement. Proof. (i) x (A ) c x A no open set containing x is wholly contained in A every open set containing x meets A c x A c. (ii) x (A) c x A some open set containing x does not meet A some open set containing x is wholly contained in A c x (A c ). Theorem 7.13. Let A be a subset of a topological space (X, τ). Then (i) A is the intersection of all the closed supersets of A; (ii) A is a closed set; (iii) A A; (iv) A is the smallest closed superset of A. Proof. (i) By the first part of the previous theorem (A) c = (A c ) = {U : U open and U A c }. Take complements: A = (A) cc = ( {U : U open and U A c }) c (by de Morgan s Laws) = {U c : U c closed and A U c } = {F : F closed and A F }. (ii) follows from (i) and [T2 ]. (iii) follows from (i). (iv) follows from (i) and (ii). 47

Isolated points of a Set Definition 7.14. Let A be a subset of a topological space (X, τ). An element a of A is an isolated point of A if there is an open set containing a which contains no point of A other than a itself. Example 7.15. Consider R with the usual topology and A = { 1, 1 2, 1 3, 1 4,...}. The point 1 is a closure point and an isolated point of A, since the open set ( 3 4, 5 4 ) contains no point of A other than 1. In fact all points of this set A are isolated. Boundary of a Set Definition 7.16. A point p is a boundary point of a set A if it lies in the closure of A and the closure of its complement A c. Definition 7.17. The set A of all boundary points of A is called the boundary of A. Thus A = A A c. Example 7.18. Consider the metric space (C, d), the complex plane C with d(z 1, z 2 ) = z 1 z 2. Let A be an open ball B(a, r), for some a C and r > 0, in the C. Then the boundary of A is the circle with centre a and radius r A = {z C : z a = r}. 48

Example 7.19. Consider R with the usual topology and A = Q R. Then, since every open interval of R contains both rational and irrational numbers, every point of R lies in the closure of Q and the closure of its complement R \ Q. Thus Q = R. Topologies on Subspaces Definition 7.20. Let (X, τ) be a topological space and let A be a subset of X. The open subsets for the induced topology on A are the intersections of A with the open subsets of X, that is, τ A = {U A : U open in X}. The topological space (A, τ A ) with the induced topology is called a subspace of (X, τ). Theorem 7.21. The family τ A forms a topology on A. Proof. [T1] Note that A = X A τ A and = A τ A. [T2] Let U λ A τ A, λ Λ. Then its union is λ A) = ( λ Λ(U U λ ) A τ A, λ Λ which belongs to τ A, since λ Λ U λ is open in (X, τ). [T3] Let U i A τ A, 1 i n. Then its intersection is n n (U i A) = ( U i ) A τ A, i=1 i=1 which belongs to τ A, since n i=1 U i is open in (X, τ). Thus τ A forms a topology on A. Theorem 7.22. The closed subsets of (A, τ A ) are the intersections of A with the closed subsets of X. 49

Proof. It is easy to see that if B X then complement of B A in A is B c A: A \ (B A) = B c A. By definition, S is a closed subset of (A, τ A ) if its complement A \ S is an open subset U A of A, where U is open in (X, τ). Thus S = A \ (A \ S) = A \ (U A) = U c A, which is the intersection of A with a closed subset of (X, τ). Conversely, if F is closed in (X, τ), then F c is open in (X, τ) and F c A = A \ (F A) is open in (A, τ A ). Therefore F A is closed in (A, τ A ). 50

8 Compactness Definition 8.1. Let S be a subset of a topological space (X, τ). An open cover of S is a family U λ, λ Λ, of open subsets of X such that S λ Λ U λ. Example 8.2. Thus the family of open intervals is an open cover of R. ( n, n) (n = 1, 2, 3,...) Definition 8.3. A subfamily of an open cover which is itself an open cover is called a subcover. Example 8.4. For example, the family is a subcover of the open cover above. ( 2n, 2n) (n = 1, 2, 3,...) Definition 8.5. A subset K of a topological space (X, τ) is compact if every open cover of K has a finite subcover. Example 8.6. Consider C with the usual topology. The subset B(1, 1) = {z C : z 1 < 1} is not compact, since U n = {z C : z 1 < 1 1 n }, n = 1, 2,..., is an open cover of B(1, 1) in C with the usual topology, but it does not have a finite subcover. 51

Proposition 8.7. Every finite subset of a topological space is compact. Proof. Let F = {x 1, x 2,..., x n } be a finite set, and let U λ, λ Λ, be an open cover of F. Then for 1 i n, there is a λ i Λ such that x i U λi. Then {U λ1, U λ2,..., U λn } is a finite subcover. Thus every finite subset of a topological space is compact. Theorem 8.8. A closed subset of a compact set is compact. Proof. Let C be a closed subset of a compact set K in a topological space (X, τ), and let U λ, λ Λ, be an open cover of C. Then, by Theorem 7.22, C = K F for some closed subset F of (X, τ). Thus F c is open in (X, τ). Therefore U λ, λ Λ, together with the open set F c, is an open cover of K. But K is compact, so we can extract a finite subcover of K. If F c occurs in this subcover, discard it. We are left with a finite subcover of C, extracted from U λ, λ Λ. Therefore C is compact. Example 8.9. Consider R with the usual topology. (i) The set R is not compact, since {( n, n) : n = 1, 2, 3,...} is an open cover with no finite subcover. (ii) The half-open interval [0, 1) is not compact, since is an open cover with no finite subcover. {( 1/n, 1 1/n) : n = 2, 3,...} Proposition 8.10. A finite union of compact sets is compact. 52

Proof. Exercise. Theorem 8.11. Let K be a compact subset of a Hausdorff space (X, τ) and let x K. Then there exist disjoint open sets U and V such that x U and K V. Proof. Since (X, τ) is a Hausdorff topological space, for each k K, there exist disjoint open sets U k and V k with x U k and k V k. Then (V k ) k K is an open cover of K, so it contains a finite subcover V k1, V k2,... V kn. Let U = n j=1u kj and V = n i=1v ki. Then U and V are open subsets x U, K V and U V = ( n j=1u kj ) ( n i=1v ki ) since, for every k, V k U k =. = n i=1 ( Vki ( n j=1u kj ) ) =, Theorem 8.12. A compact subset K of a Hausdorff space X is closed. Proof. By the previous theorem, for each x K c there is an open set U x such that x U x K c. Then K c is the union of all such U x. The union of open sets is open. Therefore K is closed. Example 8.13. Compact subsets of non-hausdorff topological spaces need not be closed: in {a, b} with the indiscrete topology both of the singletons are compact but not closed. Theorem 8.14. (Heine-Borel) Every bounded closed interval [a, b] of R with the usual topology is compact. 53