MAS3706 Topology. Revision Lectures, May I do not answer enquiries as to what material will be in the exam.

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1 MAS3706 Topology Revision Lectures, May 208 Z.A.Lykova It is essential that you read and try to understand the lecture notes from the beginning to the end. Many questions from the exam paper will be similar to what you have been asked to do during the semester. I do not answer enquiries as to what material will be in the exam. Section Open Subsets of R and Continuous Functions You need to understand all main notions from this section to be able to follow further material. Definitions: an open set, a continuous function, the inverse image. Topology on R; Theorem.9 (statement and proof) Examples:.3-.7,.20,.2. Qu. () Give the definition of an open subset in R. (2) Which of the following subsets of R are open, which are not? Justify your answer: (i) S = [0, ], (ii) S 2 = ( n, + n ), (iii) S 00 3 = ( n, + n ), (iv) S 4 = n= n= ( n, n). (3) Which of the above sets are open in R with the indiscrete topology τ ind = {R, }, which are not? Solution. () Definition 0.. 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. (2) (i) The closed interval [0, ] is not open. If you take p to be 0 or then every open interval (p h, p + h) contains points outside [0, ]. (ii) Note that if and only if r S 2 = ( n, + n ) n= r ( n, + n ) n=

2 for all n N. Thus r =. Therefore S 2 = {} is a singleton and not open in R. iii) The set S 3 = 00 n= (, + ) = (, + ) is an open interval and open. n n iv) The set S 4 = n= ( n, n) = R is open. (3) S = [0, ] is not in τ ind = {R, }, thus it is not open in (R, τ ind ). S 2 = {} is not in τ ind = {R, }, thus it is not open in (R, τ ind ). S 3 = ( 00, + 00 ) is not in τ ind = {R, }, thus it is not open in (R, τ ind ). S 4 = R is in τ ind = {R, }, thus it is open in (R, τ ind ). Section 2 Metric Spaces Definitions: a metric, a metric space, an open ball, an open set, a metric topology (Thm 2.2). Theorem 2.22 (statement and proof). Examples of metric spaces, open balls, open sets. Section 3 Topological Spaces Definitions: a topology, a topological space. Examples of topological spaces Equivalent Metrics (3.). Theorem 3.3 (statement and proof). Example 3.5. Qu. (i) Give the definition of topologically equivalent metrics on a set X. Definition 0.2. Two metrics d 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 -open if and only if it is d 2 -open. (ii) Consider the following metrics on C 3 : and d (x, y) = Σ 3 i= x i y i, d 2 (x, y) = Σ 3 i= x i y i 2 d (x, y) = sup{ x i y i : i 3} for x = (x, x 2, x 3 ) and y = (y, y 2, y 3 ) from C 3. Show that d, d 2 and d are topologically equivalent. Proof.. For all x = (x, x 2, x 3 ) and y = (y, y 2, y 3 ) C 3, we have (i) d (x, y) = max{ x i y i : i 3} Σ 3 i= x i y i = d (x, y) 2

3 and (ii) d (x, y) = Σ 3 i= x i y i Σ 3 i= max{ x i y i : i 3} = Σ 3 i=d (x, y) = 3 d (x, y). Thus, by Theorem 3.3, the metrics d and d are topologically equivalent. 2. For all x = (x, x 2, x 3 ) and y = (y, y 2, y 3 ) C 3, we have (i) d (x, y) = max{ x i y i : i 3} Σ 3 i= x i y i 2 = d 2 (x, y) and (ii) d 2 (x, y) = Σ 3 i= x i y i 2 Σ 3 i= max{ x i y i 2 : i 3} = 3 d (x, y). Thus, by Theorem 3.3, the metrics d 2 and d are topologically equivalent. 3. Consider any a subset U of C 3. We have shown that U is d -open if and only if it is d -open and U is d -open if and only if it is d 2 -open. Hence U is d -open if and only if it is d 2 -open. Thus the metrics d and d 2 are topologically equivalent. Qu. Let C[, 2] be the set of all continuous complex-valued functions defined on the closed interval [, 2], and let and d (f, g) = sup{ f(x) g(x) : x 2}, { 2 d 2 (f, g) = } /2 f(x) g(x) 2 dx be metrics on C[, 2]. Show that d and d 2 are not topologically equivalent. Proof. Let ˇ0 denote the constant function with value 0 for all t [, 2]. We shall show that the d -open ball B (ˇ0, ) is not open in the topology induced by d 2 (i.e., is not d 2 -open). For any ɛ > 0, there exists a continuous function { n(t ) + if t [, + n g ɛ (t) = ] 0 if t [ +, 2], n where < ɛ. Then it is easy to see that 3n { 2 /2 d 2 (ˇ0, g ɛ ) = 0 g ɛ (x) dx} 2 = { + n ( n(t ) + ) 2 dt } /2 3

4 and = { /2 { n = ( ns + ) ds} 2 n = (n 2 s 2 2ns + )ds 0 { n 2 3n 2n 3 2n + } /2 = 2 n 0 { 3n n + n } /2 = d (ˇ0, g ɛ ) = sup 0 g ɛ (t) = g ɛ () =. t [,2] } /2 3n < ɛ Thus, for every ɛ > 0, there is g ɛ B 2 (ˇ0, ɛ) such that g ɛ does not belong to B (ˇ0, ). Therefore B (ˇ0, ) is not d 2 -open. It follows that d and d 2 are not topologically equivalent. Section 4 Hausdorff Spaces and Limits Definitions: a Hausdorff Space, a convergent sequence in a metric space (4.7), a convergent sequence in a topological space (4.). Theorems 4.5, 4.3 (statements and proofs). Examples , 4.9, 4.2. Qu. What does it mean to say that a topological space (X, τ) is Hausdorff? Solution. A topological space (X, τ) is Hausdorff if for each pair of distinct points x and x 2 in X there are disjoint open sets U and U 2 such that x U and x 2 U 2. Qu. Show that the topological space R with the usual topology τ d is Hausdorff. Solution. Let x and x 2 be distinct points in R, let r = x x 2 /4 and let U = (x r, x + r) and U 2 = (x 2 r, x 2 + r). We know that U and U 2 are open in the usual topology τ d. It is clear that x U and x 2 U 2. For u U U 2, we have a contradiction. Hence U U 2 =. 4r = x x 2 x u + u x 2 < r + r = 2r, Qu. Show that the topological space X with the discrete topology τ is Hausdorff. Solution. Let x and x 2 be distinct points in X. Recall that the discrete topology τ is the collection of all subsets of X. Let U = {x } and U 2 = {x 2 }. By definition of the discrete topology, U and U 2 are in τ, so open. It is clear that x U, x 2 U 2 and U U 2 =. 4

5 Qu. Show that the topological space X with the indiscrete topology τ is not Hausdorff provided X contains more than one point. Solution. Let x and x 2 be distinct points in X. Recall that the indiscrete topology τ = {X, }. Every open set U τ such that x U has to be X and therefore x 2 U. Thus there are no disjoint open sets U and U 2 in the indiscrete topology such that x U and x 2 U 2. Qu. Show that the topology τ s.i. on R which consists of R,, and all the semi-infinite open intervals (a, ), a R, is not Hausdorff. Solution. Let x and x 2 be distinct points in R and let x < x 2. Every open set U τ s.i. such that x U is of the form (a, ) for some a R and a < x and therefore x 2 U since a < x < x 2. Thus there are no disjoint open sets U and U 2 in the topology τ s.i. such that x U and x 2 U 2. Qu. (i) Give the definition of a convergent sequence in a metric space (X, d). Definition 0.3. Let (X, d) be a metric space. We say a sequence (x n ) n=, x n X, converges to x X as n if for every ε > 0 there is N N such that, for all n N, that is, x n B(x, ε). d(x n, x) < ε, (ii) Consider the metric space (C 3, d ), where d ( v, w) = max v j w j, for v = (v, v 2, v 3 ), w = (w, w 2, w 3 ) C 3. j=,2,3 Show that the sequence ( v n ) n=, where v n = (3i n 2, 5, i 3 n ) C 3, i 2 =, converges to u = (3i, 5, 0) as n with respect to the metric d. Solution. (ii) Note that d ( v n, u) = d ((3i n, 5, i ), (3i, 5, 0)) 2 3n = max{ 3i i 3i, 5 5, n2 3 0 } = max{ n n, 0, 2 3 } n = n 2 0 as n. Thus lim n v n = u = (3i, 5, 0) with respect to the metric d. 5

6 Qu. (i) Give the definition of a convergent sequence in a topological space (X, τ). Definition 0.4. Let (X, τ) be a topological space. We say a sequence (x n ) n=, 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. (ii) Let (R, τ s.i. ) be the topological space with the topology τ s.i., which consists of R,, and all the semi-infinite open intervals (a, ), a R. Show that if a sequence (x n ) n=, x n X, converges to x X as n, then it converges to every y X such that y x. Solution. (ii) Every open set U τ s.i. such that y U is of the form (a, ) for some a R such that a < y x. Therefore x U and, by the definition of the limit, there is N N such that x n U for all n N. Thus lim n x n = y with respect to τ s.i. for every y x. Section 5 Closed Sets Definitions: a closed set, a closed ball. Propositions 5.4, 5.6 (statements and proofs). Examples 5.2, 5.3, 5.7. Section 6 Separation axioms Read and try to understand the matirial. Section 7 Basic Topological Concepts Definitions: an interior of a set, the closure of a set, the boundary of a set, a topology on a subset. Examples 7.4, 7.5, 7.0, 7., 7.8, 7.9. Theorems 7.2 (statement). Qu. (i) Give the definition of the interior of a subset A in a topological space (X, τ). (ii) Consider R with the usual topology. Find the interior of the following subsets of R: (a) A = [2, 0] \ {3, 5}, (b) A 2 = Q the set of rational numbers. (iii) The exterior Ext(A) of a set A is the interior of its complement Ext(A) = (A c ). Consider R with the usual topology and A = Q R. Find the exterior of A. Solution. (i) The subset A of A consisting of all the interior points of A is called interior of A. (ii) (a) A = [2, 0] \ {3, 5} = [2, 3) (3, 5) (5, 0]. Every point r of the union of open intervals S = (2, 3) (3, 5) (5, 0) is an interior point of A. One can see that r S and the open subset S A, but 2 and 0 are not interior points. Therefore A = (2, 3) (3, 5) (5, 0). 6

7 (b) No point of Q is an interior point of Q, since every nonempty open interval contains irrational points which are not in Q. Therefore Q =. (iii) For A = Q R, the complement A c = R\Q. Since every nonempty open interval contains points which are in Q, which are not in R\Q, we have Ext(A) = (R\Q) =. Qu. Give the definition of the closure of a subset A in a topological space (X, τ). (ii) Consider R with the usual topology. Find the closure of the following subsets of R: (a) A = (2, 0) \ {3, 5}, (b) A 2 = R \ Q the set of irrational numbers. Solution. (i) 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. (ii) (a) A = (2, 0) \ {3, 5}. The following points 2, 3, 5 and 0 are closure points of A since every open set containing one of those points r contains an open interval ( ɛ+r, r+ɛ) and hence a point of A. The closure of A is [2, 0]. (ii) (b) Note that every point of R is a closure point for A 2 = R\Q, since every nonempty open interval contains points which are in A 2 = R \ Q. Thus the closure of A 2 is R. Qu (i) Give the definition of the boundary of a subset A in a topological space (X, τ). (ii) Consider R with the usual topology. Find the boundary of the set of irrational numbers A = R \ Q. Proof. (i) The set A of all boundary points of A is called the boundary of A. Thus A = A A c. (ii) Then, since every open interval of R contains both rational and irrational numbers, every point of R lies in the closure of A = R \ Q and the closure of its complement Q. Thus (R \ Q) = R. Section 8 Compactness Definitions: an open cover of X, an open subcover of X, compact. Propositions/Theorems 8.7, 8.8, 8.0, 8., 8.2, 8.4, 8.5, 8.7 (statements and a number of proofs). Examples 8.2, 8.4, 8.6, 8.9, 8.6. Qu. Let (X, τ) be a topological space.. Give the definitions of an open cover of (X, τ) and a subcover of (X, τ). 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. 7

8 A subfamily of an open cover which is itself an open cover is called a subcover. 2. What does it mean to say that a subset K of a topological space (X, τ) is compact? A subset K of a topological space (X, τ) is compact if every open cover of K has a finite subcover. 3. Use the definition to show that subset {z C : z 3i < 2} is not compact in C with the usual topology. Proof. The subset S = {z C : z 3i < 2} is not compact, since U n = {z C : z 3i < 2 n }, n =, 2,..., is an open cover of S in C with the usual topology, but it does not have a finite subcover. Qu. (i) Show that a compact subspace K of the topological space R with the usual topology is bounded. Proof. Let K be compact in the topological space R with the usual topology. Note that {( n, n) : n =, 2, 3,...} is an open cover of K. Thus it has a finite subcover. The union of the subcover is of the form ( N, N), so x N for all x K, that is, K is bounded. (ii) Are the following subsets (, ) and (, 27) of the topological space R with the usual topology compact? Justify your answer. Solution. The subsets (, ) and (, 27) of R are not bounded and therefore, by Part (i), are not compact. Section 9 Continuity Definition of a continuous function (8., 8.2), a homeomorphism. Theorems 8.3, 8.8 and 8.9 (statements). Examples 8.5, 8.6. Section 0 Metric Spaces Again Theorems 0. and 0.2 (statements). Section Completeness in Metric Spaces Definitions: a Cauchy sequence, completeness of a metric space, a contraction, a fixed point of a mapping. Theorems/Propositions.2,.3,.8,.9 (statements and some proofs) Completeness of C(K) (statement and proof). Examples.4,.6,.6. 8

9 Remarks.2,.3. Theorem.5 (statement and proof) Qu. (i) Give the definition of a contraction mapping on a metric space (X, d). (ii) Give the definition of a fixed point of a mapping T : X X. State the Contraction Mapping Principle. Proof. (i) Let (X, d) be a metric space and T : X X be a mapping. T is a contraction or contraction mapping if there exists k R, 0 k <, such that d(t x, T y) kd(x, y) for all x, y X. (ii) A fixed point of a mapping T : X X is a point x X such that T x = x. Theorem. The Contraction Mapping Principle. Every contraction mapping T on a complete metric space (X, d) has a unique fixed point. Example 0.5. Let f : [a, b] [a, b] be a function such that f (x) K < for all x : a x b. Then f has a unique fixed point on [a, b]. Proof. Note that [a, b] is a closed subspace of the complete metric space R with the usual metric. Thus [a, b] is complete. By the Mean Value Theorem, for all x, x 2 [a, b], we have f(x ) f(x 2 ) = f (ξ) x x 2 K x x 2 where ξ lies between x and x 2. Since K <, f is a contraction mapping. Therefore, by the Contraction Mapping Theorem, f has a unique fixed point. Qu. Let a > 0, and define T : [ a, ) [ a, ) by T x = [ x + a ]. 2 x (i) Show that T is a contraction on [ a, ). (ii) Prove that there exists a unique fixed point x of T on [ a, ). (iii) Show that x = a. Proof. (i) Note that for every positive real number x we have T x = 2 [x 2 a + a x ] + a = 2 ( x a x ) 2 + a a. 9

10 In particular, T maps [ a, ) into itself. Note that [ a, ) is a closed subset of the complete metric space R, and so it is complete. Furthermore, for all x, y [ a, ), T x T y = ( x + a ) ( y + a ) 2 x 2 y = ( 2 (x y) + a x ) y = a(x y) 2 (x y) xy = 2 x y a xy x y. 2 Therefore T is a contraction (with k = 2 ) on [ a, ). (ii) By the contraction mapping theorem, there exists a unique fixed point x given by T x = x, i.e., x = ( x + a ). 2 x (iii) On multiplying through by 2x we find that 2x 2 = x 2 + a, and hence x 2 = a, i.e., x = a. Thus if x 0 is any number greater than or equal to a, the sequence (T n x 0 ) converges to a. Section 2 Connectedness Read and try to understand the matirial. Section 3 Picard s Theorem Picard s Theorem (try to understand the proof). Homework : Qu., Qu. 4. Homework 2: Qu., Qu. 3, Qu. 4. Homework 3: Qu., Qu. 2, Qu. 5, Qu. 0, Qu. 2, Qu. 6, Qu. 8. Homework 4: Qu., Qu. 2, Qu. 3, Qu. 9, Qu. 26, Qu

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