Topology. Xiaolong Han. Department of Mathematics, California State University, Northridge, CA 91330, USA address:

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1 Topology Xiaolong Han Department of Mathematics, California State University, Northridge, CA 91330, USA address:

2 Remark. You are entitled to a reward of 1 point toward a homework assignment if you are the first person to report a bona-fide mathematical mistake (i.e. not including language typos and grammatical errors.)

3 Contents Chapter 1. Preliminaries Sets and functions The real number system Cardinality 8 Chapter 2. Topological spaces and continuous functions Topological spaces Closed sets and limit points Hausdorff spaces Continuous functions Metric topology Continuous functions involving metric spaces Product topology 23 Chapter 3. Connectedness and compactness Connected spaces Intermediate value theorem Path connected spaces Compact spaces Maximum value theorem Uniform continuity theorem 30 Chapter 4. The metrization theorems 32 Chapter 5. The fundamental groups Homotopy of paths The fundamental groups 37 Bibliography 40 3

4 CHAPTER 1 Preliminaries 1.1. Sets and functions We speak the language of sets and functions (or mappings) in mathematics. Definition (Sets). Given a set A, x A denotes that x is an element (or member, point) of A and x / A denotes that x is not an element of A. We say that two sets A and B are equal, denoted by A = B, if they have the same elements. Given two sets A and B, we say that A is a subset of B, denoted by A B (or A B), if each element of A is a member of B; we say that A is a proper subset of B, denoted by A B, if A B and A B. Remark. Given two sets A and B, A = B if and only if (i.e. iff) A B and B A. Example. N = {1, 2, 3,...} denotes the set of natural numbers. (Notice that in some other books, N refers to the set {0, 1, 2, 3,...}.) Z = {..., 3, 2, 1, 0, 1, 2, 3,...} denotes the set of integers. Q = {p/q : p, q Z, q 0} denotes the set of rational numbers. R denotes the set of real numbers. R \ Q is called the set of irrational numbers. Definition (The empty set). The set that has no elements is called the empty set and is denoted by. A set that is not equal to the empty set is called nonempty. Definition (A singleton set). A set that contains a single element is called a singleton set. Definition (Cartesian products). The Cartesian product A 1 A m of m sets A 1,...A m is the set of all m-tuples (x 1,..., x m ) such that x i A i, i = 1,..., m. That is, A 1 A m = {(x 1,..., x m ) : x i A i, i = 1,..., m}. Definition (Relations). A relation on a set A is a subset C of the Cartesian product A A. Let C be a relation on a set A. We denote as xcy if (x, y) C and say that x is in the relation C to y. Example (Equivalence relations). An equivalence relation on a set A is a relation C that satisfies (i). [Reflexivity] xcx for every x A. (ii). [Symmetry] If xcy, then ycx. (iii). [Transitiity] If xcy and ycz, then xcz. In this case, we say x is equivalent to y if xcy and also denote by x y. Definition (Union, intersection, and difference). Let A and B be two sets. 4

5 and 1.1. SETS AND FUNCTIONS 5 The union of A and B is A B = {x : x A or x B}. The intersection of A and B is A B = {x : x A and x B}. We say that A and B are disjoint if A B =. The set difference of A and B is A \ B = {x : x A and x / B}, and is also denoted by A B. It is sometimes called the complement of B relative to A. In particular, if all the set operations are within a universal set X, then for a set A X, we simply call X \ A the complement of A, and is also denoted by A c. Remark. Given a family of sets F, we define E = {x : x E for some E F}, E F E F E = {x : x E for all E F}. In particular, if F = {E λ } λ Λ, where λ is called the index and Λ is called the index set, then we define E λ = {x : x E λ for some λ Λ}, and For example, and E F λ Λ E λ = {x : x E λ for all λ Λ}. λ Λ n E i = {x : x E i for some i = 1,..., n}, i=1 n E i = {x : x E i for all i = 1,..., n}. i=1 Theorem (De Morgan s Law). Let F be a family of sets. Then ( ) c E = ( ) c E c and E = E c. E F Definition (Functions). Let A and B be two sets. A function f from A to B, denoted by f : A B, is a correspondence that assigns to each element of A an element of B; for x A, we denote by f(x) the assigned element in B. Definition (Domain and range). Let f be a function from A to B. We call A the domain of f and B the codomain of f. Given a subset E A, we define f(e) = {y B : y = f(x) for some x E} the image of E. We call f(a) the range of f. Definition (Preimage). Let f be a function from A to B. Given a subset E B, we define f 1 (E) = {x A : f(x) E} the preimage of E. Notice that f 1 is not necessarily a function. In order to make f 1 a function, we need the following concepts. Definition (Injective and surjective functions). Let f be a function from A to B. We say that f is injective (or one-to-one) if x, y A and x y imply f(x) f(y). We say that f is surjective (or onto) if f(a) = B. E F E F

6 1.2. THE REAL NUMBER SYSTEM 6 Definition (Bijective functions). Let f be a function from A to B. We say that f is bijective if f is injective and surjective. In this case, we say that A and B are equipotent. Let f be a bijective function from A to B. Given each element y B, there is exactly one element x A such that f(x) = y and we denote by f 1 (y) = x. This defines a function f 1 : B A and we call f 1 the inverse of f. Problems i Let f : A B be a function. Suppose that B 1, B 2 B. Prove that (1). If B 1 B 2, then f 1 (B 1 ) f 1 (B 2 ). (2). f 1 (B 1 B 2 ) = f 1 (B 1 ) f 1 (B 2 ). (3). f 1 (B 1 B 2 ) = f 1 (B 1 ) f 1 (B 2 ). (4). f 1 (B 1 \ B 2 ) = f 1 (B 1 ) \ f 1 (B 2 ) ii Let f : A B be a function. Suppose that A 1, A 2 A. Prove that (1). If A 1 A 2, then f(a 1 ) f(a 2 ). (2). f(a 1 A 2 ) = f(a 1 ) f(a 2 ). (3). f(a 1 A 2 ) f(a 1 ) f(a 2 ) and the equality holds if f is injective. (4). f(a 1 \ A 2 ) f(a 1 ) \ f(a 2 ) and the equality holds if f is injective Let f : A B be a function. Suppose that A 1, A 2 A. Are f(a 1 A 2 ) = f(a 1 ) f(a 2 ) and f(a 1 \ A 2 ) = f(a 1 ) \ f(a 2 )? Prove your assertion The real number system The real number system R is an example of a complete ordered field. We define its structure in three steps: Step 1: R is a field. Definition (Fields). A field F is a nonempty set together with two operators + and, called addition and multiplication, which satisfy the following axioms. (A0). The operations + and are binary operations, that is, if a, b F, then a + b and a b are uniquely determined elements of F. (A1). Commutativity of addition: If a, b F, then a + b = b + a. (A2). Associativity of addition: If a, b, c F, then (a + b) + c = a + (b + c). (A3). The additive identity: There is an element in F, denoted by 0, such that 0+a = a+0 = a for all a F. (A4). The additive inverse: For each a F, there is an element b F, called the additive inverse of a and denoted by a, such that a + b = 0. (A5). Commutativity of multiplication: If a, b F, then a b = b a. (A6). Associativity of multiplication: If a, b, c F, then (a b) c = a (b c). (A7). The multiplicative identity: There is an element in F, denoted by 1, such that 1 a = a 1 = a for all a F. (A8). The multiplicative inverse: For each a F and a 0, there is an element b F, called the multiplicative inverse of a and denoted by a 1 or 1/a, such that a b = 1. (A9). The distributive property: If a, b, c F, then a (b + c) = a b + a c. Remark. We will simply denote a + ( b) by a b, the multiplication a b by ab, and a b 1 by a/b. i This problem shows that f 1 preserves inclusions, unions, intersections, and differences of sets. ii This problem shows that f preserves inclusions and unions of sets; if in addition f is injective, then it also preserves intersections and differences of sets.

7 1.2. THE REAL NUMBER SYSTEM 7 Question. Among N, Z, Q, R, R \ Q, which one(s) are fields, why not for N, Z, and R \ Q, and why for Q and for R? After Step 1, we can do addition, subtraction (as added by the additive inverse), multiplication, and division (as multiplied by the multiplicative inverse) on R, according to all the rules that we knew before. Step 2: R is an ordered field. Definition (Order relations). An order relation (or a simple order, or a linear order) on a set A is a relation C that satisfies (i). [Comparability] For every x, y A and x y, either xcy or ycx. (ii). [Nonreflexivity] There is no x A such that xcx. (iii). [Transitiity] If xcy and ycz, then xcz. In this case, we also denote as x < y and also y > x if xcy. Definition (Ordered field). Let F be a field. Then F is an ordered field if (A10). it has an order relation < that satisfies (i). If x < y, then x + z < y + z. (ii). If x < y and 0 < z, then xz < yz. Question. Among Q and R, which one(s) are ordered fields? After Step 2, we can compare the greatness of elements in R. We can also define the intervals in R. In fact, we can define the intervals in any set equipped with an order relation. Definition (Intervals). Given two numbers a, b R, we define an interval as one of the following sets. (a, b) = {x : a < x < b}, (a, b] = {x : a < x b}, [a, b) = {x : a x < b}, [a, b] = {x : a x b}, (, a) = {x : x < a}, (, a] = {x : x a}, (a, ) = {x : x > a}, [a, ) = {x : x a}. Definition (Absolute value). Let F be an ordered field. We define the absolute value a of an element a F by { a if x 0, a = a if x < 0. Step 3: R is a complete ordered field. There are several ways to define completeness equivalently. Here it is done through least upper bounds (or greatest lower bounds). Definition (Upper and lower bounds). Let F be an ordered field. A set E in F is said to be bounded above provided there is an element b F such that x b for all x E; the number b is called an upper bound of E. A set E in F is said to be bounded below provided there is an element b F such that x b for all x E; the number b is called an lower bound of E. A set E in F is said to be bounded if E is bounded above and is bounded below. Remark. If a set of real numbers E is not bounded above, then we denote sup E = ; if a set of real numbers E is not bounded below, then we denote inf E =. Definition (Supremum and infimum).

8 1.3. CARDINALITY 8 Let E be a set that is bounded above. We say b is the supremum (or least upper bound) of the set E, denoted by sup E, if b is an upper bound of E and b c for any upper bound c of E. Let E be a set that is bounded below. We say b is the infimum (or greatest lower bound) of the set E, denoted by inf E, if b is a lower bound of E and b c for any lower bound c of E. Theorem 1.1. (i). A number a = sup E for a set of real numbers E iff a is an upper bound of E and for any ε > 0, there exists x(ε) E such that x(ε) > a ε. (ii). A number a = inf E for a set of real numbers E iff a is a lower bound of E and for any ε > 0, there exists x(ε) E such that x(ε) < a + ε. Definition (Completeness by supremum). Let F be an ordered field. We say F is complete if it satisfies the following axiom. (A11). For any subset E of F that is bounded above, there exists sup E F. Remark. Equivalently, we can define completeness by infimum: Let F be an ordered field. We say F is complete if for any subset E of F that is bounded below, there exists inf E F. See Problem 1-5 to transform between supremum and infimum. Question. Among Q and R, which one(s) are complete ordered fields, why not for Q and why for R? Question. Let E = {x Q : x 2 < 2}. Is sup E Q? In fact, we define 2 as a new real number (which is irrational.) After Step 3, we can do limits on R. Theorem 1.2. Between any two distinct real numbers, there is a rational number and an irrational number. Question. Prove that each real number is the supremum of a set of rational number. Proof. If x is rational, then x = sup E for E = {x}. If x is irrational, then there is a rational number x 1 (x 1, x) by the above theorem; there is a rational number x 2 (x 1/2, x) by the above theorem again. Inductively, there is a set of rational numbers E = {x 1, x 2,...} such that x n (x 1/n, x). One can see that sup E = x. Problems For a nonempty set of real numbers E, prove that inf E = sup E iff E consists of a single point Use the completeness axiom to prove that every nonempty set of real numbers E that is bounded below has an infimum and that inf E = sup{ x : x E} Prove that each real number is the supremum of a set of irrational numbers Cardinality Recall that two sets A and B are equipotent if there is a bijection between A and B, in this case, we say that A and B have the same cardinality, denoted by Card (A) = Card (B). In fact, Definition. Let A and B be two sets.

9 1.3. CARDINALITY 9 If there is a function f : A B that is surjective, then we say that the cardinality of A is greater than or equal to the cardinality of B, denoted by Card (A) Card (B); If there is a function f : A B that is injective, then we say that the cardinality of A is less than or equal to the cardinality of B, denoted by Card (A) Card (B); Card (A) > Card (B) if Card (A) Card (B) and Card (A) Card (B); Card (A) < Card (B) if Card (A) Card (B) and Card (A) Card (B). Cardinality generalizes the simple concept of number of elements of a set. Definition (Finite, countable, uncountable sets). A set A is said to be finite if either it is empty or there is a natural number n for which E is equipotent to the set {1,..., n}. A set A is said to be countably infinite if it is equipotent to N. A set A is said to be countable if it is either finite or countable infinite, that is, it is equipotent with a subset of N. A set A is said to be uncountable if it is not countable. A set E is finite iff its element can be enumerated as x 1 = f(1),..., x n = f(n) by the bijection f : {1,..., n} E; E is countably infinite iff its elements can be enumerated by x 1 = f(1),..., x n = f(n),..., by the bijection f : N E. Question. Examples of countable and uncountable sets? In particular, are N, Z, Q, R, R\Q countable? Theorem 1.3. Any subset of a countable set is countable. In particular, any set of natural numbers is countable. Definition (The power set). Given a set A, the set of all the subsets of A is called the power set of A and is denoted by P(A). Example. Let A = {1, 2, 3}. Then Remark. N. P(A) = {, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} }. If A has N elements, then P(A) has 2 N elements, which is strictly greater than Theorem 1.4. (i). The Cartesian product of countable sets is countable. (ii). The countable union of countable sets is countable. (iii). For any set A, Card (P(A)) > Card (A). Here, P(A) is the power set of A. Example. Card (P(N)) = Card (R). Theorem 1.5. A nonempty interval of real numbers is uncountable. uncountable. In particular, R is Problems Let a, b, c, d R, a < b, and c < d. Prove that the two intervals (a, b) and (c, d) are equipotent.

10 CHAPTER 2 Topological spaces and continuous functions 2.1. Topological spaces Definition (Topological spaces). A topology T on a set X is a collection of subsets of X that satisfies (i)., X T. (ii). The union of the elements in any subcollection of T is in T. That is, for any F T, E T. E F (iii). The intersection of the elements of any finite subcollection of T is in T. That is, for any E 1,..., E n T, n E i T. i=1 A set X for which a topology T has been specified is called a topological space. The elements in T are called the open sets. Example (Discrete topology and trivial topology). Let X be a set. The discrete topology is its power set P(X); the trivial (or indiscrete) topology is {, X}. Question. Provide a topology on X = {a, b, c}. Observe the many topologies on such a set. Example (Finite complement topology). Let X be a set. Let T f be the collection of all subsets U X such that U c is finite or is all of X. Then T f is a topology on X, called the finite complement topology. Definition. Let T 1 and T 2 be two topologies on a set X. We say that T 2 is finer (resp. strictly finer) than T 1 if T 2 T 1 (resp. T 2 T 1 ). In this case, we also say that T 1 is coarser (resp. strictly coarser) than T 2. We say that T 1 and T 2 are comparable if T 1 T 2 or T 2 T 1. Definition (Basis). Let X be a set. A basis for a topology on X is a collection B of subsets of X (called the basis elements) such that (i). For each x X, there is at least one basis element B containing x. (ii). If x belongs to the intersection of two basis elements B 1 and B 2, then there is a basis elements B 3 containing x such that B 3 B 1 B 2. If B satisfies these two conditions, then we define the topology T generated by B as follows: A subset U X is said to be open in X (that is, to be an element of T ) if for each x U, there is a basis element B B such that x B and B U. Note that each basis element is itself an element of T. Example (Topology generated by basis). The discrete topology is generated by the basis B as the collection of all singleton subsets of X. 10

11 2.1. TOPOLOGICAL SPACES 11 Example. Let B 1 be the collection of all circular regions (interiors of circles) in the plane. Then B 1 is a basis and generates a topology T 1 on the plane. That is, a subset U of the plane is open if every x U lies in some circular region contained in U. Let B 2 be the collection of all rectangular regions (interiors of rectangles) in the plane. Then B 2 is a basis and generates a topology T 2 on the plane. That is, a subset U of the plane is open if every x U lies in some rectangular region contained in U. In fact, T 1 = T 2 (by Proposition 2.3). This shows that the basis of a topology is not unique. Proof of T generated by a basis X is a topology. (i). It is vacuously true that ; it is also true that X T by Condition (i) in the definition of basis. (ii). Let F T and x U F U. Then there exists U F such that x U. Since U is open, there is a basis element B B such that x B and B U. Hence, U F U B and therefore is open. (iii). We prove by induction. First we show the intersection of two open sets is open. Let x U 1 U 2 for some U 1, U 2 T. Then there exist basis elements B 1 U 1 and B 2 U 2 such that x B 1 and x B 2. Hence, x B 1 B 2. By Condition (ii) in the definition of basis, there is an basis element B 3 B 1 B 2 such that x B 3. Therefore, U 1 U 2 B 3 and is then open. Suppose that the intersection of n 1 open sets is open. Since U 1 U n = (U 1 U n 1 ) U n, in which U 1 U n 1 is open by induction, U 1 U n is open by the first step. Remark (Principle of mathematical induction). For each natural number n, let S(n) be some mathematical assertion. Suppose that S(1) is true and S(k) is true implies that S(k + 1) is also true for all natural numbers (or the statement that S(1),...,S(k 1) are all true implies that S(k + 1) is true.) Then S(n) is true for all natural numbers. Proposition 2.1. Let B be a basis on a set X. Then the topology T generated by B is the collection of all unions of elements of B. Proof. Since B T, it is then clear that any union of basis elements is in T. On the other hand, let U be open and x U, there is a basis element B x such that B x x and B x U. Observe that x U B x = U (Both inclusions are immediate.) and the proof is finished. Proposition 2.2. Let T be a topology on a set X. Suppose that C is the collection of open sets such that for each open set U of X and each x U, there is an element C of C such that x C U. Then C is a basis of T. Proof. See Problem 2-1. Proposition 2.3. Let B 1 and B 2 be two bases for topologies T 1 and T 2 on a set X, receptively. Then the following statements are equivalent. (i). T 2 is finer than T 1. (ii). For each x X and each basis element B 1 B 1 containing x, there is a basis element B 2 B 2 such that x B 2 B 1. Proof.

12 2.1. TOPOLOGICAL SPACES 12 (ii) (i). Let U T 1. For any x U, there exists B 1 B 1 such that B 1 U since B 1 generates T 1. By Condition (ii), there exists B 2 B 2 such that x B 2 B 1. Hence, U T 2 since B 2 generates T 2. (i) (ii). For each x X and each basis element B 1 B 1 containing x, B 1 T 1 and hence B 1 T 2 since T 2 is finer than T 1. Because B 2 generates T 2, there is B 2 B 2 such that x B 2 B 1. Example (Topologies on R). We define several topologies on R: (1). Let B 1 = {(a, b) : < a < b < }. The topology generated by B 1 is called the standard topology on R and is denoted as R (if no confusion is caused). Check that only the intervals of the forms (a, b), (a, ), and (, a) are open in this topology. (2). Let B 2 = {[a, b) : < a < b < }. The topology generated by B 2 is called the lower limit topology on R and is denoted as R l. (It is sometimes called the Sorgenfrey line.) (3). Let K = {1/n : n N} and B 3 = {(a, b) : < a < b < } {(a, b) \ K : < a < b < }. The topology generated by B 3 is called the K-topology on R and is denoted as R K. (4). Let B 4 = {(a, b] : < a < b < }. The topology generated by B 4 is called the upper limit topology on R and is denoted as R u. We have that R R l. First, R R l : For each x R and each basis element (a, b) B 1 containing x, there is a basis element [x, b) B 2 such that x [x, b) (a, b). Second, R l R: For each x R and the basis element [x, b) B 2 containing x, there is no basis element (a, b) B 1 such that x (a, b) [x, b). We have that R R K. First, R R K : For each x R and each basis element (a, b) B 1 containing x, there is a basis element (a, b) B 3 such that x (a, b) (a, b). Second, R K R: For the basis element ( 1, 1) \ K B 3 containing 0, there is no basis element (a, b) B 1 such that 0 (a, b) ( 1, 1) \ K. We have that R l and R u are not comparable. First, R u R l : For each x R and each basis element (c, x] B 4 containing x, there is no basis element [a, b) B 2 such that x [a, b) (c, x]. Second, R l R u : For each x R and the basis element [x, c) B 2 containing x, there is no basis element (a, b] B 4 such that x (a, b] [x, c). We have that R l and R K are not comparable. First, R K R l : For the basis element ( 1, 1) \ K B 3 containing 0, there is no basis element [a, b) B 2 such that 0 [a, b) ( 1, 1) \ K. Second, R l R K : For the basis element [2, b) B 2 containing 2, there is no basis element (c, d) B 3 such that 2 (c, d) [2, b) and there is no basis element (c, d) \ K B 3 such that 2 ((c, d) \ K) [2, b). Definition (Subspace topology). Let X be a topological space with topology T. If Y is a subset of X, then the collection T Y = {U Y : U T } is a topology on Y, called the subspace topology. With this topology, Y is called the subspace of X; the open sets consist of all intersections of open sets of X with Y. We say U is open in Y (or open relative to Y ) if U T Y. Remark. It is straightforward to see that if Y is open in X, then the open sets in Y are also open in X. Otherwise, it is not necessarily true. For example, let R be equipped with the standard topology and Y = [0, ) R. Then [0, b) is open in Y (since [0, b) = (, b) Y in which (, b) is open in R) but is not open in R. Proposition 2.4. If B is a basis for the topology T of X, then the collection B Y = {B Y : B B} is a basis for the subspace topology T Y on Y.

13 2.2. CLOSED SETS AND LIMIT POINTS 13 Proof. Let U Y be an open set in T Y, in which U is open in X. For each y U Y, y U and there exists a basis element B B such that x B U, since B is a basis for T. Hence, B Y B Y and y B Y U Y. By Proposition 2.2, B Y = {B Y : B B} is a basis for T Y. Problems Prove Proposition Let {T α } be a family of topologies on a set X. Prove that α T α is a topology on X Let {T α } be a family of topologies on a set X. Is α T α is a topology on X? Prove your assertion On R, let T 1 be the standard topology, T 2 = R l, T 3 = R K, T 4 = R u, and T 5 be the finite complement topology. Determine for each of these topologies, which of others it contains Let T be a topology on a set X and A Y X. Prove that (T Y ) A = T A, that is, the topology A inherits as a subspace of Y is the same as the topology it inherits as a subspace of X Let T 2 be a strictly finer topology than T 1 on a set X. Suppose that Y X and Y. Is (T 2 ) Y strictly finer than (T 1 ) Y? Prove your assertion Closed sets and limit points Definition (Closed sets). Let X be a topological space. A subset V X is said to be closed if X \ V is open. Example. (1). In any topology on a set X, the sets and X are both open and closed. i (2). In the discrete topology T = P(X), every set is both open and closed. In fact, every set in a topology T is both open and closed iff T = P(X). (3). In the finite complement topology, the closed sets consists of X and all the finite subsets of X. (4). In the standard topology on R, the intervals [a, b], (, a], and [a, ) are closed; the intervals (a, b] and [a, b) are neither open nor closed. (5). In the lower limit topology R l, classify all the intervals as open or/and closed sets. See Problem 2-7. Using De Morgan s law, we can establish Theorem 2.5. Let X be a topological space. Then (i). and X are closed. (ii). Arbitary intersections of closed sets are closed (iii). Finite unions of closes sets are closed. Remark. Instead of using open sets, one could just as well specify a topology on a space by giving a collections of sets (to be called closed sets ) satisfying the three properties of the above theorem. One could then define open sets as the complements of closed sets and proceed just as before. The two procedures using open sets or closed sets are equivalent, however, most mathematicians prefer to use open sets to define topologies. Theorem 2.6. Let X be a topological space and Y X be equipped with the subspace topology. Then A Y is closed iff A = V Y for some closed set V X. In this case, we say that A is closed in Y. i A set that is both open and closed are sometimes called a clopen set. This terminology is however not widely adapted.

14 2.2. CLOSED SETS AND LIMIT POINTS 14 Proof. Sufficiency: Let A = V Y in which V X is closed. Then X \ V is open and Y \ A = Y \ (V Y ) = Y (X \ V ) is open in the subspace topology. Necessity: Let A be closed in Y. Then Y \ A is open in the subspace topology. There exists an open set U X such that Y \ A = U Y. Hence, A = Y \ U = Y (X \ U), in which X \ U is closed. Remark. It is straightforward to see that if Y is closed in X, then the closed sets in Y are also closed in X. Otherwise, it is not necessarily true. For example, let R be equipped with the standard topology and Y = (0, ) R. Then (0, b] is closed in Y (since (0, b] = (, b] Y in which (, b] is closed in R) but is not closed in R. Definition (Interior and closure). Let X be a topological space and A X. Define The interior of A Int A = U. U A, U open That is, the interior of A is the union of all open sets contained in A. the closure of A A = V. V A, V closed That is, the closure of A is the intersection of all closed sets containing A. Remark. It is clear that x Int A iff there is an open set U x such that U A. Int A is open and A is closed. Int A A A. A is open iff Int A = A; A is closed iff A = A. Example. Notice that the interior and closure of a set in a topological space X depend on the given topology. For example, (1). Let R be the standard topology on R. Then Int [a, b) = (a, b) and [a, b) = [a, b]; Int (a, b] = (a, b) and (a, b] = [a, b]. (2). Let R l be the lower limit topology on R. Then Int [a, b) = [a, b) and [a, b) = [a, b); Int (a, b] = (a, b) and (a, b] = [a, b]. Theorem 2.7. Let X be a topological space and Y X be equipped with the subspace topology. Then for any A Y, the closure of A in Y equals A Y, in which A is the closure of A in X. Proof. Denote B the closure of A in Y. We need to show that B = A Y. B A Y : Since A is closed in X, A Y is closed in Y by Theorem 2.6. Notice that A A Y. So B A Y since the closure of A in Y is the intersection of all closed sets containing A. A Y B: Since B is closed in Y, B = V Y for some closed set V X by Theorem 2.6. Notice that A V. So A V since the closure of A in X is the intersection of all closed sets containing A. Therefore, A Y V Y = B. The definition of the interior if A implies that x Int A iff there is an open set U such that x U A. While the definition of the closure of a set A does not provide a convenient way to find the closure, since all closed sets containing A are too big to work with. Hence, the following theorem gives two equivalent statements for the closure.

15 2.2. CLOSED SETS AND LIMIT POINTS 15 Theorem 2.8. Let X be a topological space and A X. Then (i). x A iff every open set U x intersects A, i.e. U A. (ii). Suppose that the topology of X is given by a basis B. Then x A iff every B B and B x intersects A. Remark. We say an open set U x is a neighborhood of x. Then (i) can be rephrased as x A iff every neighborhood of x intersects A. Proof. (i). Let (P ) be x A and (Q) be every open set U x intersects A. Then it suffices to prove that (P ) (Q). If (P ), i.e. x A, then x X \ A, which is an open set since A is closed, i.e. (Q). If (Q), i.e. there is an open set U x such that U A =, then A (X \ U), in which X \ U is closed. But A is the intersection of all closed sets containing A. Hence, (X \ U) A so x A, i.e. (P ). (ii). If every open set containing x intersects A, then every basis element containing x intersects A; if every basis element containing x intersects A, then every open set U x there is a basis element B U and B x, so U A B A. Example. Let R be equipped with the standard topology. Then (a, b) = [a, b] and {1/n : n N} = {0} {1/n : n N}. Definition (Limit points). Let X be a topological space and A X. We say that x is a limit point (or cluster point, point of accumulation) of A if every neighborhood U of x intersects A in some point other than x, i.e. (U A) \ {x}. The set of limit points of A is denoted by A. Example. Let R be equipped with the standard topology. Then the set of limit points of (a, b) is [a, b]; the only limit point of {1/n : n N} is 0. Theorem 2.9. Let X be a topological space and A X. Then A = A A. Proof. A A A: A A follows from the definition of closure, while A A follows Theorem 2.8, that is, if x A, then every neighborhood of x intersects A so x A. A A A : Let x A. If x A, then x A A ; if x A, since every neighborhood U of x intersects A, U A. But x A, (U A) \ {x} so x A. Since a set is closed iff it contains its closure, we have the following corollary. Corollary Let X be a topological space and A X. Then A = A iff A contains all its limit points, i.e. A A. Definition (Convergence of a sequence). Let X be a topological space and {x n } n=1 X be a sequence. We say that the sequence {x n } converges to x X, denoted by x n x, if for every neighborhood U of x there is N N such that x n U for all n N. In this case, we say that x is a limit of the sequence. It is not always true that convergent sequence has a unique limit point. For example, i let X = {a, b, c} be equipped with the topology {, {a, b}, {b, c}, {b}, X}. Then the sequence {x n } i In the trivial topology, every sequence is convergent and all the points are its limit point. In the discrete topology, a sequence x n x iff x n = x with finite exceptions of n N.

16 2.2. CLOSED SETS AND LIMIT POINTS 16 for x n = b converges to b (since all the points in the sequence are in the neighborhood {b} of b), to a (since all the points in the sequence are in the neighborhood {a, b} of b), and to c (since all the points in the sequence are in the neighborhood {b, c} of c). i To avoid such situation, we need the following concept Hausdorff spaces. Definition (Hausdorff space). We say that a topological space X is a Hausdorff space if for every x, y X and x y, there are open sets U x and W y such that U W =. Remark. Notice that X = {a, b, c} equipped with the topology {, {a, b}, {b, c}, {b}, X} is not a Hausdorff space since there are no disjoint open sets containing b and c, respectively. Example. Among R, R l, R u, R K, R f, which ones are Hausdorff spaces? Lemma Every finite set in a Hausdorff space is closed. Proof. It suffices to show that the singleton set A = {x 0 } is closed in a Hausdorff space X, since finite union of closed sets is closed. Let x X and x x 0, i.e. x A. Then there are open sets U x 0 (i.e. U A) and W x such that U W. Hence, W A W U =. By Theorem 2.8, x A. This is true for all x X and x x 0 so A = A = {x 0 }, which means that {x 0 } is closed. Theorem Let X be a Hausdorff space and A X. Then x is a limit point of A iff every neighborhood of x contains infinitely many points of A. Proof. Sufficiency: If every neighborhood of x contains infinitely many points of A, then x is a limit point of A by the definition. Necessity: Let x be a limit point of A. Suppose that there is a neighborhood U of x intersecting A at finitely many points. Let U (A\{x}) = {x 1,..., x n }. Then U \{x 1,..., x n } x and is open by the previous lemma. But (U \ {x 1,..., x n }) A =, contradicting with the fact that x is a limit point of A. Theorem Let X be a Hausdorff space. Then a sequence of points of X converges to at most one point in X. Proof. See Problem Problems In the lower limit topology R l, classify all the intervals as open or/and closed sets Let X be a topological space and Y X be equipped with the subspace topology. Suppose that Y is closed in X and A Y. Prove A is closed in Y iff A is closed in X Let X be a topological space and Y X be equipped with the subspace topology. Suppose that A Y. Then is that A is closed in Y iff A is closed in X? Prove your assertion Let X be a topological space and A, B, A α be the subsets of X. Prove that (1). If A B, then A B. (2). A B = A B. (3). α A α α A α ; provide an example that the equality fails. (4). A B A B; provide an example that the equality fails. i It is interesting to ask if it is possible for a constant sequence not to converge to the constant. The answer is no: Let {x n } be a constant sequence in a topological space X such that x n = c for all n N. Suppose that x n c. Then there is a neighborhood U of c such that for every N N, there exists some m N such that x m U. But this means that c = x m U, contradicting with the fact that U is a neighborhood of c.

17 2.3. CONTINUOUS FUNCTIONS 17 (5). A \ B A \ B; provide an example that the equality fails Let X be a Hausdorff space. Prove that a sequence of points of X converges to at most one point in X Let X be a topological space and Y X be equipped with the subspace topology. Suppose that X is a Hausdorff space. Prove that Y is also a Hausdorff space Let X be a topological space and A X. Define the boundary of A by A = A X \ A. (1). Prove that Int A A = and A = Int A A. (2). Prove that A = iff A is both open and closed. (3). Prove that U is open iff U = U \ U. (4). If U is open, is U = Int (U)? Prove your assertion Continuous functions Definition (Continuous functions). Let X and Y be two topological spaces. function f : X Y is continuous if f 1 (U) is open in X for every open set U in Y. We say a Remark. Continuity of a function f : X Y of course depends on the function itself, but also on the topologies on the domain X and on the codomain Y. For example, let R and R l be the standard and lower limit topologies on R. Define the function f : R R be f(x) = x. Then f is continuous relative to the topologies R and R. f is not continuous relative to the topologies R and R l, since the preimage of the open set [a, b) in R l, f 1 ([a, b)) = [a, b), is not open in R. f is continuous relative to the topologies R l and R, since the preimage of any open set in R is itself and is open in R l (because R R l ). Therefore, any discussion about the topological properties (open, closed, continuous, etc) is under the premise of topologies. If one wishes to emphasize the dependence of continuity on topologies, one can say f : X Y is continuous relative to the topology T X on X and the topology T Y on Y, and denote the function by f : (X, T X ) (Y, T Y ). Theorem Let X and Y be two topological spaces and f : X Y be a function. Then the following statements are equivalent. (i). f is continuous. (ii). For every closed set B Y, f 1 (B) is closed in X. (iii). For every A X, f(a) f(a). (iv). For every x X and every neighborhood W of f(x), there is a neighborhood U of x such that f(u) W. (v). Suppose that B Y generates the topology on Y. For every B B Y, f 1 (B) is open in X. Proof. We follow the program that (i) (ii), (i) (iii), (iii) (ii), (i) (iv), and (i) (v). (i) (ii). Recall that f 1 preserves differences of sets. Since f 1 (Y ) = X, X \ f 1 (B) = f 1 (Y ) \ f 1 (B) = f 1 (Y \ B). So if B is closed, then Y \ B is open and f 1 (Y \ B) is open since f is continuous. It thus follows that f 1 (B) is closed. (ii) (i). Similarly as before, if B is open, then Y \ B is closed and f 1 (Y \ B) is closed by assumption. It follows that f 1 (Y \ B) = f 1 (Y ) \ f 1 (B) = X \ f 1 (B) is closed. Hence, f 1 (B) is open so f is continuous. (i) (iii). Let y f(a). We need to show that y f(a). Since y f(a), there is x A such that y = f(x). Let W be a neighborhood of y in Y. f 1 (W ) := U is open and contains

18 2.3. CONTINUOUS FUNCTIONS 18 x. Hence, U is a neighborhood of x in X and so U A since x A. It then follows that W f(a) f(u) f(a) f(u A) so y f(a). (iii) (ii). Let B Y be closed and denote A = f 1 (B). We need to show A is closed and it suffices to prove that A = A. Notice that f(a) = f(f 1 (B)) B. Let x A. Then f(x) f(a) f(a) by assumption. But f(a) B = B since B is closed. Then f(x) B so x f 1 (B) = A, completing the proof. (i) (iv). Let x X and W be a neighborhood of f(x) in Y. Then U := f 1 (W ) is open by (i) and contains x. Hence, f(u) W and U is a neighborhood of x in X. (iv) (i). Let W Y be open and U = f 1 (W ). For each x U, f(x) f(u). By (iv), there is a neighborhood U x of x such that f(u x ) W. But this means that U x U and U = x U x is therefore open. (i) (iv). It is straightforward since any basis element in B Y is open. (iv) (i). It is also straightforward since any open set in Y is a union of basis element in B Y and f 1 preserves unions. Definition (Homeomorphisms). Let f : X Y be a bijection between two topological spaces. We say f is a homeomorphism if f and f 1 are both continuous. Two topological spaces are said to be homeomorphic if there is a homeomorphism between them. Remark. One can say that f : X Y is a homeomorphism between (X, T X ) and (Y, T Y ) if one wishes to emphasize the dependence of a homeomorphism on the topologies. Example. (1). It is easy to see the identity map f : (X, T ) (X, T ) defined by f(x) = x is a homeomorphism. (2). Let R be equipped with the standard topology. Then any linear function f : R R with nonzero slope is a homeomorphism. (3). Let (a, b) and (c, d) be equipped with the subspace topology from the standard topology R. Then the linear function from (a, b) to (c, d) is a homeomorphism. (4). Let R be equipped with the standard topology and ( π/2, π/2) R be equipped with the subspace topology. Then f : ( π/2, π/2) R by f(x) = tan x is a homeomorphism. (5). The identity map f : R R l is not a homeomorphism. In fact, R and R l are not homeomorphic, i.e. there are no homeomorphism between them. Theorem Let X, Y, Z be topological spaces. (i). The constant function f : X Y defined by f(x) = y 0 for all x X and some y 0 Y is continuous. (ii). Let A X be equipped with the subspace topology. Then the inclusion function i : A X defined by i(x) = x for all x A is continuous. (iii). Let f : X Y and g : Y Z be continuous. Then the composition g f : X Z defined by g f(x) = g(f(x)) for x X is continuous. (iv). Let f : X Y be continuous and A X be equipped with the subspace topology. Then the restriction function f A : A Y defined by f A (x) = f(x) for all x A is continuous. Proof. (i). Let f(x) = y 0 for all x X and some y 0 Y. Let U Y be open. If y 0 U, then f 1 (U) = X is open in X; if y 0 U, then f 1 (U) = is open in X. This means that f is continuous. (ii). Let i(x) = x be the inclusion map from A to X. Notice that i 1 (U) = U A is open in A from the definition of subspace topology. This means that i is continuous.

19 2.3. CONTINUOUS FUNCTIONS 19 (iii). Notice that (g f) 1 = f 1 g 1. Let U Z be open. Then (g f) 1 (U) = f 1 (g 1 (U)), in which g 1 (U) is open in Y since g is continuous. So f 1 (g 1 (U)) is open in X since f is continuous. This means that g f is continuous. (iv). Let U Y be open. Then (f A ) 1 (U) = f 1 (U) A, in which f 1 (U) is open since f is continuous. It follows that f 1 (U) A is open in A from the definition of subspace topology. This means that f A is continuous. Alternatively, f A = f i, in which i : A X is the inclusion function. Then f A is continuous because both i and f are continuous. Theorem 2.16 (The Pasting Lemma). Let X be a topological space. Suppose that X = A B with two closed sets A, B X. Let f : A Y and g : B Y be continous. If f(x) = g(x) for all x A B, then f and g combine to define a continuous function h : X Y, defined by { f(x) for x A, h(x) = g(x) for x B. Proof. Let W Y be closed. It suffices to show that h 1 (W ) is closed in X. Notice that h 1 (W ) = f 1 (W ) g 1 (W ), in which f 1 (W ) is closed in A and g 1 (W ) is closed in B since f and g are continuous. It then follows that f 1 (W ) and g 1 (W ) are closed in X since A and B are closed. Hence, h 1 (W ) is closed. Remark. The Pasting Lemma is also valid if A and B are open. It is a special case of Problem Note that f(x) = g(x) on the A B is only required so that h is well defined. Example. (1). The function f : R R defined by f(x) = { x for x 0, x/2 for x 0, is continuous, by The Pasting Lemma since both pieces are continuous. (2). The function f : R R defined by { x 2 for x < 0, f(x) = x + 2 for x 0, is not continuous even though both pieces are continuous. Problems A function f : R R is continuous iff for every x R and ε > 0, there exists δ > 0 such that f(y) f(x) < ε if y (x δ, x + δ) Let X and Y be two topological spaces and f : X Y be continuous. Suppose that A X and x A. (Recall that A is the set of limit points of A.) Then is f(x) f(a)? Prove your assertion Let T 1 and T 2 be two topologies on a set X. Let f : (X, T 1 ) (X, T 2 ) be the identity map. Prove that (1). f is continuous iff T 1 is finer than T 2. (2). f is a homeomorphism iff T 1 = T 2.

20 2.4. METRIC TOPOLOGY Let X and Y be two topological spaces and f : X Y be a function. Suppose that X = α U α in which U α X are open sets. If f Uα is continuous for all α, then prove that f is continuous Let X be a topological space and f, g : X R be two continuous functions. Define h(x) = min{f(x), g(x)}. Prove that h is also continuous Metric topology One of the most important and frequently used ways of imposing a topology on a set is to define the topology via a metric on the set, called the metric topology. Metric topology lie at the heart of modern analysis, the examples of which include the Euclidean (or standard) topology on R n. Definition (Metric). A metric on a set X is a function d : X X [0, ) that satisfies (i). d(x, y) = 0 iff x = y, (ii). d(x, y) = d(y, x) for all x, y X, (iii). d(x, y) + d(y, z) d(x, z) for all x, y, z X (triangle inequality). Given a metric d on X, d(x, y) is often called the distance between x and y. Example. In R n, denote a point x = (x 1,..., x n ) R n. Then ) 1 d(x, y) = ( x 1 y x n y n 2 2 is a metric on R n. In fact, for any 1 p, ) 1 d p (x, y) = ( x 1 y 1 p + + x n y n p p is a metric on R n. In particular, and is called the square metric. d 1 (x, y) = x 1 y x n y n ; ρ(x, y) := d (x, y) = sup { x i y i } i=1,...,n Definition (Metric topology). Let X be a set equipped with a metric d. For x X and ε > 0, denote B d (x, ε) = {y X : d(x, y) < ε} the ε-ball centered at x. Then the metric topology induced by the metric d on X is generated by the basis B d = {B d (x, ε) : x X, ε > 0}. We say X is a metric space if it is a topological space with the topology induced by some metric d on X. The above definition implies that on a set X equipped with a metric topology induced by a metric d, a set U X is open iff for every x U, there is δ > 0 such that B d (x, δ) U. Remark. We show that B d is a basis. (i). For each x X, there is B d (x, ε) B d that contains x.

21 2.4. METRIC TOPOLOGY 21 (ii). Let x B d (x 1, ε 1 ) B d (x 2, ε 2 ). First, x B d (x 1, ε 1 ) implies that d(x, x 1 ) < ε 1. Choose δ 1 = ε 1 d(x, x 1 ). For every y B d (x, δ 1 ), d(y, x) < δ 1 so triangle inequality implies that d(y, x 1 ) d(y, x) + d(x, x 1 ) < δ 1 + d(x, x 1 ) = ε 1. Hence, y B d (x 1, ε 1 ) so B d (x, δ 1 ) B d (x 1, ε 1 ). Similarly, we can choose δ 2 = ε 2 d(x, x 2 ) so B d (x, δ 2 ) B d (x 2, ε 2 ). Let δ = min{δ 1, δ 2 }. Then B d (x, δ) B d (x, δ 1 ) B d (x 1, ε 1 ) and B d (x, δ) B d (x, δ 2 ) B d (x 2, ε 2 ). This means that x B d (x, δ) B d (x 1, ε 1 ) B d (x 2, ε 2 ). Example. (1). The Euclidean topology on R n is induced by the metric d. In fact, the metric topologies induced by the metrics d p are identical for all 1 p. (2). Let X be a nonempty set. Define the metric { 0 if x = y, d(x, y) = 1 if x y. Then this metric induces the discrete topology on X. It is because the metric balls B d (x, ε) = {x} if ε 1 and B d (x, ε) = X if ε > 1. (3). Denote R ω = Π i=1x n with X n = R for all n N. Define d(x, y) = min{ x y, 1} for x, y R as the standard bounded metric on R. Then for x = (x 1, x 2,...) and y = (y 1, y 2,...), (4). ρ(x, y) := sup{d(x i, y i )} i N defines a metric on R ω. We call ρ the uniform metric on R ω and the topology induced by ρ the uniform topology. defines a metric on R ω. D(x, y) := sup i N { } d(xi, y i ) The following discussion is on the interpretation of topological statements (continuity, convergence, etc) in metric spaces Continuous functions involving metric spaces. Theorem Let X and Y be equipped with metric topologies T X and T Y, induced by metrics d X and d Y, respectively. Then f : X Y is continuous iff for any x X and ε > 0, there exists δ > 0 such that d Y (f(x), f(y)) < ε if d X (x, y) < δ. Proof. See Problem Lemma Let X be a topological space and A X. (i). If there is a sequence x n x for {x n } A and some x X, then x A. (ii). If in addition X is a metric space, then x A implies that there is a sequence x n x for {x n } A. Proof. (i). Suppose that there is a sequence x n x for {x n } A and some x X. Then for every neighborhood U of x, there is N N such that x n U for all n N. This means that {x n : n N} U A and x A. (ii). Suppose that X is a metric space. Let x A. Then for each n N, B d (x, 1/n) A. So there is x n A such that d(x n, x) < 1/n. Notice that x n x. i

22 2.4. METRIC TOPOLOGY 22 Theorem Let X and Y be two topological spaces. Let f : X Y. (i). If f is continuous, then for every convergence sequence x n x, we have that f(x n ) f(x). (ii). Suppose that in addition X is a metric space. If for every convergent sequence x n x, we have that f(x n ) f(x), then f is continuous. Proof. (i). Suppose that f is continuous. Let x n x. For every W f(x) open in Y, f 1 (W ) is open in X since f is continuous. Now f 1 (W ) is a neighborhood of x in X because f(x) W. Since x n x, there is N N such that x n f 1 (W ) for all n N. This means that f(x n ) W for all n N so f(x n ) f(x). (ii). By (iii) in Theorem 2.14, it suffices to show that f(a) f(a) for each A X. Let x A, by (ii) in the previous lemma, there is {x n } A such that x n x. By the assumption, f(x n ) f(x). By (i) in the previous lemma, that f(x) f(a) since f(x n ) f(a). Hence, f(a) f(a). Notice that here we do not need to require that Y is a metric space. Definition (Uniform convergence). Let X and Y be two topological spaces. Let f n : X Y be a sequence of functions and f : X Y be a function. We say that f n f pointwisely if f n (x) f(x) for all x X. Let Y be equipped with a metric topology induced by metric d. We say that f n f uniformly if for every ε > 0, there exists N N such that for all x X and n N. d(f n (x), f(x)) < ε Theorem Let f n : X Y be a sequence of continuous functions from a topological space X to a metric space Y. If f n f uniformly for some function f : X Y. Then f is continuous. Proof. Let W be open in Y. We need to show that f 1 (W ) is open in X. Choose x 0 f 1 (W ). It suffices to prove that there is a neighborhood U of x 0 such that U f 1 (W ) (i.e. f(u) W ) so f 1 (W ) is open in X. Denote y 0 = f(x 0 ). Then y 0 W since x 0 f 1 (W ). First choose ε > 0 such that B(y 0, ε) W. By uniform convergence of f n f, there exists N N such that d(f n (x), f(x)) < ε/3 for all x X and n N. Since f N is continuous, f 1 N (B(y 0, ε/3)) := U is open in X. Hence, d(f N (x), f N (x 0 )) < ε/3 if x U. Now we claim that f(u) W. Compute for x U that d(f(x), f(x 0 )) d(f(x), f N (x)) + d(f N (x), f N (x 0 )) + d(f N (x 0 ), f(x 0 )) < ε. So f(x) B(y 0, ε) W if x U. Problems Prove that every metric space is a Hausdorff space Prove Theorem Let X be a topological space and A X. Let x A. Is there a sequence x n x for {x n } A? Prove your assertion Let f n : X Y be a sequence of continuous functions from a topological space X to a metric space Y. If f n f pointwisely for some function f : X Y, then is f continuous? Prove your assertion.

23 2.5. PRODUCT TOPOLOGY Product topology Definition (Product topology). Let X i be topological spaces generated by the basis B i for i = 1,..., n. Then the product topology is defined to be a topology on X 1 X n that is generated by the basis B := {B 1 B n : B i B i, i = 1,..., n}. Notice that X 1 X n = if X i = for some i = 1,..., n. To avoid the trivial cases, we assume all the sets X i are nonempty. Theorem B in the above definition is a basis. Proof. Denote X = X 1 X n. Let x = (x 1,..., x n ) X. Then x i X i and there is a basis element B i B i such that x i B i. Hence, x B 1 B n B. Let C 1 = B 11 B 1n, C 2 = B 21 B 2n, and x = (x 1,..., x n ) C 1 C 2. Then x i B 1i B 2i. Since B i is a basis, there is B 3i B i such that x i B 3i B 1i B 2i. Now C 3 := B 31 B 3n B and x C 3 C 1 C 2. Remark. Let X i be topological spaces and X = X 1 X n be equipped with the product topology. Then the boxes U = U 1 U n X for which U i X i open are open in X. But there are other open sets different with the boxes, e.g. the union of two boxes. Definition (Projection maps). Let X = X 1 X n. The projection map π i : X X i is defined as π i (x) = x i for x = (x 1,..., x n ) X. Theorem Let X i be topological spaces and X = X 1 X n be equipped with the product topology. Then the projections maps π i : X X i are continuous for all i = 1,..., n. Proof. One only needs to observe that π 1 i (U) = X 1 U X n if open in X if U is open in X i. The product topology on R n are generated by the basis {(a 1, b 1 ) (a n, b n ) : < a i < b i <, i = 1,..., n}. Theorem The product topology on R n equals its Euclidean topology. Proof. Notice that in the ρ-metric on R n, B ρ (x, ε) = (x 1 ε, x 1 + ε) (x n ε, x n + ε) for x = (x 1,..., x n ) R n and ε > 0. It immediately follows that the product topology is finer than the ρ-metric topology since B ρ (x, ε) is a basis element for the product topology. For every x = (x 1,..., x n ) (a 1, b 1 ) (a n, b n ), x i (a i, b i ). So there is ε i > 0 such that (x i ε i, x i + ε i ) (a i, b i ). Choose ε = min{ε 1,..., ε n }. Then (x i ε, x i + ε) (a i, b i ) for all i = 1,..., n. This means that x B ρ (x, ε) (a 1, b 1 ) (a n, b n ). Hence, the ρ-metric topology is finer than the product topology. Problems Let X i be Hausdorff spaces and X = X 1 X n be equipped with the product topology. Prove that X is also Hausdorff Let X and Y be topological spaces. Let X Y and Y X be equipped with the product topology. Prove that X Y is homeomorphic to Y X.

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