Topology. December 7, 2004

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1 Topology December 7, Basic Definitions Definition 1.1 A topological space is a set X together with a family O of subsets of X satisfying the following properties: 1. φ, X O; 2. O 1, O 2 O O 1 O 2 O; 3. S O {O : O S} O. Members of O are called open sets. It follows immediately from the definition that intersection of finitely many open sets is open. Example For any set X, O = {φ, X} defines a topology, called the indiscrete topology on X. 2 For any set X, O = P(X) the power set of X, defines a topology, called the discrete topology on X. 3. Let X = {0, 1, 2} together with O = {φ, {1}, {0, 1}, {1, 2}, X} is a topological space. 4. R n with the usual open sets is a topological space. Exercise 1.1 Prove that a set O in R is open if and only if it is a disjoint union of countably many open intervals.[ For x, y O, define an equivalence relation on O by x y if and only if the interval [x, y] lies in O. Prove that each equivalence class determined by is an open interval. Then show that there are countably many equivalence classes.] Definition 1.2 Let X be a topological space. A subset C of X is called closed if its complement is open. Proposition 1.1 Let C be the family of closed sets of a topological space X. Then: 1. φ, X C; 2. C 1,..., C n C C 1... C n C; 3. S C {C : C S} C. 1

2 Definition 1.3 Let X be a topological space. A family B of open sets is called a base for the topology of X if for every open set O and x O, there exists B B such that x B O. A topological space is called second countable if it has a countable base. Example In R n with the usual topology, the family of open balls is a base for the topology. 2. In example??, the family {{0, 1}, {1, 2}} is not a base. Proposition 1.2 A family B of open sets of a topological space X is a base if and only if every open set of X is a union of some members of B. Proposition 1.3 Let X be a set. A family B of subsets of X is a base for some topology for X if and only if (i) union of sets in B is X, (ii) for every two members B 1, B 2 of B and each x B 1 B 2 there is B B such that x B B 1 B 2. Definition 1.4 Let X be a topological space and x X. A subset N of X is called a neighborhood of x if there exists an open set O with x O N. Note that by definition every open set is a neighborhood of each of its members. Exercise 1.2 Prove that a set U in a topological space is open if and only if every point in U has a neighborhood contained in U. Definition 1.5 Let X be a topological space and x X. A family B of neighborhoods of x is called a neighborhood base at x if for every neighborhood Nof x there exists B B such that B N. Proposition 1.4 Let X be a topological space and B a base for the topology. Then for each x X, the family {B B : x B} is a neighborhood base at x. Example 1.3 In R n, the family of open balls centered at x is a neighborhood base at x. Definition 1.6 A topological space X is called first countable if every point in X has a countable neighborhood base. Proposition 1.5 Every second countable topological space is first countable. Proof. Let X be a second countable topological space and let T = {O 1, O 2,...} be a countable base. Let x X. The (countable) family {O T : x O} forms a neighborhood base at x. So X is first countable. Exercise Prove that R n with the usual topology is second countable. 2

3 Definition 1.7 A subset A of a topological space is called dense if for every nonempty open set O, O A φ. A topological space is called separable if it has a countable dense subset. Exercise Prove that every second countable topological space is separable. 2. Find a countable dense subset of R n. Definition 1.8 Let X be a topological space and A a subset of X. The interior of A, denoted by int(a) is the largest open set contained in A, i.e. int(a) = {O : O open, O A} Dually, the closure of A, denoted by A, is the smallest closed set containing A, i.e. A = {C : C closed, C A} Exercise Let A be a subset of a topological space X. A point x is called an interior point of A if there exists an open set O such that x O A, i.e. A is a neighborhood of x. Prove that int(a) is equal to the set of interior points of A. 2. Prove that A = {x X : N x A φ for every neighborhood N x of x}. 3. Prove Proposition in textbook. 4. Prove Proposition in textbook. Let X is a topological space. Suppose to each x X there associates a neighborhood base B x at x containing solely of open sets. Then: (H1) each B x is nonempty and x B for every B B x ; (H2) B 1, B 2 B x B B x with B B 1 B 2 ; (H3) if B B x and y B, there exists C B y such that C B. Conversely if to each x X there associates a family B x of sets satisfying H(1)- (3), then there is a unique topology on X such that each B x is a neighborhood base at x. Moreover, each set in B x is open. Exercise 1.6 Prove the above statement. Let X be a topological space with topology O, and Y X. The family {O Y : O O} defines a topology on Y, called the relative topology on Y induced by (the topology of ) X. Example The relative topology on {0, 1, 2} induced by R with the usual topology is discrete. Proposition 1.6 Let X be a topological space and Y a subspace of X with the relative topology. Let A be a subset of Y. The closure of A relative to the relative topology of Y is equal A Y. Definition 1.9 A topological space X is connected if it is not a union of two disjoint nonempty open sets. A subset Y of X is connected if it is connected in its relative topology. 3

4 The following follows immediately from the definition. Proposition 1.7 Let X be a topological space. The following are equivalent: 1. X is connected. 2. X is not a union of two disjoint nonempty closed sets. 3. The only subsets of X that are both open and closed are φ and X. Proposition 1.8 A subset Y of X is connected if and only if there do not exist disjoint open sets O 1, O 2 such that O i Y φ, i = 1, 2 and Y O 1 O 2. Proposition 1.9 Let X be a topological space and {C i : i Λ} a family of connected subsets of X. If i Λ C i φ, then i Λ C i is connected. Example Any discrete space with more than one point is disconnected (not connected). 2. The set Q of rational numbers with the relative topology induced by R is disconnected. 3. A nonempty subset of R is connected if and only if it is a singleton or an interval. 4. An open subset O of R n is connected if and only if every two distinct points in O can be joined by a polygon inside O. Exercise Prove Example 3 above. 2. Prove Example 4 above. 3. Prove that a subset of R n is open if and only if it is a disjoint union of countably many open connected sets. Proposition 1.10 If A is a connected subset of a topological space and A B A, then B is also connected, in particular, A is connected. Proof. Let O 1, O 2 be disjoint open subsets of X such that B O 1 O 2. We must show that either B O 1 or B O 2. Since A B O 1 O 2 and A is connected, we have A O 1 or A O 2. Without loss of generality, assume A O 1. Then A O 1 B. Since the closure of A with respective to B is A B = B and O 1 B is closed relative to B (because the complement of O 1 B relative to B is O 2 B), we have B O 1 B, which is the same as B O 1. This proves that B is connected. Example 1.6 We will see later that continuous image of connected set is connected and therefore A = {(x, sin 1/x) : 0 < x 1} is connected. Since A = A {(0, y) : 1 y 1}, the above proposition implies that A any subset of {(0, y) : 1 y 1} is connected. This is somewhat counter intuitive. Let X, Y be topological spaces with topologies τ X and τ Y, respectively. The unique topology on X Y determined by the base {O Q : O τ X, Q τ Y } 4

5 is called the product topology on X Y. The product topology on the product of a finitely many topological spaces is defined similarly. Let X be a topological space. A family F of open sets of a topological space X is called an open cover if X = {O : O F}. X is called compact if every open cover of X has a finite subcover. A family F of subsets of X is said to have finite intersection property if the intersection of any finitely many members of F is nonempty. A family F of subsets of X is called a filter if (i) φ / F ; (ii) A, B F A B F ; (iii) A F and B A B F. A family F of subsets of X is called a filter base if (i) φ / F ; (ii) A, B F C F such that C A B. By definition every filter is a filter base. It is easy to prove that a filter base generates a filter in the following sense: Proposition 1.11 If F is a filter base, then {B : B A for some A F} is a filter. An x X is called a limit point of a filter base F if x {F : F F}. Theorem 1.1 Let X be a topological space. The following are equivalent: 1. X is compact; 2. Every family of closed sets of X with finite intersection property has a nonempty intersection; 3. Every filter base in X has a limit point. If X is second countable, each of these conditions is also equivalent to: 4. Every sequence in X has a convergent subsequence. Proof. 1. 2: If F is a family of closed sets with finite intersection property, then the family G = {F : F F} is a family of open sets any finite number of its members do not cover X. Consequently G does not cover X since X is compact. By DeMorie s theorem, the family F has a nonempty intersection. 2. 3: Let B be a filter base. Then the family {B : B B} is a family of closed sets satisfying condition 2 (why?) and hence has a nonempty intersection. This means B has a limit point (Exercise) 2 1. (Exercise) Suppose X is second countable (we need only X be first countable for this part of the proof.) Let x n, n = 1, 2,... be a sequence in X. Let B n = {x k : k n}. The family {B n : n = 1, 2,...} is a filter base, so it has a limit point x. Let {O 1, O 2,...} be a neighborhood base at x. We may assume that O 1 O Since x B 1, there exists n 1 1 such that x n1 O 1. Since x B n1+1, there exists n 2 > n 1 such that x n2 O 2. Continuing this process, we obtain a subsequence 5

6 of {x n : n = 1, 2,...} that converges to x Suppose X is second countable. Then every open cover of X has a countable subcover (why). Let this countable subcover be O 1, O 2,.... Now we need to show that this countable subcover has a finite subcover. Suppose not. Then for each n, there exists x n in the complement of O 1 O 2... O n. By 4, the sequence {x n, n = 1, 2,...} has a convergent subsequence with limit x. Since O 1, O 2,... cover X, x O k for some integer k 1. Since x is a subsequential limit of x n, n = 1, 2,,..., there exists l > k such that x l O k. This is a contradiction since x l is in the complement of O 1 O 2... O k... O l. Exercise Define compact subset of a compact topological space. 2. Prove that every closed subset of a compact topological space is compact. 3. Prove that product of two (and hence finitely many) compact spaces is compact. 4. Prove that product of two (and hence finitely many) connected spaces is connected. 5. (Heine-Borel) A subset of R is compact if and only if it is closed and bounded. 6. A subset of R n is compact if and only if it is closed and bounded. Definition 1.10 A topological space X is called Hausdorff if every two distinct points x, y in X can be separated by two disjoint open sets, i. e. there exist disjoint open sets O 1, O 2 such that x O 1 and y O 2. A topological space X is called Normal if every two disjoint closed sets C 1, C 2 in X can be separated by two disjoint open sets, i. e. there exist disjoint open sets O 1, O 2 such that C 1 O 1 and C 2 O 2. Proposition Every singleton in a Hausdorff space is closed. 2. Every compact subset of a Hausdorff space is closed. Definition 1.11 A mapping (or function) f : X Y from a topological space X to another topological space Y is continuous if f 1 (O) is open for every open set O of Y. Theorem 1.2 Let f : X Y be a continuous function from a topological space X to another topological space Y. Let A be a subset of X. 1. If A is compact, so is f(a). 2. If A is connected so is f(a). Theorem 1.3 (Extreme Value Theorem) Let X be a compact topological space and f : X R be a continuous map. Then f attains its maximum and minimum values. Theorem 1.4 (Intermediate Value Theorem) Let X be a connected topological space and f : X R be a continuous map. Then f(x) is either a singleton or an interval. 6

7 Definition 1.12 Let X be a set. A function d : X X R + is called a metric on X if it satisfies the following properties: 1. d(x, y) = 0 if and only if x = y; 2. d(x, y) = d(y, x) x, y X; 3. d(x, z) d(x, y) + d(y, z) x, y, z X. (X, d) is called a metric space. Example Let X be a nonempty set. Define d(x, y) = 1 if x y, = 0 if x = y. d is a metric on X. 2. Let X = R n. The Euclidean distance is a metric on X. If (X, d) is a metric space, and x X. For each real number r > 0, the open ball centered at x with radius r is defined to be the set B(x, r) = {y X : d(x, y) < r} A subset O of X is called open if for every x O, there exists r > 0 such that B(x, r) O. The collection of open sets of X defines a topology on X, called the topology induced by the metric d. Exercise 1.9 Let (X, d) be a metric space. Prove the following: 1. Every open ball is open. 2. The function d : X X R is continuous. 3. Let A be a nonempty subset of X. For each x X, define d(x, A) = inf{d(x, y) : y A}. Then d(x, A) is a continuous function on X. [ d(x, A) d(y, A) d(x, y)]. d(x, A) = 0 if and only if x A. 4. X is normal. [If A, B are disjoint closed sets, O A = {d : d(x, A) < d(x, B)} and O B = {x : d(x, B) < d(x, A)} are disjoint open sets separating A and B.] 5. X is first countable. It is second countable if and only if it is separable. [For an example of a separable first countable space that is not second countable, see Kelley, chapter 5, problem M.] Let (X, d) be a metric space. A sequence {x n : n = 1, 2,...} in X is called Cauchy if for every ɛ > 0 there exists an integer N > 0 such that d(x m, x n ) < ɛ for all m, n N. X is called complete if every Cauchy sequence in X converges. A subset A of a metric space (X, d) is called bounded if there exists M > 0 such that d(x, y) M for all x, y A. If A is bounded, the diameter of A is defined to be diam A = sup{d(x, y) : x, y A}. Let ɛ > 0. A subset N of X is called an ɛ-net if the family {B(x, ɛ) : x N} covers X. X is called totally bounded if it contains a finite ɛ-net for every ɛ > 0. Every totally bounded metric space is bounded. Theorem 1.5 Let (X.d) be a metric space. The following are equivalent: 1. X is compact. 2. X is complete and totally bounded. 3. Every sequence in X has a convergent subsequence. 7

8 Proof. We shall prove that 1 3, 3 2, 2 3 and then 2 and is the proof of 3 4 in Theorem??. 3 2: Since every Cauchy sequence with a convergent subsequence converges, the proof for completeness is immediate. If X is not totally bounded, then there exists ɛ > 0 such that no finite ɛ-nets exist. Let x 1 X be an arbitrary point. Since {x 1 } is not an ɛ-net, there exists x 2 X such that d(x 1, x 2 ) ɛ. Since {x 1, x 2 } is not an ɛ-net, there exists x 3 such that d(x 3, x i ) ɛ, i = 1, 2. Continuing in this way, we find a sequence x n, n = 1, 2,... such that d(x i, x j ) ɛ for all distinct positive integers i, j. This sequence has clearly no convergent subsequences. 2 3: Let x n, n = 1, 2,... be a sequence in X. Since X is covered with finitely many balls of radius 1, some subsequence of x n is contained in a ball of radius 1. Again, for the same reason, this subsequence has a subsequence that is contained in a ball of radius 1/2. Continuing in this way, and using the standard diagonal process, we obtain a subsequence whose tail starting from nth term is contained in a ball of radius 1/n, making it a Cauchy subsequence. By completeness, the subsequence converges. 2 and 3 1: For each positive integer n, let B n be a finite 1/n-net of X. Then the union of these B n, n = 1, 2,... is clearly a dense subset of X. Thus, X is separable and hence second countable. By 3 and Theorem??, X is compact. A subset A of a topological space X is called nowhere dense if the closure of A contains no nonempty open set, i.e. has empty interior. We say X is of first category if it is a countable union of nowhere dense sets, otherwise, X is said to be of second category. Theorem 1.6 (Baire category theorem) Every complete metric space is of second category. For proof, see p.356 in textbook. Let (X, d) be a metric space. A mapping f : X X is called a contraction if there exists 0 λ < 1 such that d(f(x), f(y)) λd(x, y) for all x, y X. A point x X is called a fixed point of f if f(x) = x. Theorem 1.7 (Banach contraction principle) Let (X, d) be a complete metric space and f : X X a contraction. Then f has a unique fixed point p. Moreover, for every x X the sequence {f n (x) : n = 1, 2,...} converges to p. 8