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General Topology §1. Topological Space 1.1 Definitions of topology & topological space Definition 1. Topology on set : T is a collection of subsets in X, satisfied that a) Ø, X T ; b) intersection of any finite number of elements of T belongs to T ; c) union of elements of any subfamily of T belongs to T . Remark: T includes Ø, X, and closed under finite intersection & arbitrary union . Example 1 Anti-discrete topology : {Ø, X } (trivial topology). Ex 2 Discrete topology : 2 X (Exp X) , all subsets of X. Ex 3 Sierpinski topology: X={0,1}, T ={Ø, {0}, X }. Definition 2. Topology space are called open sets. ( X ,T ); the elements of T For any set, normally we can define several different topologies Definition 3 Let T , T 2 be the topologies on set X, if T 1 T 2 , T 1 is weaker or coarser than T 2 . T 2 is stronger or finer than T 1 . 1 Let T : A be a collection of topologies on X. T = T : A=U X :U T for all A is weaker than all the T . 1.2 Sub-bases & bases LetE be a collection of subsets of a set X . LetBE be the smallest collection of subsets of X which contains all the elements of E together with the empty set and the whole set X and which is closed under finite intersection. Let T E { : BE } , then T E is a topology on X , the smallest topology containing E . The collection E is called a subbase of the topology T E and we say that the topology T E is generated by the subbase E . Definition 3. A collection B of open subsets of a topology space X is called a base of the space if each open set in X is a union of some elements of B . Since T is closed under arbitrary union . T itself is also a base. We want to get a base of smaller cardinality. Theorem B is a base of (X ,T ) if and only if all the elements of B are open and for any x∈X, open set U containing x, there exists V ∈B such that x ∈V U. Proof : For any x∈X , open set U containing x , U is the union of some elements B , i.e. B ' B such that U = V :V B ' , x∈U , then V B ', x ∈V. It is clear that V U . For any open set U and any x U , xVx U , thus U xU Vx. U. Vx B such that Topology T is uniquely determined by its base B , and T = : B As is known, single topology T can have different bases . Uncountable T can have countable base. Proposition 1. A collection B of subsets of a set X is a base of a topology on X if and only if, B X and for any U, V B and any x U V , there exists W B such that x W U V. Proof : ( ) X T , hence X B . U ,V B U V . T Therefore, for x U V , W B , x W U V . ( ) Let T U X : BU B s.t. U = BU .;, X T and T is closed under arbitrary union. For U , V T . we prove that U V T . For any x U V , since U BU V = BV , U 1 ,V1 B such that x U 1 V1 , then Wx B such that xWx U1 V1 . Thus U V = xU V Wx . We let BU V Wx , xU V , then U V = BU V. This shows that T is closed under finite intersection. T is a topology. It is easy to see that B is a base for T . Note: (1) A collection B of subsets of a set X is a base of a topology on X if B X , and B is closed under finite intersection. (2) BE in definition of sub-base is a base of T E . Example 4. Given a set R. B . ( a,b , a x isba base for the usual topology T . One can show that U T if and only if U is a union of pairwise disjoint intervals . 2 Example 5. Given a plane R 2 , B for natural topology on plane, where B x {(x,y): x 2 + y 2 } . B (x):x R is a base Example 6 Bs =a,b : a x b . It is clear that Bs is closed under finite intersection & Bs R , form a base for T S . The topological space ( R ,Bs ) is called the Sorgenfrey line. 1 (a, b) a , b , so T S is stronger than the n usual topology T on the line R . T T S and T T S . T S does not have a countable base. n =1 B0 B B0 P B0 Proof : Let t= U,V , U , V P . Let :P P B , t t t W such that V W U if such W exists; otherwise (t)= . Let B0 (t) : (t) . B0 B and B0 c P P c P . B0 is a base for T . How to prove that φ is surjective and B is a bsae? We can prove it as follows: for any open set O and xO , there exists U P such that xU O . Choose W B with xW U Choose V P with xV W U .We can do this since both P and B are bases. By definition of . Let t= U ,V and W =(t)B0 , then x (t) O .Thus B0 is a base since x , O are arbitrarily chosen. Corollary No countable subset of Bs can be a base for X,T s . Let A {a : b R , a,b } , if it is countable . Choose c,d cR\ A and c d , then c,d Bs can not be written as union of elements in . If there is some a < c, then…. Otherwise, … 1.3 Neighborhood of points. Nearness of a point to a set and the closure operator Definition 4 A set O X is called a neighborhood of a point x in the topological space X if there exists an open subset U of X such that xO U . Remark 1. Open set is neighborhood of every point it contains. Remark 2. Intersection of two nbds is again a neighborhood. Neighborhood is a “measure” of nearness to the point , also it is an approximation to the point. There is no distance in topological spaces. Definition 5. Let X,T be a topological space and let A X and x X .We say that the point x is near to(or at distance zero from) the set A and write x A (at distance 0) if for every open nbhd U of x , U A . Otherwise we say x is far from A ,write x A (or x,A 1 ) . is said to be induced by topology T , to emphasize this we write δ= δ(T ) . Properties of : x X , 1) x ( x, 1 ) 3) x A B x A x B A X , 2) B X x A x A 4) x B yB y A x A (weak form of triangle axiom) In condition 3, if a point far from A B , then it is also far from A and B . In condition 4 , the transitivity of the nearness relation 0+0=0 . The set of all points near to a subset A X is called the closure of A in the topological space X. Thus , A x X : x A .The map A A is called the closure operator in the topological space X. The closure operator possesses the following properties 1) 2) A A for all A X 3) A B A B 4) A A Definition : The set which contains all points near to it is said to be closed. A is closed if and only if A A . Proposition 3 A set A is closed in a topological space X,T if and only if X\A is open in X,T Proof : A= A {x : x A} For y X \ A , there exists a nbhd U of y,such that U A= , hence y U X \ A every point y in X\A is its interior point X\A is open. By De Morgan law, closedness is closed under finite union & arbitrary intersection. Note that: A is a closed set ( A A). and is the smallest closed set containing A. Example 7. Let X,T be an anti-discrete space. T =,X . For any nonempty A, A X . In discrete space, T 2 X , A= A . In Sorgenfrey line R ,T s , R R . Proposition 4. If U and V are disjoint open subsets of X , then U V ( and V U ) Proof: For any xV , V itself is an open neighborhood of x and V U = . Hence xU by the definition, U V . 1.4 Definition of a Topology using a Nearness relation or a closure operator An (abstract) closure operator on X is a rule which assigns to each subset A of X . Map 2 X 2 X, A A , satisfies 1) 2) A A for all A X 3) A B A B 4) A A Closure operator can be obtained via (abstract) nearness relation by defining A x X : x.A Let F A X A= A ,then T X \ A AF is a topology on X . Proposition 5. The topology of a space X is uniquely determined by any one of the following objects: the nearness relation , the closure operator , or the collection of all closed sets. Proof: Only for closure operator. Let T X \ A AF 1) X \ = X T ; X X hence X X , X\ X T . 2) A,B T X \ A X \ A X \ B X \ B X \ A B X \ A X \ B X \ A X \ B X \ A X \ B = X \ A B . Hence A BT . 3) Suppose that G T . For any AG , X \ A X \ A . On the other hand for any A1 G , we have X \ AG A AG X \ A X\ A1 . Hence X \ AG A X\ A1 X\ A1 and X \ AG A It is clear that X \ AG A X \ A1G X \ AG X \ A1 X \ AG A A X \ , AG A1G AG A AT A1 X \ AG A , therefore . Thus T is a topology on X , and - is the closure operator. uniquely determines the closed subsets in X, hence uniquely determines its topology . Definition. For any set A X , the derived set of A is denoted by d(A) , it consists of all its limit points (accumulation point). A point x is called to be a limit point of A if for any neighborhood O of x , O A\x . We have that A A d A by definition of the derived set. Four properties of derived set 1) d 3) d A B d A d B 2) A B d A d B 4) d d A A d A A The above four axioms uniquely determined a topology T such that in this topological space, the derived set of A is exactly d(A) . The relation of interior , closure and complement , -,' IntA A x A| an open nbhd O of x , O A , which is an open set Remark: (1) A is open iff A = Int(A); (2) Int(A) is the union of all open sets contained in A. 1) 3) A , X X B A B 2) A A 4) A A Theorem 1. A A' - ' , A- A' ' . Proof : For any x A , A is a neighborhood of x. Since A A' , then x A' - x A' - ' . Thus A A' - ' . On the other hand , for any x A' - ' , we have x A' - . Then there exists a neighborhood V of x , such that V A' V A . That is A' - ' A . Theorem 2 (C.Y.Yang) d(A) is closed d{x} is closed. P 77 6. Û Theorem 3 (Kuratowski) P 81 4. Definition 2 Boundary set of A, Bd(A): Bd (A)= ¶ (A)= A \A . Bd ( A) = A \ A = A Ç ( X \ A ) = A Ç A'- = Bd ( A¢). Bd ( A) = A Ç ( X \ A)- . Definition 3. The neighborhood system U x of point x : the collection of all x's neighborhoods. It satisfies 1) For any U x , U U . xU 2) Closed under finite intersection: U,V U x , U V U . 3) Closing up : U U x , U V V U . 4) Contains an open neighborhood V (V is a neighborhood of all its points) U Ux V Ux V U y V V U y Neighborhood system uniquely determine a topology T and in (X,T ) , the neighborhood system coincide with pre-assigned one. 1.5 Subspaces of topological space To each subset Y of a topological space (X, T ) associated a new topological space Y , T |Y where T | U Y : U T is the set of all “traces” in Y of the open subsets of X. T |Y is said to be the topology generated(or induced) by T and Y , T |Y is called a subspace of (X, T ) . Y Properties of subspace can be very different from the whole space. Any figure on the plane: disk, circle, disk with a hole. Q, J rational & irrational number with induced usual topology. Thm. If B is a base of (X, T ), then B ' = base of the subspace (Y, T Y ). {B Ç Y : B Î B } is a Proof: (1) Obviously, each U Y B is open in Y. (2) Suppose x U and U is open in Y, we have U = V Y where V is open in X. By the definition of bases, B B such that x B V. It follows that x B Y V Y = U. Thm. Let Y be a subspace of X. If U is open in Y and Y is open in X, then U is open in X. Relative nature of closed set . A topological space (X, T ) , (Y , T Y ) is a subspace. (1) A Ì Y can be closed in Y, but not in X. (2) clY (A)= clX (A)Ç Y for A Ì Y Ì X . A = cl X (A) Proof: (1) Let x cly(A), we prove that x clx(A). For each nbhd U of x in X, U Y is a nbhd of x in Y. So, (U Y) A = . That is U A . Therefore, x clx(A) Y. Let x clx(A) Y. For each nbhd V of x in Y, nbhd U of x in X s.t. V = U Y. From x clx(A), U A . This implies that (U Y) A , i.e., V A . Thus, we conclude that x cly(A). Thm. Closed family in Y := {F Ç Y : F is closed in X }, using subspace we can construct many interesting examples. Proof: (1) Let A be closed in Y, then Y \ A is open in Y. So, open set U in X s.t. Y \ A = U Y. It is easy to check that A = Y (X \ U), and X \U is closed in X. (2) Assume that F is closed in X, then X \ F is open in X. So, (X \ F) Y is open in Y. Since (X \ F) Y = Y \ (F Y), we know that F Y is closed in Y. Example 8. (The Cantor perfect set) We use the word segment to refer to a set of the form [ a, b] and interval to refer to a set of the form (a, b) . A1= segment [0,1] , An is a union of a finite number of pairwise disjoint segment. An + 1denote set obtained by removing from each segment of An the middle 13 interval. Then we will get a decreasing sequence An : n Î N . C = Ç¥n= 1 An of the line R is called the Cantor perfect set . C is closed, has no isolated points, but does not contain any interval , uncountable and measure 0. Remark: A point of the set A \ d(A) is called an isolated point of A. A point x is an isolated point of X iff the one-point set {x} is open. 1.6 The Product Topology Def. Let ( X , T X ) and (Y , T Y ) be topological spaces. The product topology on X ´ Y is the topology T having as base the collection {U ´ V : U Î T X ,V Î T Y } Prop: If B is a base of the space ( X , T X ) andC is a base of the space (Y , T Y ) , then the collection {B ´ C : B Î B , C Î C } is a base of the product space (X, T ) . Def: Let p1 : X ´ Y ® X be defined by the equation p1 ( x, y) = x Let p 2 : X ´ Y ® Y be defined by the equation p 2 ( x, y) = y The map p1 and p 2 are called the projections of X ´ Y onto its first and second factors, respectively. Thm. The collection S { 11 (U ) : U T X } { 2 1 (V ) : V T Y } is a subbase of the product topology on X Y. Proof: Let T be the product topology and T the topology generated by S. (1) Because S T , we have T T. (2) On the other hand, every base element U V of the topology T has the form U V 11 (U ) 2 1 (.V ). Thus, U V T . So that T T as well. 1.7 The metric topology Def: A metric on a set is a function d: X Y R having the following properties: (1)d(x, y) 0 for all x, y X; equality holds iff x = y. (2)d(x, y) = d(y, x) for all x, y X. (3)d(x, y) + d(y, z) d(x, z), for all x, y, z X. (triangle inequality) Def: Given a metric d on X, the number d(x, y) is called the distance between x and y. Given > 0, the set {y: d(x, y) < } is called the –ball centered at x. It is usually denoted by B(x, ). Def: If d is a metric on the set X, then the collection of all balls B(x, ), for x X and > 0, is a base of a topology on X, called the metric topology induced by d. Def: If X is a topological space, X is called to be metrizable if there exists a metric d on the set X that induces the topology of X. Remark: A metric space is a metrizable space. Def: Let X be a metric space with metric d. A subset A of X is said to be bounded if there is some number M such that d(x, y) M for every pair x, y of A. Thm: Let X be a metric space with metric d. Define d: X X R by the equation d(x, y) = min {d(x, y), 1}. Then d is a metric that induces the same topology as d. 1.8 Continuous Functions Def: Let X and Y be topological spaces. A function f: X Y is said to be continuous if for each open set V of Y, the set f- -1 (V) is an open set of X. Prop.: Let X and Y be topological spaces, and B a base of the topology T of Y. A function f: X Y is continuous iff the inverse image of every base element is open. Proof: The arbitrary open set V of Y can be written as a union of base elements: V = {B : J}, and we have that f- -1 (V) = {f- -1 (B ): J}, Prop.: Let X and Y be topological spaces, and S a subbase of the topology T of Y. A function f: X Y is continuous iff the inverse image of every subbase element is open. Proof: The arbitrary base element B of Y can be written as a finite intersection of subbase elements: B = S1 S2 … Sn , and we have that f- -1 (B) = f- -1 (S1 ) f- -1 (S2 ) … f- -1 (Sn ). Ex.: Let T be the usual topology on R, and T l the lower limit topology on R. Then, the identity function f: x x is not a continuous function from (R, T ) to (R, T l ). On the other hand, the identity function g: x x is a continuous function from (R, T l ). To (R, T ) Thm.: Let X and Y be topological spaces and f: X Y is a function . Then, the following are equivalent: (1) (1) (2) (3) f is continuous. For every subset A of X, we have f- (cl(A)) cl( f (A)). For every closed subset B of Y, the set f- -1 (B) is closed in X. For each x X and each nbhd. V of f (x), there is a nbhd. U of x such that f (U) V. Proof: (1) (2) Assume that x cl(A). If V is an open nbhd. of f (x), then f -1 (V) is an open nbhd. of x. So, it must intersect A in some point y. It follows that V intersects f (A) in the point f (y). This implies that f (x) cl( f (A)). (2) (3) Let B be closed in Y and A = f- -1 (B). If x cl(A), then f (x) f (cl(A)) cl(f (A)) cl(B) = B. Therefore, x f- -1 (B) =A. (3) (1) Let V be an open set in Y and B = Y – V. Then f- -1 (B) = f- -1 (Y) - f- -1 (V) = X - f- -1 (V) . This means that f- -1 (V) is open in X. (1) (4) Let x X and let V be an open nbhd. of f (x). Then U = f- -1 (V) is an open nbhd. of x, and f (U) V. (4) (1) Let V be an open set in Y. For each x f- -1 (V), we have f (x) V. There is an open nbhd. U of x such that f (U) V, This implies that U f- -1 (V). Therefore, f- -1 (V) can be written as the union of open sets. Def: Let X and Y be topological spaces, f: X Y be a bijection . If both the function f and its inverse f- -1 : Y X are continuous, then f is called a homeomorphism. Remark 1: Another way to define a homeomorphism is to say that f: X Y is a bijective correspondence such that f (U) is open U is open. Remark 2: A homeomorphism f: X Y gives us a bijective correspondence not only between X an Y, but between the collection open sets of X and Y. So, any property of X that is entirely expressed in terms of the topology of X yields the corresponding property of Y. Ex: A bijective function f: X Y can be continuous without being a homeomorphism. Let S1 denote the unit circle, S1 = {(x, y): x2 + y 2 =1}, considered as a subspace of the plane R2 , let f : [0, 1) be the map defined by f (t) = (cos2t, sin2t). (1) f is bijective and continuous. (2) The image of the open set U = [0, 0.2) [0, 1) under f is not open in S1. Local formulation of continuity: The map f: X Y is continuous if X can be written as the union of open sets U such that f U is continuous for each B . Proof: Let X be the union of open sets U and each f U is continuous. For each open set V in Y, it is easy to check that f- -1 (V) U = (f U) -1 (V). Since f U is continuous, the set f- -1 (V) U is open in U, and thus open in X. So, f- -1 (V) = (f- -1 (V) U ) is open in X. The pasting lemma: Let X = A B, where A and B are closed in X. Let f: A Y and g: B Y be continuous. If f (x) = g (x) for every x A B, then f and g combine to give a continuous function h: X Y, defined by setting h (x) = f (x) if x A, and h (x) =g (x) if x B. Proof: Let C be closed subset of Y, then we have h -1 (C) = f- -1 (C) g -1 (C) by elementary set theory. Since f is continuous, f- -1 (C) is closed in A, and thus closed in X. Similarly, we can prove that g -1 (C) is closed in B, and therefore closed in X. It follows that h -1 (C) is closed in X. 1.8 Sequential Spaces.The Sequential Closure Operator Definition 9. A topological space X is called sequential if , for every set A X which is not closed in X , there exists a sequence xn n 1 of points of A converging to a point of the set A \A . Sequential closure : Aseq x X : sequencexn n1 in A , xn x; A Aseq is a map from 2 X 2 X , it is called the sequential closure operator. It can be defined in any topological spaces. Properties : 0 seq 1seq 2 seq Remark: . seq . A Aseq A . A B A B A A seq seq seq seq seq in general. seq Example. Let X be the set consisting of 3 different type points: isolated points : xmn . limit points : yn of sequence xmn : m N point z : with most complicated local structure The collection B = xmn : n N , m N Vk yn : n, k N is a base of a topology on X , where Vk yn yn xmn : m k . and W : z W , p N ,Vk yn \ W is finite and yn W for all n p . 1. X is sequential space (Arens Fort space, Modified) Because only those yn ' s and z could be limit point of sequence with distinct points . If yn ' s or z in G \ G , there must be a sequence in G converging to it. 2. Let A xmn : m N , n N , B yn : n N B seq B z ,therefore z Aseq seq A A seq seq seq , then Aseq A , but z A . Thus B seq . Definition 10 . A topological space X is called a Frechet-Uryshon space if the closure of every subset A X in X coincides with the sequential closure of A : A Aseq . It follows that Frechet-Uryshon space is sequential and from the above example, it shows that not every sequential space is Frechet-Uryshon space. Example 14. Consider the subspace Y A z of the space X in example. It is easy to verify that no sequence of points in A converges to the point z. But A A z . Consequently, Y is not a sequential space. Thus a subspace of a sequential space need not be sequential. Proposition 6. A topological space X is a Frechet-Uryshon space if and only if each subspace Y of X is sequential. Proof: Let Y X with subspace topology. A Y , if A is not closed in Y.Then clY A \ A , clY A cl X A Y Aseq Y , For any point x in( Aseq Y ) \ A , there is a sequence in A converging to x. Therefore Y is sequential. For any subset , let y A \ A , consider Y A y , then Y is sequential. Therefore in A sequence xn n 1 converging to y, this shows that A Aseq hence A Aseq . X is Frechet-Uryshon space. A X 1.9. The First Axiom of Countability and Bases of a Space at a Point (and at a Set) Definition11. A collection of open neighborhoods of a point x in a topological space X is called a base of the space X at the point x if each neighborhood of x contains an element of . First Axiom of countability: each point has a countable base. C Example: Space with countable base, all discrete spaces satisfies C . Note that: the discrete space X has countable base X itself is countable. Hence not every C space is C space. Proposition 7. If X satisfies C X is a Frechet-Uryshon space. Proof : For any set A X ,we show A Aseq . Let x A , suppose is a countable base at x, say On : n , WLOG we may assume that Om On when m n , otherwise we can simply let Om mi 1 Oi , then replace Om by Om . It is easy to see that A Om for any m , pick xm A Om , then xn : n is a sequence in A converging to x. For any neighborhood U of x, Om such that x Om U , then xn Om U for all n m . Proposition 8. A countable space X is base. C it has countable Proof : Trivial. For each point x, Let Bx be a countable neighborhood base S Bx : x X is a countable collection, use it as subbase, the topology generated by S is the original topology T . B S1 Sn : Si S is a countable base for T . Example 15. X in Example 13, Y in Example 14 are countable , but do not have C .They are not even Frechet-Uryson space. Thus , not every countable space has a countable base. Note that all single point sets (singletons) in Y are closed and only one point (the point z) is not isolated. 1.10 Everywhere Dense Sets and Separable Space. Definition 13. A subset A of a topological space X is called everywhere dense in X if its closure is equal to X: A X . : equivalent to for every nonempty open U , U A : enough for every nonempty base element U B U A X is separable if it contains a countable dense subset. Example 20. Every countable space is separable. is dense in ,T as well as , T S . ,T has countable base, , T S does not. Proposition 10. Not every separable space satisfying the first axiom of countability possesses a countable base. Proposition 11. Every space with a countable base is separable. Proof : Choose a point in each nonempty base element B . A of all such points is countable and dense in X . C is hereditary property. Note 1. Separability is not hereditary. Example(P149): X , T is a topological space , X ,X * X , T * A {}| A T , X * , T * is a space. 1 X * , T * is separable . belongs to any nonempty open set , then is a dense set in X * , T * . 2 X , T can have any properties we want . For example choose any nonseparable space , say uncountable discrete space. X , T is a subspace of X * , T * . Note 2. Even every subspace of X is separable(X is hereditarily separable ). X need not be C . Sorgenfrey line: not C but hereditarily separable. Think of how to prove it. Countable space without a countable base Example 14 or Example 19 , countable Frechet-Uryshon fan.