GeneralTopology22011.03.12.ppt

1、Definition 1. Topology on set : T is a collection of subsets in X, satisfied that a) , 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 .,T includes: , X, and closed under finite intersection hence , .,. Hence .,3) Sup

2、pose that . For any , . On the other hand for any , we have . Hence and It is clear that , therefore , .,Thus T is a topology on X , and - is the closure operator. - uniquely determined the closed subsets in X, hence uniquely determined its topology . P73 4,Supplementary fact P75,Definition 1. For a

3、ny set ,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 , .,We have that by definition of the derived set.,Four properties of derived set,2) 3) 4),The above four axioms uniq

4、uely 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 (Theorem 2.5.1 p 74),Theorem 1. , . Proof : For any , A is a neighborhood of x. For , then . Thus . On the other hand , for any ,if and only if .

5、Then there exists a neighborhood V of x ,such that . That is .,Theorem 2 (C.Y.Yang) d(A) is closed dx is closed. P 74 6,Theorem 3 (Kuratowski) P 78 4,Neighborhood system uniquely determine a topology T and in (X,T ) , the neighborhood system coincide with pre-assigned one.,Definition 2 Boundary set

6、of A, Bd(A):,Definition 3. The neighborhood system of point x : the collection of all neighborhoods. It satisfies For any , . 2) Closed under finite intersection: , . 3) Closing up : , . 4) Contains an open neighborhood V (V is a neighborhood of all its points),Interior point , Interior point of a s

7、et A, 2) 3) 4),1.5 Subspaces of topological space P93,To each subset Y of a topological space (X, T ) define a new topological space where is the set of all “traces” in Y of the open subsets of X. is said to be the topology generated(or induced) by T and is called a subspace of (X, T ) .,Properties

8、of subspace can be very different from the whole space.,Any figure on the plane, disk, circle, disk with a hole.,rational is a map from , it is called the sequential closure operator. It can be defined in any topological spaces. Properties : . . Remark: in general.,Example 12 . A as Example 10. is t

9、he functions of first Baire class. is the functions of second Baire class. Since ,we have .It is not idempotent. This contrasts to the closure operator , we have .,X is not sequential. Even in sequential spaces in general.,Example 13. Let X be the set consisting of 3 different type points: The colle

10、ction is a base of a topology on X , where .,and .,1. X is sequential space (Arens Fort space, Modified) Because only those and z could be limit point of sequence with distinct points , if or z in .There must be a sequence in G converging to it,2. Let , , then ,therefore , but . Thus .,It follows th

11、at Frechet-Uryshon space is sequential and from Example 13 , it shows that not every sequential space is Frechet- Uryshon space.,Definition 10 . A topological space X is called a Frechet-Uryshon space if the closure of every subset in X coincides with the sequential closure of A : .,Example 14. Cons

12、ider the subspace of the space X in example 13. It is easy to verify that no sequence of points in A converges to the point z. But . 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

13、 if and only if each subspace Y of X is sequential.,1.9. The First Axiom of Countability and Bases of a Space at a Point (and at a Set),Proof: Let with subspace topology. , if A is not closed in Y.Then , For any point x in , there is a sequence in A converging to x. Therefore Y is sequential.,For an

14、y subset , let , consider , then Y is sequential. Therefore in A sequence converging to y, this shows that hence . X is Frechet-Uryshon space.,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

15、 x contains an element of .,First Axiom of countability: each point has a countable base.,Example: Space with countable base, all discrete spaces satisfies .,Note that: the discrete space X has countable base X itself is countable. Hence not every space is space.,Proposition 7. If X satisfies X is a

16、 Frechet-Uryshon space.,Proof : For any set ,we show . Let , suppose is a countable base at x, say , WLOG we may assume that when , otherwise we can simply let , then replace by . It is easy to see that for any , pick , then is a sequence in A converging to x. For any neighborhood U of x, such that

17、, then for all .,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.,Proposition 8. A countable space X is it has countable base.,Proof : Trivial.,For each point x, Let be a countable ne

18、ighborhood base is a countable collection, use it as subbase, the topology generated by is the original topology . is a countable base for .,Example 15. X in Example 13, Y in Example 14 are countable , but do not have .They are not even Frechet-Uryson space.,Space of countable pseudo-character: each

19、 point is an intersection of countable many open subsets.,Example: Y as in example14 , ,it is not and not sequential.,Proposition 9. Not every space of countable pseudo-character is sequential, not every such space satisfies the first axiom of countability.,Fact (consequence) countable open collecti

20、on such that , but X does not have countable base at x.,Example 16. The Sorgenfrey line satisfies the first axiom of countability, at all points: the countable collection of open sets in is a base of at the point a.,Definition 12. A family of open subsets of a topological space X is said to be a bas

21、e of X at the set if and , for each open set containing A, there exists such that .,Example 17. Let X be the plane with the usual topology and .The space X does not possess a countable base at A.,In connection with Proposition 7 , we remark that not every Frechet-Uryshon space satisfies the first ax

22、iom of countability.,Example 18. Let A be an uncountable set and .Define a top- ology on the set by setting . Thus , all subsets of A are open in X and a set containing the point is open if and only if its complement is finite.The space X is called the Alexandrov supersequence of the length .It is c

23、lear that every sequence of pairwise distinct points of X converges to .,P147 5,6. Y is not of countable pseudocharacter the argument is similar to Example 18 .,But X does not satisfy the first axiom of countability. In fact, if is a countable collection of open sets in X and , then the set is count

24、able(this follows from the way neighborhoods of were defined) and hence , . Fix . The neighborhood of contains no element of .,Example 19. A countable Frechet-Uryshon space does not have countable neighborhood base. , , , define T on Y such that points in X are isolated. Neighborhood of a is form V

25、such that is finite for each . At point a , there is no countable neighborhood base . is countable Frechet-Uryshon space. It is called the countable Frechet-Uryson fan . Each sequence converges to the point a .,Proposition 11. Every space with a countable base is separable.,1.10 Everywhere Dense Set

26、s 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: .,: equivalent to for every nonempty U , : enough for every nonempty base element X is separable if it contains a countable dense subset.,Example 20. Every countabl

27、e space is separable. is dense in as well as . has countable base, does not.,Proposition 10. Not every separable space satisfying the first axiom of countability possesses a countable base.,Proof : Choose a point in each nonempty base element B . A of all such points is countable and dense in X .,is

28、 hereditary property.,Note 1. Separability is not hereditary.,Example(P149): is a topological space , , , , is a space.,can have any properties we want . For example choose any nonseparable space , say uncountable discrete space. is a subspace of .,is separable . belongs to any nonempty open set , t

29、hen is a dense set in .,Note 2. Even every subspace of X is separable(X is hereditarily separable ). X need not be .,Sorgenfrey line: not but hereditarily separable. Think of how to prove it.,Countable space without a countable base Example 14 or Example 19 , countable Frechet-Uryshon fan.,Propositi

30、on 12. If A is an everywhere dense subset of a topolo- gical space and if U is open in X, then,Proof: hence . For , and any neighborhood W of x , this shows .,1.11 Nowhere Dense Sets.The Interior and Boundary of a Set,Complement of dense set can be also dense. For example , but is also dense, i.e to

31、o.,What kind of set can represent “small” set .,Definition 14. A set is said to be nowhere dense in the topological space if , for each nonempty open set U, there exists a nonempty open set V such that and . An equivalent condition is that the complement to the closure of A be everywhere dense in X:

32、 that is .,Example 2. A straight line on the plane is a nowhere dense set. An important example of a nowhere dense set is the boundary of an open set.,It is equivalent to say the complement of closure of A is dense in X . i.e.,Proof: For any open set U , open , then . This shows is dense in X .,For

33、any nonempty open set U, then is open in X , as well as in , i.e. , , hence .,Example 1. X has no isolated point , finite subset is closed . Any finite set A is nowhere dense.,Definition 15 . The boundary Bd(A) of a set in a topological space X is the set of all points which are near to both A and i

34、ts complement . Thus, Bd(A) and the set Bd(A) is always closed.,Note 1. Closed set: =contains all its boundary 2. Open set: =contains no point of boundary 3. Clopen set: =boundary is empty 4. Bd(A)=Bd 5. “Large ” set A , dense Bd(A) Bd 6. Boundary operator Bd: Bd(A) satisfies 1.Bd 2.Bd Bd(A) Bd(B) 3

35、.Bd(Bd(A) Bd(A) 4.Bd A=Bd,Proposition 13. The boundary of any open set U in a topological space X is a nowhere dense set.,Proof: By contradiction. Suppose not , is not nowhere dense , then is not dense in X . There exist nonempty open set V . hence . Then and . For any point , is an open neighborhoo

36、d of p with . Hence , , contradiction to , .,Definition 16. Let A be a subset of a topological space . The interior of A in X is the set Int(A) of all points of A for which A serves as a neighbourhood. Thus, Int(A) : there exits Such that .,Proposition 14. A set is nowhere dense in a topological space X if and only if the