緊空間外文文獻翻譯

1、<p><b>　　河南理工大學</b></p><p>　　本科畢業(yè)設計（論文）</p><p><b>　　外文文獻資料翻譯</b></p><p>　　院（系部） 數(shù)學與信息科學學院 </p><p>　　專業(yè)名稱 數(shù)學與應用數(shù)學專業(yè) </p>

2、<p>　　年級班級 0902班 </p><p>　　學生姓名 陳勇 </p><p>　　學生學號 310911010214 </p><p><b>　　2013年6月1日</b></p><p

3、>　　Compact Spaces</p><p>　　The notion of component is not nearly so natural as that of connectedness.From the beginnings of topology,it was clear that the closed interval of the real line had a certain p

4、roerty that was crucial for proving such thorrems as the maximum value theorem and the uniform continuity theorem .But for a long time,it was not clear how this property should be formulated for an arbitrary topological

5、space.It used to be though that the crucial property of was the fact that every infinte subset of has</p><p>　　Definition. A collection of subsets of a space is said to cover ,or to be a covering of ,if

6、the union of the elements of is equal to .It is called an open covering of if its elements are open subsets of .</p><p>　　Definition. A space is said to be compact if every open covering of contains a f

7、inite subcollection that also covers .</p><p>　　EXAMPLE1. The real line is not compact,for the covering of by open intervals </p><p>　　contains no finite subcollection that covers .</p>

8、<p>　　EXAMPLE2. The following subspace of is compact:</p><p><b>　　.</b></p><p>　　Given an open covering of ,there is an element of containing 0.The set contains all but

9、finitely many of the points ;choose,for each points of not in ,an element of containing it.The collection consisting of these elements of ,along with the element ,is a finite subcollection of that covers .</p>

10、<p>　　EXAMPLE3. Any space containing only finitely many points is necessarily compacts,because in this case every open covering of is finite.</p><p>　　EXAMPLE4 The interal is not compact;the open c

11、overing </p><p>　　contains no finite subcollection covering .Nor is the interal compact;the same argument applies.On the other hand,interal is copact;you are probably already familiar with this fact from a

12、nalysis.In any case,we shall prove it shortly.</p><p>　　In general,it takes some effort to decide whether a given space is compact or not.First we shall prove some general theorems that show us how to constr

13、uct new compact spaces out of existing ones. Then in the next section we shall show certain specific spaces are compact.The spaces include all closed interals in the real line,and all closed and bounded subsets of </p

14、><p>　　Let us first prove some facts about subspaces.If is a subspace of ,a collection of subsets of is said to cover if the union of its elements contaons .</p><p>　　Lemma 1. Let be a subspac

15、e of .Then is compact if and only if every covering of by sets open in contains a finite subcollection covering .</p><p>　　Proof . Suppose that is compact and is a covering of by sets open in .Then the

16、 collection </p><p>　　is a covering of by sets of ;hence a finite subcollection </p><p>　　covers .Then is a subcollection of that covers .</p><p>　　Coversely,suppose the given co

17、ndition holds:we wish to prove compact.Let be a covering of by sets open in .For each ,choose a set open</p><p>　　In such that </p><p>　　The collection is a covering of by sets open in .B

18、y hypothesis,some finite subcollection covers .Then {} is a subcollection of that covers .</p><p>　　Theorem1 Every closed subpace of a compate space is compact.</p><p>　　Proof. Let be a clos

19、ed subspace of the compact space .Given a covering of by sets open in ,let us form an open covering of by adjoining to the single open set ,that is,</p><p><b>　　.</b></p><p>　　S

20、ome finite subcollection of covers .If this subcollection contains the set ,discard ;otherwise,leave the subcollection alone.The resulting collection is a finite subcollection of that covers .</p><p>　　Theo

21、rem2 Every compact subspace of a Hausdorff space is closed.</p><p>　　Proof Let be a compact subspace of the Hausdorff space .Ww shall prove that is open,so that is closed.</p><p>　　Let be

22、 a point of .We show there is a neihborhood of that is disjoint from .For each point of ,let us choose disjoint neighborhoods and of the points and ,respectively (using the Hausdorff condition).The collection is a c

23、overing of by sets open in ;therefore,finitely many of them cover .The open set</p><p>　　contains ,and it is disjoints from the open set </p><p>　　formed by taking the intersection of the corr

24、esponding neighborhoods of .For if is a point of ,then for some ,hence and so .</p><p>　　Then is a neighborhood of disjoint from ,as desired.</p><p>　　The statement we proved in the course

25、of the preceding proof will be useful to us later,so we repeat it here for reference purpose:</p><p>　　Lemma2 If is a compact subspace of the Hausdorff space and is not in ,then there exist disjoint open

26、 sets and of containing and ,respectively.</p><p>　　EXAMPLE5. Once we prove that the interval in is compact,it follows from theorem1 that any closed subspace of is compact .On the other hand,it follow

27、s from theorem2 that the intervals and in cannot be compact (which knew already) because they are not closed in the Huasdorff space .</p><p>　　EXAMPLE6 One needs the Hausdorff condition in hypothesis of

28、Theorem2. Consider,for eample,the finite complement topology on the real line.The only proper subsets of that are closed in this topology are the finite sets.But every subset of compact in this topology,as you can chec

29、k.</p><p>　　Theorem3 The image of a compact space under a continuous map is compact.</p><p>　　Proof. Let be continuous:let be compact. Let be a covering of the set by sets open in . The co

30、llection </p><p>　　is a collection of sets covering ; these sets are open in because is continuous. Hence finitely many of them ,say </p><p><b>　　, </b></p><p>　　cover

31、 . Then the sets cover .</p><p>　　One important use of the preceding theorem is as a tool for verify that a map is a homeomorphism:</p><p>　　Theorem 4. Let be a bijective continuous function.

32、 If is compact and is hausdorff, then is a homeomorphism.</p><p>　　Proof. We shall prove that images of closed sets of under are closed in ;this will prove continuity of the map .If is closed in , the

33、n is compact, by theorem1. Therefore, by the theorem3 just proved, is compact. Since is Hausdorff, is closed in , by Theorem 2.</p><p>　　Theorem5. The product of finitely many compact spaces is compact

34、.</p><p>　　Proof. We shall prove that the product of two compact spaces is compact;the theorem follows by induction for any finite product.</p><p>　　Step 1. Suppose that we are given spaces a

35、nd , with compact. Suppose that is a point of , and is an open set of containing the “slice” of . We prove the following:</p><p>　　There is a neighborhood of in such that contains the entire set .&

36、lt;/p><p>　　The set is often called a tube about .</p><p>　　First let us cover by basis elements (for the topology of ) lying in . The space is compact, being homeomorphic to . Therefor, we can

37、 cover by finitely many such basis elements</p><p><b>　　.</b></p><p>　　(We assume that each of the basis elements actually intersects , since otherwise that basis element would be

38、superfluous ; we could discard it from the finite collection and still have a covering of .) Define</p><p><b>　　.</b></p><p>　　The set is open, and it contains because each set in

39、tersects .</p><p>　　We assert that the sets , which were chosen to cover the slice ,actually cover the tube . Let be a point of . Consider the point of the slice having the same as this point. Now belon

40、gs to for some , so that . But for every (because ). Therefore we have , as desired.</p><p>　　Since all the sets lie in ,and since they cover , tube lies in also.</p><p>　　Step 2. Now we

41、prove the theorem. Let and be compact spaces. Let be an open covering of . Given , the slice is compact and may therefore be covered by finitely many elements of . Their union is an open set containing ; by step 1,

42、 the open set contains a tube about , where is open in . Then is covered by finitely many elements of .</p><p>　　Thus, for each in ,we can choose a neighborhood of such that the tube can be covered

43、by finitely many elements of . The collection of all the neighborhoods is an open covering of ; therefore by compactness of , there exsits a finite subcollection</p><p>　　covering . The union of the tubes &

44、lt;/p><p>　　is all of ;since each may be covered by finitely many elements of ,so many be covered. </p><p>　　The statement proved in Step 1 of the preceding proof will be useful to us later, so w

45、e repeat it here as a lemma, for reference purposes:</p><p>　　Lemma 3 (The tube lemma). Consider the product space , where is compact. If is an open set of containing the slice of , then contains some t

46、ube about , where is a neighborhood of in</p><p><b>　　.</b></p><p>　　EXAMPLE 7 The tube lemma is certainly not true if is not compact. For eample, let be the in , and let </

47、p><p>　　Then is an open set containing the set , but it contains no tube about . </p><p>　　There is an obvious question to ask at this point. Is the product of infinitely many compact spaces space

48、s compact? One would hope that the answer is “yes”, and in fact it is. The result is important(and different) enough to be called by the name of the man who proved it; it is called the Tychonoff theorem.</p><p

49、>　　In proving the fact that a cartesian product of connected spaces is connected, one proves it first for finite products and derives the general case from that. In proving that cartesian products of compact spaces ar

50、e compact, however, there is no way to go directly from finite products to infinite ones. The finite case demands a new approach, and the proof is a difficult one. Because of its its difficulty, and also to avoid losing

51、the main thread of our discussion in this chapter, we have decided</p><p>　　There is one final criterion for a space to be compact, a criterion that is formulated in terms of closed sets rather than open set

52、s. It does not look very natural nor very useful at first glance, but it in fact proves to be useful on a number of occasion. First we make a definition.</p><p>　　Definition . A collection of subsets of i

53、s said to be have the finite intersection property if for every subcollection</p><p>　　of ,the intersection is nonempty.</p><p>　　Theorem6. Let be a topological space. Then is compact if an

54、d only if for every collection of closed sets in having the finite intersection property, the intersection of all the elements of is nonempty.</p><p>　　Proof. Given a collection of subsets of , let <

55、;/p><p>　　be the collection of their complements. Then the following statements hold:</p><p>　　(1) is a collection of open sets if and only if is a collection of closed sets.</p><p>

56、　　(2) The collection covers if and only if the intersection of all the elements of is empty.</p><p>　　(3) The finite subcollection of covers if and only if the intersection of the corresponding elem

57、ent of is empty.</p><p>　　The first statement is trivial, while the second and third follow from DeMorgan’slaw.</p><p>　　The proof of the theorem now proceeds in two easy steps: taking the con

58、trapositive(of the theorem), and then the complement (of the sets)!</p><p>　　The satement that is compact is equivalent to saying : “Given any collection of open subsets of , if covers , then some finite

59、 subcollection of covers ”. Letting be , as earlier, the collection and applying (1)---(3), we see that this atatement is in turn equivalent to the following : “Given any collection of closed sets , if every finite i

60、ntersection of elements of is nonempty , then the intersection of all the element of is nonempty.” This is just the condition of our theorem.</p><p>　　A special case of this theorem occurs when we have a n

61、ested sequence of closed sets in a compact space . If each of the sets is nonempty, then the collection automatically has the finite intersection property. Then the intersection</p><p>　　is nonempty.</

62、p><p>　　Local Connectedness</p><p>　　Connectedness is a useful property for a space to possess .But for some purproses,it is more important that the space satify a connectedness condition locally .

63、Roughly speaking ,local connectedness means that each point has “arbitrarily small”neighborhoods that are connected.More precisely,one the following definition:</p><p>　　Definition: A space is said to be l

64、ocally connected at if for every neighborhood of ,there is a connected neighborhood of contained in</p><p>　　If is locally connected at each of its ponts ,it is said simply to be locally connected .Simi

65、larly,a space is said to be locally path connected at if for every neghborhood of ,there is a path-conneted neighborhood of contained in .If is locally path connected at each of its points,then it is said to be lo

66、cally path connected.</p><p>　　EXAMPLE 1 Each interval and each ray in the real line is both connected and locally connected.The subspace of is not connected.But it is locally connected.The rationals are

67、 neither connected nor locally connected.</p><p>　　Theorem 1 A space is locally connected if and only if for every open set of ,each component of is open in .</p><p>　　Proof . Suppose that i

68、s locally connected ;let be an open set in ;let be a component of .If is a point of ,we can choose a connected neighborhood of such that .Since is connected ,it must lie entirely in the component of .Therefor , is

69、open in .</p><p>　　Conversely,suppose that componnents of open sets in are open .Given a piont of and a neighborhood of ,let be the coponent of containing .Now is connected;since it is open in by hyp

70、othesis, is locally connected at . </p><p>　　A similar proof holds for the following theorem:</p><p>　　Theorem 2 A space is locally path connected if and only if for every open set of ,each pa

71、th component of is open in .</p><p>　　The relation between path components and components is given in the following theorenm:</p><p>　　Theorem 3 If is a topological space ,each path component

72、of lies in a component of .If is locally path connected ,then the components and the path components of are the same.</p><p>　　Proof Let be a component of ;let be a point of ;let be the path component

73、 of containing .Since is connected, .We wish to show that if is locally path connected, .Suppose that .Let denote the union of all the components of that are different from and intersect ;each of them necessarily li

74、es in ,so that </p><p><b>　　.</b></p><p>　　Bcause is locally path connected,each path component of is open in .therefore, (which is a path component) and (which is a union of path

75、 components ) are open in ,so they constitute a separation of .This contradicts the that is connected.</p><p>　　Compact Subspaces of the Real Line</p><p>　　The theorem of the preceding section

76、enable us to constuct new compact spaces from existing ones, but in order to get very far we have to find some compact spaces to start with. The natural place to beigin is the real line; we shall prove that every closed

77、interval in is compact. Applications include the extreme value theorem and the uniform continuity theorem of calculus, suitably generalized. We aslo give a characterization of all compact subspaces of , and a proof of t

78、he uncountability of th</p><p>　　It turns out that in order to prove every closed interval in is compact, we need only one of the order properties of the real line –the least upper bound property. We shall p

79、rove the theorem using only this hypothesis; then it will apply not only to the real line, but to well-ordered sets and other ordered sets as well.</p><p>　　Theorem1. Let be a simply order set having the

80、least upper bound property. In the order topology, each closed interval in is compact.</p><p>　　Proof. Step 1. Given , let a covering of by sets open in in the subspace topology (which is the same as th

81、e order topology). We wish to prove the existence of a finite subcollection of covering . First we prove the following : If is a point of different from , then there is a point of such that the interval can be cover

82、ed by at most two elements of .</p><p>　　If has an immediate successor in , let be this immediate successor. Then consists of the two points and , so that it can be covered by at most two elements of .

83、If has no immediate successor in , choose an element of containing . Because and is open , contains an interval of the form , for some in . Choose a point in ; then the interal is covered by the single element

84、of .</p><p>　　Step 2. Let be the set of all points of such that the interval can be covered by finitely many element of . Applying Step1 to the case , we see that there exists at least one such , so is

85、not empty. Let be the least upper bound of the set : then .</p><p>　　Step 3. We show that belongs to ; that is, we show that the interval can be covered by finitely many elements of . Choose an elements

86、 of containing c; since is open, it contains an interval of the form for some in . If is not in , there must be a point of lying in the interval , because otherwise would be a smaller upper bound on than . Sinc

87、e is in , the interval can be covered by finitely many, say , element of . Now lies in the single element of , hence can be covered eleme</p><p>　　Step 4. Finally, we show that , and our theorem is p

88、roved. Suppose that . Applying Step 1 to the case , we conclude that there exists a point of such that the interval can be covered by finitely many elements of . We proved in Step 3 that is in , so can be covered by

89、 finitely many elements of . Therefore ,the interval</p><p>　　can also covered by finitely many elements of . This means that is in , contradicting the fact that is an upper bound on .</p><p>

90、　　Corollary1 Every closed interval in is compact.</p><p>　　Now we characterize the compact subspaces of :</p><p>　　Theorem2. A subspace of is compact if and only if it is closed and is boun

91、ded in the euclidean metric or the square metric .</p><p>　　Proof. It will suffice to consider only the metric ; the inequalities</p><p>　　impply that is bound under if and only if it is boun

92、ded under .</p><p>　　Supose that is compact. Then , by theorem2 (in Compact Spaces) it is closed. Consider the collection of open sets </p><p>　　whose union is all of . Some finite subcollecti

93、on covers . It follows that for some . Therefore , for any two points and of , we have . Thus is bounded under .</p><p>　　Conversely ,supposed that is closed and bounded under ; suppose that for every

94、pair of points of . Choose a point of , and let . The triangle inequality implies that for every in . If ,then is subset of the cube , which is compact. Being closed, is also compact.</p><p>　　Students

95、often remember this theorem as stating that the collection of compact sets in a metric space equals the collection of closed and bounded sets. This statement is clearly ridiculous as it stands, because the question as to

96、 which sets are bounded depends for its answer on the metric , whereas which sets are copact depedends only on the topology of the space.</p><p>　　EXAMPLE1 . The unit sphere and the closed unit ball in ar

97、e compact because they are closed and bounded. The set </p><p>　　is closed in , but it is not compact because it is not bounded. The set </p><p>　　is bound in , but it is not compact because it

98、is not closed.</p><p>　　Now we prove the extreme value theorem of calculus, in suitably generalized form.</p><p>　　Theorem3 (Extreme value theorem) Let be continuous, where is an ordered set

99、in the order topology. If is compact, then there exist points and in such that for every .</p><p>　　The extreme value theorem of calculus is the special case of this theorem that occurs when we take to

100、 be a closed interval in and to be .</p><p>　　Proof. Since is continuous and is compact, the set is compact. We show that has a largest element and a smallest element . Then since and belong to , we

101、 must have and for some points and of .</p><p>　　If has no largest element, then the collection </p><p>　　forms an open covering of . Since is compact, some finite subcollection </p>