General Topology 22M 132
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22M132 Fall 07 J Simon Metrization Theorems Relates to text Sec 34 Introduction What properties of a topological space X T are enough to guarantee that the topology actually is given by some metric The space has to be normal since we know metric spaces are normal And the topology has to have a countable local basis at each point since metric spaces have that property In Chapter 6 of the text there are theorems saying X T is metric if and only if it has the following topological properties 77 The conditions are l the space is regular and 2 there is some countability property stronger than saying there is a countable local basis at each point but a little weaker than 2nd countable but still strong enough that regular the property gt normal In the current text section the theorem is less general we characterize separable metric spaces but this is a good introduction to the ideas How does one prove that some topology on a space is given by a metric There are two choices either explicitly de ne the metric and prove the metric topology is the same as T or show that X is homeomorphic to a subspace of a known metric space We used the rst approach when we de ned a metric on R that generates the product topology and now we will see a good example of the other approach which is also how the most general metrization theorems are proven Theorem Urysohn metrization theorem If X T is a regular space with a courztable basis for the topology therz X is homeomorphic to a subspace of the metric space Rw The way I stated the above theorem it is ambiguous we have studied two inequivalent metrics for Rw the product space metric and the uniform metric The theorem is true with either metric but it is an if and only if 7 for the product metric Recall that in the product topology R has a countable dense subset the set S all vectors Q1412 where each q E Q and all but nitely many q are 0 Since R in the product topology is metrizable and has a countable dense subset it must be 2nd countable Each subspace of space with a countable basis also has a countable basis And of course each subspace of a metric space is metric We conclude that each subspace of Rw product metric is metric and has a countable basis The above paragraph combines ideas from various parts of our course So let us take this as one bunch of of our sample problems for the Final Exam Your task s to organize the various facts ideas etc into a coherent proof and to be able to ll in details if asked as always an exam problem might involve lling in the details of one particular part of a longer argument J Simon all rights reserved page 1 Problem sample for nal epam If X77 is homeomorphic to a subspace of RR product topology then X77 is regular and has a countable basis Key steps There is a metric on R that gives the product topology R in the product topology has a countable dense subset c A metric space with a countable dense subset has a countable basis for the topology Each subspace of a 2nd countable space is 2nd countable Each subspace of a metric space is metrizable Each metric space is regular Put the pieces together F7 thDFDFL On the other hand7 the metric space RR uniform metric is not second countable so not separable since it has an uncountable discrete subspace K the set of all vectors b17b27 where each ti 0 or 1 The set of all sequences of 07s and 17s is uncountable7 and the distance between any two elements of K is 1 So each subspace of RR uniform metric topology is a metric space7 but it need not be separable We really should state the Urysohn metrization theorem as two theorems Theorem X77 is regular with a countable basis ltgt X77 is homeomorphic to a subspace of RR product topology metric Theorem X77 is regular with a countable basis gt X77 is homeomorphic to a subspace of RR uniform metric topology Proving the metrization theorems The text gives the details7 so I will focus on the gestalt and some highlights Our goal is to de ne an embedding of X into RR We want to assign to each point z E X a point 6 RR that is a countably in nite list of coordinates 17 27 How can we nd numbers that measure how a point z E X is related topologically to all the other points of X This is the bit of magic in this theorem We will use Urysohn7s lemma in nitely many times to de ne a sequence of functions fn X a 07 1 these will be the coordinate functions The space X77 has a countable basis 8 and it it regular7 so it is normal Given any closed set A and open neighborhood UA7 there exists a Urysohn function for the disjoint closed sets X 7 U and A That is7 there exists f X a 071 such that fd 0 for all z U and fa 1 for all a E A In particular7 for any pair BmBm of elements of B that happen to have En Q Em7 there exists a function f X a 071 with f 1 on En and f 0 outside Bm J Simon all rights reserved page 2 Since 8 is countable the set of such pairs B B is countable Number these pairs in any order and let f1 f2 be the Urysohn functions de ned in the preceding paragraph Then de ne F X a 01 by FW f1967f2967 We need to prove that the function F is 1 1 continuous and has a continuous inverse From FX a X The questions of continuity have to depend on what topology we use for Rw But we can check injectivity before worrying about the topology Proposition The function F X a R is injectiue Proof Suppose my 6 X with z 31 y Since X is Hausdorff there exist disjoint neighborhoods Ux Since 8 is a basis there exists some B with z E B Q U Since X is regular there exists a neighborhood U z such that U Q Bm And again since 8 is a basis there exists a basis set 8 with z 6 En Q U Since 7 Q Bm we thus have En Q Bm The Urysohn function f associated to the pair B B has z a 1 and y a 0 so 31 The text goes on to show that in the product topology F is continuous and has a continuous inverse The proof that F is continuous is easy because each coordinate function is continuous the proof that F is an open map takes more work see the text for the details To use the uniform topology we need to change F Recall that in the product topology if we are studying a function from a space into a product space ie some G Y a Had Xa and we want to show that G is continuous it is suf cient to check that each component function GD Y a X0 is continuous But in the uniform topology this is not suf cient Example Page 127 problem 4a The function G R a R de ned by Gt t2t3t4t is not continuous in the uniform topology on R In particular there is no neighborhood of0 that is mapped by G into a uniform e neighborhood of 0 0 0 To make the coordinate function F topologically well behaved77 for the uniform metric on R we need to eliminate the dif culty suggested by the above example We do this by making the coordinate functions f get smaller as j gets larger Speci cally de ne c X a R by Gx f1x gnu gnu ism There is one more part of the proofs that is cute clever or annoyingly slick depending on your tastes The uniform metric topology is ner than the product topology on Rw SO once we know F is an open map in the product topology that takes work it is easy to see that F hence G is an open map in the uniform topology Conversely once we know G is continuous in the uniform topology that takes work it is easy to see that F is continuous in the product topology J Simon all rights reserved page 3 Here are some sample problems for the Final Exam that you can use to solidify your understanding of these proofs The rst is an easier special case the others are the standard77 Urysohn rnetrization theorern Problem sample for nal epam Prove IfX is a compact Hausdor space with a countable basis then there epists an embedding ofX into R where R has the product topology Problem sample for nal epam Write a one to two page proof IfX is a regular space with a countable basis then there epists an embedding ofX into R where R has the uniform topology Problem sample for nal epam Write a one to two page proof IfX is a regular space with a countable basis then there epists an embedding ofX into R where R has the unifom topology end of handout J Simon all rights reserved page 4 22M132 Fall 07 J Simon Notes and Homework on Locally Compact Spaces Compact spaces especially compact Hausdorff spaces are extremely nice77 as we have already studied optimization problems have solutions continuous functions are uniformly continuous integrals exist There is a more general class of spaces that are important for example7 they include R and that arise a lot in analysis see7 for example7 the Riesz representation theorem These spaces are too big to be compact7 but they are compact when looked at from close up More precisely De nition A space X is locally compact iffor each x E X there epists an open neighborhood U ofz with closure U compact When X is also Hausdorff7 the property of local compactness becomes much stronger Let7s state this as a theorem Theorem 1 IfX is locally compact and Hausdor z E X and U is any neighborhood of x then there epists a neighborhood V ofz such that the closure V is compact and V Q U Remark So not only does z have some neighborhood with compact closure it has many in fact it has arbitrarily small neighborhoods with compact closure The text proves this theorem by rst embedding X in its one point compacti caton lnstead7 let7s prove the theorem more directly7 and then use this tool to help understand the one point compacti cation space Ultimately7 we are all doing the same dirty work 7 just changing the order in which we encounter various issues And I think the approach in these notes makes the issues clearer Lemma 11 IfX is Hausdor z E X and C is a compact subset ofX with z 0 then there eccist disjoint neighborhoods Uz and VC Proof This is stated as Lemma 264 in the text The technique for this proof is something you should know well7 useful for other theorems7 so here is the proof Since X is Hausdorff7 for each point y 6 07 there are disjoint neighborhoods of z and y let7s call these Uyx and The set C is covered by y E C and7 since 0 is compact7 there is a nite subcover W177VW So U Uyl Uy and V Vyl U U Vy are disjoint neighborhoods of z and C respectively B J Simon all rights reserved page 1 Lemma 12 In a IIausdor space X suppose U is a neighborhood of a point z and de is compact Then there epists a neighborhood V ofx such that the closure V Q U Proof By assumption de is compact Then by Lemma 11 there existidisjoint neighborhoods W of z and W of bd U Note this implies that the closure W is disjoint from bd U Let V U W Then VgU WUUde WU WUde VTU WU QU D Remark The idea in the preceding lemma is that if we can separate z from the boundary of a neighborhood Uz then we can shrink U to a neighborhood that is deep within U that is the closure of the new neighborhood is contained in U Proof of Theorem 1 We have x E U where U is a given neighborhood of x By de nition of local compactness there exists alnother neighborhood W of x such that the closure W is compact This makes any closed set contained in W also compact Consider the set V1 U W We might hope that V1 is the desired neighborhood of x it certainly is contained in U But its closure is in general not contained in U So we have to trim it down77 a little The set de1 is closed and contained in Vl Q W which is compact so de1 is compact By Lemma 12 there exists a neighborhood Vz such that the closure V E V1 but since V1 U W this says V Q U D Remark for the future Along with nding neighborhoods of a point that lie deep within a given one we also can use the same kind of thinking separate points from compact sets or separate compact sets from each other to get large families of nested neighborhoods In fact we can construct inductively a countable family of neighborhoods ofx inside a given U where the countable family is indecced by rationals of the form for all positive integersj and n such that the containment relations between the neighborhoods is the same as for the intervals 0 This ultimately lets us construct continuous functions from X to R that quotseparate points or quotseparate points from closed sets In a locally compact IIausdor space given two points A B or a point A and a closed set E missing A there epists a continuous function f X a R such that u 0 and fa 1 for all a E A This property is sometimes called completely regular We ll see more theorems like this in later sections J Simon all rights reserved page 2 22M132 Fall 07 J Simon Handout 2 Correction of Handout 1 plus continued discussionHW Comments and Homework in Chapter 1 Chapter 1 contains material on sets functions relations and cardinality that most all of you have studied before We will do a quick review of this material Sections 1 7 just to give you a chance to re lmaster some of the details practice writing careful proofs learn the texts de nitions that will be used throughout the course and get re lminded of some foundational theorems that we will need during the course We will skip Section 8 and discuss Sections 9 11 later when we need that material I dont want us to spend to much time and energy worrying about details of set theory and logic But there are some connections between topology and set theorylogic that should not be ignored For example the topology Tychonoff theorem on products of compact sets Chapter 5 is logically equivalent to the Axiom of Choice Section 1 Sets Unassigned practice problems If you can work these quickly that7s good If you have problems get help from the TA Page 14 2 acegikmoq 37 4 Section 2 Functions Note de nitions of injective 1 1 surjective onto bijectiive 1 1 correspondence Assigned HW problems give you a chance to practice with images and inverse images of sets very important for later that you have these ideas well mastered and various kinds of functions Homework Due Wednesday Sept 5 Page 20 1 2 feh It would be good also to practice all of 234 and 5 if you feel ambitious Section 3 Relations Key ideas 0 Key ideas equivalence relation and partition of a set Be able to prove the following which is partly done in the text but not as a clearly stated theorem Theorem 1 If is an equivalence relation on a set A then the set of equivalence classes is a partition ofA 2 Suppose A Aahe is a partition of a set A De ne a relation on A by z N y ltgt there epistsj E J such that z E A andy E Aj Then N is an equivalence relation J Simon all rights reserved page 1 0 Key idea ordered set The usual relations of lt77 and g for numbers can be generalized to various kinds of orderings of sets sets Find the text de nitions of partial ordering strict partial ordering linear ordering The important distinctions are a whether each element of the set can be compared to each other element and b whether an element can be related to itself You may have to re read the de nitions each time you work on some theorems but keep these main distinctions in mind and you should be ne Homework Due Wednesday Sept 5 Page 28 1 3 Find the fallacy AND give a counterexample 4 Probably will be easy for you but this is a very important special case 11 An exercise in being careful with the de nitions 13 May take some thought Remember that ordered77 set means page 24 there is a strict linear ordering Section 4 The Integers and the Real Numbers Just skimbrowse this section We will happily assume the existence and all the usual properties of Z Z1 Q R and C In particular we will do induction proofs without worrying about any set theorylogic technical complications Remark Just to keep us humble in thinking we understand these familiar sets See the discussion in httpenwikipediaorgwikiContinuumhypothesis about the so called quotcontinuum hypothesis We know Z1 Z and Q have the same size as each other and we know that R C and R 7 Q have the same size as each other and are strictly larger than the sets in the rst group but we do not know and in a sense can not know if there is a set whose size is properly in between those two End of handout Discussion of Chapter 1 to be continued Section 5 Cartesian products Once you de ne an indecced family of sets you can de ne unions intersections and products UAjalHj Jsuchthata AJ jeJ Ajale J7 QEAJ39 jeJ HAjp J UJEJAJ39 le 6 J7 6A7 J Simon all rights reserved page 2 If the index set J is 17 27 7n or even all of Zo we can write elements of the cartesian product as n tuples or sequences7 eg a17a27a3 or a17a27a37 But if the index set is larger7 eg all of R then we have to use the function notation to de ne and analyze the cartesian products We often will be dealing with cartesian products that involve crossing one set with itself some number of times For example7 R3 or R or RR Another example that you have seen7 or will see7 in analysis courses is RW and the subset Ca7b f a7b a R l f is continuous Homework Due Wednesday Sept 5 Page 38 2 might help to read 1 rst 3d 4 d7 e It would be good for you to readunderstand all of 1 7 4 Section 6 Finite sets We will accept this material without fussing You should readskim to be sure you are comfortable with these ideas In particular7 a set is nite if it is empty or in 1 1 correspondence with 17 27 7n for some n E Zr A nite set cannot be put in 1 1 correspondence with a proper subset of itself It is easy to see bijections between Z and proper subsets7 so Z is not nite There is no assigned HW in Section 6 Section 7 Countable and Uncountable sets This section is very important In addition to further honing your abilities to write proofs7 the facts themselves will be used often in the course Propositions such as o a subset of a countable set is countable7 or o a countable union of countable sets is countable7 or o the caitesian product ofn countable sets is countable7 or 0 any in nite product of sets is uncountable unless nearly all the sets are empty or singletons are things you should believe and understand7 as well as being able to prove them There are two main ways to show that a set X is countable One is to de ne explicitly a bijection between X and a set previously known to be countable the other is to use some version of the following theorem J Simon all rights reserved page 3 Here is a slightly specialized version of text Theorem 71 Theorem a IfX Q Z then X is countable b IfX is a set such that there em39sts afunctlon f Z a X that is surjectlue then X is countable Note This proof in the text involves some set theory fussing that I would like to avoid So for this theorem you should understand the theorem well but I wont ask you to reproduce the detailed proof Proof Remark Both parts of this proof use the fact that the usual size ordering of Z1 is a well ordering ie a linear ordering such that each nonernpty subset has a srnallest elernent Part a De ne a bijection between X and some initial segment of Z1 either 1 2 n or all of Z1 inductively as follows Let 1 denote the smallest element of X and de ne fz1 1 Let X1 Xz1 lf X1 is empty we are done if X1 74 Q then let x2 the smallest element of X1 and de ne fz2 2 Continue inductively that is where the set theory fussing happens to get the desired bijection f between X and a set 1 2 n or all of Z1 Partb For each x E X let Am be the preiirnage f 1z The sets Awhex are a partition of Z1 We know page 28 Exercise 4 that this set of equivalence classes is in 1 7 1 correspondence with X Form a set W Q Z1 by picking one element from each set Am For example again using the well ordering of Z1 we can pick am to be the smallest element of Am Because the sets A1 are pairwise disjoint the elements am are all distinct from each other So the set W amhex is in 1 7 1 correspondence with X But W is a subset of Z1 so W is countable and therefore X is countable D Corollary Theorem 71 as stated in the teat Proof Exercise for the reader Use D Corollary 74 Z1 gtlt Z1 is bljectluely equivalent to Z1 Proof Note This illustrates the second main technique for proving that sets are countable Exhibit a bijection that does the job The function f nm a 2 3 is a bijection between Z1 gtlt Z1 and a subset B of Z1 Corollary call it 741 If XY are countable sets then X gtlt Y is countable Proof Exercise for the reader This follows from Corollary 74 and Theorem 71 J Simon all rights reserved page 4 Corollary A nite product of countable sets is countable Proof Exercise for the reader This follows by induction and your earlier homework page 38 2a from the previous Corollary D Theorem 75 A countable union of countable sets is countable Proof Slightly different from text7s proof The set Z1 gtlt Z1 is a countably in nite union of copies of Z17 so any other countable union of countable sets can be injectively mapped into Z1 gtlt Z1 Roughly7 use the rst Z1 coordinate to represent the index and the second Z1 factor to capture the set elements So any countable union of countable sets is bijectively equivalent to a subset of the countable set Z1 gtlt Z1 Theorem Q is countable Proof Let7s try to do this in careful steps 1 Z is countable Exhibit a speci c bijection between Z and Z1 2 Z1 Z0 is countable This follows from step 1 and Theorem 74 3 Z gtlt Z1 is countable This follows from steps 17 27 and Corollary 741 above 4 The function 77171 a g is a surjection of Z gtlt Z1 onto Q Now invoke Theorem 74 again D Homework Due Wednesday Sept 5 Page 51 3 4 5 C d e f g h You can justify your answers77 by citing theorems or earlier problems If you want some more problems to think about7 try showing 0 There is a bijection between R and R1 R0 0 There is a bijection between R gtlt R and R END OF HANDOUT ON CHAPTER 1 J Simon all rights reserved page 5 22M132 Fall 07 J Simon Urysohn s Lemma Relates to text Sec 32 36 Introduction Saying that a space X is normalturns out to be a very strong assumption In particular normal spaces admit a lot of continuous functions Theorem Urysohn7s Lemma IfAB are disjoint closed sets in a normal space X then there epists a continuous function f X a 01 such that Va 6 A fa 0 andVb E B fb 1 This lemma has several big77 applications 1 Urysohn Metrization Theorem Section 34 IfX is a normal space with a countable basis ie second countable then we can use the abundance of continuous functions from X to 01 to assign numerical coordinates to the points ofX and obtain an embedding ofX into R From this we see that every second countable normal space is a metric space Tietze Extension Theorem Section 35 This theorem is a useful technical tool rather than a big conceptual climax like the Metrization theorem Suppose A is a subset of a space X and f A a 01 is a continuous function IfX is normal and A is closed in X then we can nd a continuous function from X to 01 that is an extension off Embedding manifolds in RN Section 36 A space X is called a topological n manifold if each point z E X has an open neighborhood Uz such that U is homeomorphic to an open n ball Usually people insist that X be Hausdorff sometimes 2nd countable sometimes paracompact de ned later in the text Right now our text assumes X is Hausdorff and has a countable basis so we will assume that as part of the de nition of manifold The spaces R S any open subsets of R are n manifolds the cartesian product of an n manifold with a k manifold is a nk manifold so there are lots of manifolds If f R a R is a smooth function see 22Ml33 then for almost all a E R the solution set W w E R fw a is a n l manifold So solution sets of equations often are manifolds What about the other way around ls every manifold actually realizable as a solution set in some R Maybe an easier question ls every manifold homeomorphic to a subspace of some R Using Urysohn7s lemma to develop the tool called partitions of unity we obtain the following theorem A D V A 0 V Each compact n manifold is homeomorphic to a subspace of some RN J Simon all rights reserved page 1 Of course the N77 depends on X it is impossible to embed a high dimensional manifold in a low dimensional RN in fact7 generally N Z n The embedding theorem in our text does not attempt to control the dimension N however7 the machinery of smooth manifolds 22M133 or 22M200 allows one to embed each compact n dimensional smooth manifold in R2 Proof of Urysohn7s Lemma The text gives a detailed proof So this discussion offers a different approach to some parts7 and includes more comments and motivation for what we do But the actual proof is close to the texts Somehow7 we have to associate to each point z E X a number7 Furthermore7 our function f has to be continuous otherwise the proof would be trivial and the theorem would have no meaningful content7 send set A to 07 and B to 1 All we know about X is the hypothesis that X is normal Here is the plan We are going to de ne a certain large collection of open sets in X then we will decide for each x E X what f should be by looking at which of these open sets do7 or do not7 contain x Let D be the set of dyadic rationals in 017 that is D 07 La 3 g g We will construct a sequence of open sets Uq in X7 indexed by q E D First7 let U1 X Now we get to the real work Since X is normal7 there exist disjoint open neighborhoods UA andjB Note that the existence of V disjoint from U tells us that U B Q ie U Q X 7 B Let U0 be this neighborhood UA For all the subsequent sets Uq that we will de ne7 we will have A Q Uq and for all q lt1Uq B Q The closed set U0 is contained in the open set X 7 B Since X is normal7 there exists an open set call it Ulg such that U0 Q U12 Q 712 Q X B We continue inductively interpolate U14 between U0 and U12 interpolate U34 between U12 and X 7 V then de ne Uls UgS etc We get a sequence of open sets Uq such that 1 For each q 6 D7 A Q Uq 2 BQ U1 andforeachqlt1B Uq 3 For each pq E D with p lt q we have Up Q Uq We now de ne f X 7 Q1 by fx mfq l 96 E Uq for each x E X J Simon all rights reserved page 2 The function f is de ned because each point of X is contained in some Uq at least in U1 X By condition 1 f is 0 on set A And by condition 2 f is identically 1 on set E Note We are not claiming that f is 0 only on A or 1 only on B In general the zero set and 1 set of f will be larger than just A and B It remains to show that f is continuous This is Step 4 in the texts proof We rst establish two mini lemmas A If fx gt q then z Uq B If fx lt q then x E Uq For each x E X let Dz q l x E Uq So fx inf The numbers q and the sets Uq are ordered the same way So if q E Dz and q gt q then 1 E The in mum f might be the smallest element of Dz or it might be a lower limit point that it not itself in Proof of A If f gt q then there must be some gap between q and D in particularLthere exists some 1 such that q lt q lt But 1 lt fx gt z W and then Uq Q Uq gt z Uq Proof of B If f lt q then there exists 1 E D such that fx lt q lt q in which case q E D so z E U We now can show that f is continuous We need to show that the pre image of each sub basic set a1 or 0a is open in X Suppose rst that f E a1 Pick some q with a lt q lt We claim that the open set W X 7 Uq is a neighborhood of x that is mapped by f into a1 First by A f gt q gt z E W so W is a neighborhood of x If y is any point of W then fy must be 2 q gt 1 otherwise if fy lt 17 then by B y E Uq Q Uq The argument is simpler for f 10b Suppose f lt b and pick q such that f lt q lt b By B x E Uq We claim that the neighborhood Uq is mapped by f into 0b Suppose y is any point of Uq Then q E Dy so fy S q lt b HOMEWORK in Section 33 Due Wednesday Dec 12 Page 212 1 ie do you understand the proof of Urysohn7s Lemma 2 cute see sample Final Exam problems below for another way to ask the same questions 3 practice with metrics 7 a reView of LCH spaces To warm up for this you might rst prove that a compact Hausdorff space is completely regular Also remember our old friend the pasting lemma J Simon all rights reserved page 3 Problem Sample problem for nal epam Write a one page summary outline sketch of a proof of Urysohn s Lemma Problem Sample problem for nal epam a Suppose X is a countable space Proue X has the Lindelo39f property b There epists free gift not done in our class a space X that is countable Hausdor and connected Use this space to show that the theorem quotregular Lindelo39f gt normal cannot be modi ed to say that quotHausdor Lindelo39f gt regular end of handout J Simon all rights reserved page 4 22M132 Fall 07 J Simon Some highlights on Countability and Separation Properties Relates to text Sec 30736 Introduction There are two basic themes to the next several sections a What properties of a topology allow us to conclude that the topology is given by a metric b What properties of a space allow us to conclude that the space actually is homeomorphic to a subspace of R or at least a subspace of R Countability Properties Here are several properties of spaces7 all saying that the topology7 or some key feature of it7 can be described in terms of countably many pieces of information The names are historical they are not very descriptive or otherwise useful7 but you should know them since they are used in the literature 1 First axiom of countability77 The space X77 is called rst countable if the topology has a countable local basis at each point z E X 2 77Second axiom of countability77 The space X77 is called second countable if the topology 7 has a countable basis 3 77Separable77 The space X77 is called separable if X contains a countable dense subset Recall a subset A Q X is called dense in X if the closure A is all of X7 ie each open set contains at least one point of A 4 Lindelof property The space X77 is called a Lindela39f space if each open cover of X has a countable sub cover The familiar space R72 with the standard topology has all of the above properties proof below For more general spaces7 we can ask many questions 0 Do any of these properties imply others o If a space X has one of the properties7 do all subspaces of X have the property In that always happens7 we would call the property hereditary o If we have a family of spaces with one of these properties7 does the cartesian product have the property o If f X a Y is a continuous surjection7 and X has one of the properties7 must Y also have the property o If X77 has one of these properties7 and T is a coarser resp ner topology7 must X7T have the property We will focus on just some highlights J Simon all rights reserved page 1 Theorem text 303 Countable basis gt all the other countability properties Proof Suppose B is a countable basis for the topology on X a Countable local basis Let x E X and let U be any neighborhood of x Since 8 is a basis for the topology U is a union of elements of 8 Thus there exists an element B E B such that z E B Q U So the set 8 is a countable local basis for each point z E X Separable For each nonempty set E E 8 pick a point 3 6 B Since 8 is countable the set 3 l B E B is countable Since each open set is a union of elements of 8 each nonempty open set U contains at least one of the sets B and so x3 6 U Thus 3 l B E B is dense in X Lindelo39f Let Uahe be an open cover of X We want to prove there exists a countable subcover by somehow using the existence of a countable basis 8 for the topology For convenience to make the exact argument a little simpler assume that one of the sets U0 is actually the empty set Or adjoin one additional set U0 Q to the covering The idea of the proof is to use the elements of B to point to77 certain special UDs Speci cally for each set E E B we will select one set UB from among the UDs as follows First ask if there exists at least one of the open sets U0 containing that set B If not let UB U0 Q If the set E is contained in some Ua then pick one such U0 and call it UB We might pick the same U0 corresponding to several B7s because a given U0 usually contains many basis sets but we have at most one U0 chosen for each B so the set UB B E B is countable We now show that UB B E 8 covers X Let x E X We shall prove that at least one of the sets UB contains x Since the UDs cover X there is some U0 containing x Since 8 is a basis there exists B E B with z E B 6 U0 Since that basis set E is contained in some Ua B is one of the basis sets for which we chose a set U3 2 B So z E B Q UB in particular z 6 U3 F7 0 D The previous theorem says that having a countable basis for the topology is the strongest of the countability properties The next example shows that it is strictly stronger that is the other properties do not imply it Example ex 3 page 192 The space RI is rst countable separable and Lindelo39f but not second countable Proof The details are given in the text you should be able to prove RI is rst countable and separable and that it is not second countable You are not required to know the proof that RI is Lindelo39f D The previous example shows some of the independence of the properties However in metric spaces the rst countable separable and second countable properties are equivalent J Simon all rights reserved page 2 Theorem Exercise 5 page 194 Suppose X is a metric space Then i X has countable local bases at each point ii X separable gt X has a countable basz39s iii X Llndelo39f gt X has a countable basz39s Proof i The idea of countable local basis is precisely a generalization of the balls of radius 171 in metric spaces The set Bx is a local basis at x ii Let znheN be a countable dense set in X For each x let 8 be the set of all open balls centered at xn with rational radius Then the set 8 is countable for each n so the set 8 UBn n E N is countable We claim this set 8 is a basis The proof is an exercise in using the triangle inequality that the metric satis es You can work out the details here is the idea Take any open set U Q X We want to show that U is a union of our alleged basis elernents Let y be any point of U we shall show that there is one of our alleged basis sets B such that y E B Q U The point y is contained in some 6 ball inside U Now look at an 6100 ball around y This ball must contain some point xn from our countable dense subset So the distance from xn to y is less than 6100 Then by picking a rational radius slightly larger than 6100 we can nd a rational radius ball centered at z containing y and contained in U iii For each n consider the open covering of X consisting of all balls of radius 171 The Lindelo39f property says there exists a countable subcover B Let B LAB n E N This is a countable union of countable sets hence countable Check that it is indeed a basis D Remark The the preceding proofs it might seem that all we need is separable countable local basis or Lindelo39f countable local basis to conclude that X has a countable basis But remember the previous example of a space that IS separable DOES have the Lindelo39f property DOES have a countable local basis at each point but does not have a countable basis for the topology The triangle inequality property of metrics provides an extra amount of niceness for the topology so we can connect from one property to the other Now let us consider subspaces and cartesian products Theorem 302 The properties countable local basis and countable basis are preserved for subspaces and for countable cartesian products Proof Proofs are given in the text D J Simon all rights reserved page 3 Theorem The Lindelo39f property is inherited by closed subspaces analogous to compactness But it is not necessarily inherited by arbitrary subspaces see Example 4 on page 193 Proof This is Problem 9 on page 194 Consider it one of your possible problems for the Final Exam D Example The property of being separable need not be inherited by subspaces having the subspace be closed seems irrelevant to this question The space RI gtlt RI is separable but the diagonal line d7 7x z E R is a closed subspace that is homeomorphic to R with the discrete topology This uncountable closed discrete subspace makes R a useful counteredample for several questions It shows that being separable is not hereditary the Lindelo39f property is not always preserved by nite products and nept section the property of being normal is not always preserved by nite products Homework in Section 30 Page 194 Problems 27 37 47 117 13 Additional problems it would be good to study consider these among the sample problems for the nal exam Problems 57 6 just R07 77 97 107 14 end of handout J Simon all rights reserved page 4
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