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MA4266 Topology Lecture 18. Friday 16 April 2010 Wayne Lawton Department of Mathematics S17-08-17, 65162749 [email protected] http://www.math.nus.edu.sg/~matwml/ http://arxiv.org/find/math/1/au:+Lawton_W/0/1/0/all/0/1 Bounded Continuous Functions Definition For a topological space ( X , T ) we denote by C ( X , R) the vector space of bounded, continuous, real-valued functions with domain X and topology determined by the metric ( f , g ) lub {| f ( x) g ( x) | : x X }. Exercise 8.6 Problem 4 (proof similar to Example 3.7.6 on pages 93-94) that (C ( X , R), ) is complete so it is a Banach space (a complete normed vector space). Prove this ! Complete Regularity ( X , T ) is called completely regular (page 247) if it is T1 and T3 1 . Definition A topological space 2 The latter condition means that for every closed C X and p X \ C there exists a continuous f :X R with f ( p) 0, f (C ) 1. Prove that then we can choose f C ( X , R ). Definition A topological space ( X , T ) is a Urysohn space if for every pair of distinct points a continuous p, q X there exists f : X R with f ( p) 0, f (q) 1. Prove that completely regular Urysohn. Complete Regularity ( X , T ) is completely regular, then the weak topology for X generated by C ( X , R) is T . Theorem 8.20 If T be the weak topology generated by C ( X , R). Clearly T T Why ? For every U T and x U the set C X \ U is a Proof Let closed set satisfying of X x C so the complete regularity implies there exists a continuous f : X [0,1] f ( x) 0, f (C ) 1. Therefore 1 x f ([0, 12 )) T so U is a union of sets in T hence U T . Therefore T T . such that Stone-Cech Theorem Theorem 8.21 If X is completely regular then there exists a compact Hausdorff space ( X ) which contains X as a dense subspace and for which every f C ( X , R) ~ can be extended to f C ( ( X ), R ). Proof: For f C ( X , R) let I f be the smallest closed interval containing f ( X ) and let Y I f and define e : X Y f C ( X , R ) by p f (e( x)) f ( x), x X , f C ( X , R), and ( X ) e( X ) Clearly e is continuous Why ? If x y X there exists f C ( X , R ) such that f ( x) f ( y ) Why ? Hence e is one-to-one Why ? Stone-Cech Theorem To show that e : X Y is an embedding it suffices to show that it is open. Since it is one-to-one it suffices To show that there exists a subbasis S for the topology of X such that e maps every element of S to an open set in e( X ). Theorem 8.20 implies that 1 S { f (U ) : f C ( X , R), U open R } is such a subbasis for the topology of 1 X. 1 f e ( f (U )) p (U ) e( X ) is an open set in the subspace topology on e( X ) Y . Furthermore ~ Finally, the extension is given by f ( y) p f ( y). Why ? Answers to Homework #3 1. Let X be a nonempty set. A subset F P( X ) is called ‘frisbee’ (on X) if it satisfies the following: F, ( A F and A B) B F, ( A F and B F ) A B F. Frisbee’s are Filters in Mathematical Language http://en.wikipedia.org/wiki/Filter_(mathematics) is called a ‘freeflyer’ if AF A , Freeflyers are Free Filters, and is called a ‘highflyer’ it is a frisbee that is not a proper subset of another frisbee. Highflyers are Ultrafilters. Answers to Homework #3 (a) For nonempty S X define FS { A X : S A}. Prove that FS is always a frisbee and that it is a a highflyer if and only if S is a singleton set. F and ( A F and A B) B F ( A F and B F) A B F Choose any Then Fp pS and define is a filter, FS This shows that if FS and are obvious. Fp { A X : p A}. Fp , and FS Fp iff S {p}. an ultrafilter S is a singleton. The converse follows since Fp is clearly an ultrafilter. Answers to Homework #3 (b) For | X | show that F { A X : | X \ A | } is a freeflyer. F follows since | X \ | | X | ( A F and A B) X \ B X \ A | X \ B | | X \ A | B F. ( A F and B F) X \ ( A B) ( X \ A) ( X \ B) | X \ ( A B) | A B F. F is a filter. Furthermore, for every p X X \ { p} F AF A so F is a free filter. therefore Answers to Homework #3 (c) Use the Hausdorff Maximal Principle to prove that every frisbee is contained in a highflyer. X and let P be the set of filters that contain F. Then P s partially ordered by inclusion. For any linearly ordered L P the set U L P and is an upper bound for L . Let F be any filter of a set LL Hence by Zorn’s Lemma (equivalent to the HMP) there M P. ultrafilter that contains F. exists a maximal filter Then M is an Answers to Homework #3 F is a highflyer if and only if for every A X , either A F or X \ A F. Only If Part: Assume that F is an ultrafilter and let A X and A F. Construct the set G {G : G F ( X \ A) for some F F}. Then every set in G is nonempty since F ( X \ A) F A A F. Hence G is clearly a filter and F G . Since F is an ultrafilter F G hence X \ A X ( X \ A) G X \ A F. (d) Show that a frisbee Answers to Homework #3 F is a highflyer if and only if for every A X , either A F or X \ A F. If Part: Assume that F is a filter satisfying the property and F G where G is a filter. If A G then X \ A F else we obtain the contradiction X \ A G A ( X \ A) G . Therefore A F so F G and F is an ultrafilter. (d) Show that a frisbee Answers to Homework #3 2. Let X be a nonempty topological space and x X . F (on X) ‘lands on’ x if every open set that contains x belongs to F. A frisbee A filter lands on x if it converges to x. A filter hovers near x if x is a limit point of the filter. X is a topological space and x X the set Nx { A X : x int A } of neighborhoods of x is a filter called the neighborhood filter of x. Result A filter F converges to x (F x) iff Nx F. Definition If Answers to Homework #3 (a) Let X and a function Y be topological spaces. Show that f : X Y is continuous at x X F that converges to x, the filter G { G : G f ( A) for some A F } converges to f (x ). if and only if for every filter f ( x) O open and filter F x. since f : X Y is continuous there exists open U x such that f (U ) O. Since F x U F f (U ) G O G so G f (x). Only If Part: Let Answers to Homework #3 (a) Let X and a function Y be topological spaces. Show that f : X Y is continuous at x X F that converges to x, the filter G { G : G f ( F ) for some F F } converges to f (x ). if and only if for every filter If Part: Let f ( x) O open F be the filter x. Then F x and let consisting of all neighborhoods of G f(x) O G. Therefore there exists a neighborhood A of x with f ( A) O so x int A and f (int A) O f is continuous . hence Answers to Homework #3 2. Let X be a nonempty topological space and A filter F (on X) converges to if every open x belongs to X A X be a product space and let set that contains (b) Let x F. x X . F be a filter on X. Prove that F x if and only if G { G : G p ( A) for some A F } p ( x) for every A. Fx iff every subasic open set 1 x p (O) F so the conclusion follows from part (a). Answers to Homework #3 (c) Prove that a space X is compact if and only if every ultrafilter converges to some point in X . Only If Part. Let F be a filter on X . First we show that F hovers near some point. Otherwise for every p X there exists open O p p and Fp F with O p Fp . Clearly {O p : p X } is an open cover of X hence since X is compact there exists a finite subcover O p ,..., O p . m Then F j 1 Fp F satisfies F F X . This contradiction shows that there exists p X such that F hovers near p. Then the set below is a filter 1 m j G {G : G A B, for some A Np , B F} containing F and Np . So F ultrafilte r F G Np F p. Answers to Homework #3 (c) Prove that a space X is compact if and only if every ultrafilter converges to some point in X . X is not compact so O of X that does not If Part. Assume to the contrary that that there exists an open cover have an open subcover. Define the set FO { G : G X \ O for some O O }, observe that it is a filter, and let F be an ultrafilter that contains F so by assumption there exists p X O such that F p. Choose a O O with p O so X \ O FO X \ O F. Now F p O F hence O ( X \ O) F - a contradiction. Answers to Homework #3 (d) Prove The Tychonoff Theorem using ultrafilters. Theorem (Tychonoff) Let { X : A} be a family of compact topological spaces. Then the product space X A X F is compact. X . By 2 (c) it suffices to show F x for some x X . Define G { G : G p ( A) for some A F }, A Proof Let be an ultrafilter on the set and observe they are ultrafilters so 2(c) implies there exist x X such that G x , A. Define x X by p ( x) x , A. Therefore 2 (b) implies that F x. Exam Suggestions Knowledge Definitions: Basis, Hausdorff, Connected, Compact, etc. Results: Statements of Theorems, Lemmas, Corollaries Examples: Construction and Properties of Spaces Skills Derive: Proofs Construct: Examples Explain: Proofs of Major Theorems in Outline Form