Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
MA4266 Topology Lecture 15. Tuesday 30 March 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 Chains and Maximal Elements Assume that ( X , ) is a partially ordered set. Definition A subset C X is called a chain if (C , ) is a linearly (or totally) ordered set. Definition An element m X is a maximal element if n X , n m n m. Remark If X has a maximum m max ( X ) then m is a maximal element. However a maximal element need not be a maximum element. Example X P({a, b, c}) \ {{a, b, c}}, U V U V { , {a},{a, b}} and {{b},{b, c}} are chains, { ,{a},{b, c}, X } and {{a},{b}} are not chains, {a, b}, {b, c},{a, c}} are (all 3) maximal elements. Zorn’s Lemma ( X , ) is a partially ordered set. If every chain C X has an upper bound then X has a maximal element. Lemma (Zorn) Assume that Proof This is equivalent to the axiom of choice, the well ordering principle, and the Hausdorff maximal principle. http://en.wikipedia.org/wiki/Zorn%27s_lemma Lemma Hausdorff Maximal Principle (HMP) Every chain in X is contained as a subset of a maximal chain. Proof It suffices by Zorn’s lemma to show that every chain of chains has a maximal element. This follows since the union of a chain of chains is itself a chain which is also an upper bound for the chains in the chain. Alexander Subbasis Theorem X is compact if and only if there exists a subbasis S for X with the following Property F: Every cover of X by members of S has a finite subcover. Proof If X is compact then every subbasis has prop. F. To prove the converse assume that S is a subbasis with property F and assume that X is not compact. It suffices to obtain a contradiction. Let O be an open cover of X that has no finite subcover. Covers are Theorem A space partially ordered by inclusion and the union of any chain of covers, each having no finite subcover, is an upper bound of the covers in this chain. So the HMP implies Alexander Subbasis Theorem that there exists a maximal open cover of X with the property that it has no finite subcover. So we can and will choose O to be a maximal open cover with this property. (remark: this required the Axiom of Choice) Lemma If U ,...,U n are open sets and there exists O O such that I U1 U n O then there exists i {1,..., n } such that U i O. Proof Else j { 1,..., n }, U j O O j ,1 ,..., O j ,m j O such that X O j ,1 O j ,m U j . This gives the j 1 contradiction n X O j 1 O j ,1 O j ,m j Alexander Subbasis Theorem Now for every U1 ,...,U n S OO p O there exist x U1 U n O. and every such that Question Why does this statement hold ? Then the preceding lemma implies that there exists U S such that x U O. Therefore x ,O {U x ,O : x O O } is an open cover of X of S by members and hence admits a finite subcover {U x1 ,O1 ,..., U xL ,OL }. Since U xi ,Oi Oi it follows that {O1 ,..., OL }is a finite subcover of X by members of O. This contradiction completes the proof. The Tychonoff Theorem Theorem7.11The product of compact spaces is compact Proof Let { X : A} be a collection of compact X be the product space and spaces, let X let A S { p (O ) : A, O open X } be the 1 ‘standard’ subbasis. It suffices, by the AST to show that every cover of X by elements of S has a finite subcover. Let U be such a cover and assume to the U did not have a finite subcover. For each 1 A let U { O open X : p (O ) U }. No finite subset of U covers X , so U doesn’t cover X , so x X that is not in any member of U . If x X , x( ) x , A then x U U U . contrary that Assignment 15 Read pages 234-237, 237-241, 243-251. Written Homework #3 Due Friday 9 April 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 is called a ‘freeflyer’ if AF A , and is called a ‘highflyer’ it is a frisbee that is not a proper subset of another frisbee. Assignment 15 (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. (b) For | X | show that F { A X : | X \ A | } is a freeflyer. (c) Use the Hausdorff Maximal Principle to prove that every frisbee is contained in a highflyer. (d) Show that a frisbee for every F is a highflyer if and only if A X , either A F or X \ A F. Remark (b) and (c) ensure the existence of freeflying highflyers, and (d) shows that they are mind boggling ! Assignment 15 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) Let X and Y be topological spaces. Show that a function f : X Y is continuous at x X if and only if for every frisbee F (on X) that lands on x, f (F ) (clearly a frisbee on Y) lands on f (x ). A frisbee (b) Let X A X be a product space and let F lands on x X if and only if p (F) lands on p ( x) X , A. F be a frisbee on X. Prove that Assignment 15 (c) Prove that a space X is compact if and only if every highflyer lands on some point in X. (d) Prove The Tychonoff Theorem using highflyers.