MA4266 Topology Wayne Lawton Department of Mathematics S , Lecture 15. Tuesday 30 March 2010

Chains and Maximal Elements is a partially ordered set. Definition A subset are (all 3) maximal elements. is called a chain if Example is a linearly (or totally) ordered set. are chains, Assume that Definition An elementis a maximal element if Remark Ifhas a maximumthen is a maximal element. However a maximal element need not be a maximum element. are not chains,

Zorn’s Lemma Proof This is equivalent to the axiom of choice, the well ordering principle, and the Hausdorff maximal principle. has an upper bound thenhas a maximal element. Lemma (Zorn) Assume thatis a partially ordered set. If every chain Lemma Hausdorff Maximal Principle (HMP) Every is contained as a subset of a maximal chain. Proof It suffices by Zorn’s lemma to show that every since the union of a chain of chains is itself a chain chain of chains has a maximal element. This follows which is also an upper bound for the chains in the chain. chain in

Alexander Subbasis Theorem by members ofEvery cover of for Theorem A spaceis compact if and only if there exists a subbasis has a finite subcover. Proof If is a subbasis is compact then every subbasis has prop. F. with property F and assume that To prove the converse assume that with the following Property F: is not compact. It suffices to obtain a contradiction. Letbe an open cover ofthat has no finite subcover. Covers are 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 will choose that there exists a maximal open cover of property that it has no finite subcover. So we can and property. Lemma If to be a maximal open cover with this (remark: this required the Axiom of Choice) with the are open sets and there exists such thatthen there such thatexists Proof Else such thatThis gives the contradiction

Alexander Subbasis Theorem Question Why does this statement hold ? Now for every there exist such that Then the preceding lemma implies that there exists and every such that is an open cover of Therefore by members ofand hence admits a finite subcover Sinceit follows that is a finite subcover ofby members of This contradiction completes the proof.

The Tychonoff Theorem Proof Let Theorem7.11The product of compact spaces is compact be a collection of compact be the product space and spaces, let be the ‘standard’ subbasis. It suffices, by the AST to show by elements ofhas a finitethat every cover of be such a cover and assume to thesubcover. Let let contrary thatdid not have a finite subcover. For each let No finite subset ofdoesn’t coversocovers sothat is not in any member of If then

Assignment 15 Read pages , , Written Homework #3 Due Friday 9 April 1. Letbe a nonempty set. A subset is called ‘frisbee’ (on X) if it satisfies the following: and is called a ‘highflyer’ it is a frisbee that is not a proper subset of another frisbee. is called a ‘freeflyer’ if

Assignment 15 (a) For nonempty (c) Use the Hausdorff Maximal Principle to prove show that define Prove thatis always a frisbee and that it is a a highflyer if and only if S is a singleton set. (b) For is a freeflyer. that every frisbee is contained in a highflyer. (d) Show that a frisbeeis a highflyer if and only if for everyeitheror Remark (b) and (c) ensure the existence of freeflying highflyers, and (d) shows that they are mind boggling !

Assignment Let ‘lands on’ be a nonempty topological space and A frisbee(on X)if every open set that contains (a) Letbe topological spaces. Show thatand (on X) that lands onif and only if for every frisbee belongs to a functionis continuous at (clearly a frisbee on Y) lands on (b) Letbe a product space and let be a frisbee on X. Prove thatlands on if and only iflands on

Assignment 15 (c) Prove that a spaceis compact if and only if every highflyer lands on some point in (d) Prove The Tychonoff Theorem using highflyers.