1. MA4266 Topology Wayne Lawton Department of Mathematics S17-08-17, 65162749 matwml@nus.edu.sgmatwml@nus.edu.sg http://www.math.nus.edu.sg/~matwml/ http://arxiv.org/find/math/1/au:+Lawton_W/0/1/0/all/0/1 Lecture 14

2. Test 2: Question 1 1. Assume It suffices to show that show that is open and We need to is openAssume that and Sinceit follows that hence and sinceis open 2. Let Then

3. Test 2: Question 2 1. are locally connected there existSince with open. and open connected Then is open connected and hence 2. The countable infinite product of the locally connected is locally connected. (with the discrete topology) is the Cantor set which is not locally connected.

4. Test 2: Question 3 Assume thatis compact and locally connected. Ifis a component ofand then since is open there exists an open henceis open hence everyTherefore component is open so since the set of components forms a partition ofit is an open cover. Since is compact there exists a finite subcover so there are only a finite number of components.

5. Test 2: Question 4 Assume thatis compact and let a Cauchy sequence in be Sincehas the and a subsequence Bolzano-Weierstrass property there exists such that It suffices to prove that choose Let so that and Then if

6. Test 2: Question 5 Assume thatis the Sorgenfrey line with the topologygenerated by the basis was locally compact then there would existIf a nonempty open setwith then there would exist a nonempty basic open set and hence compact. is compact - why? However - why? Since the set is not compact – why ?

7. Test 2: Question 6 Exercise 8.3 Question 3. Prove that every locally compact Hausdorff space is closed and with means if there exist disjoint open Proof Let Is regular. This be the one point compactification of then Now let so there exist disjoint openwith

8. Test 2: Question 6 We may assume that all The result on the hence are closed. Then Also preceding page implies there exist disjoint open sets Similarly there Since exist disjoint open sets hence We may also chooseso thatis compact. Clearly is compact

9. Supplement: Partial Orderings Definition A partially ordered set is a pair and a relation Question Compare this with equivalence relations. that satisfies the following three properties: consisting of a seton 2. (anti-symmetry) 1. (reflexivity) 3. (transitivity) Example Real numbers with usual order. Example Definition

10. Supplement: Linear Orderings Definition A linearly (or totally) ordered set is a is called a chain if that satisfies:partially ordered set If 1. (comparability) is a partially ordered set then is linearly ordered. Definition Rays are subsetsand Definition Ifis a linearly ordered set then the order topologyonis obtained from the subbasis consisting of rays. Example Ifis the real numbers with the usual ordering thenis the usual topology.

11. Supplement: Linear Orderings Theorem Let Proof Let be the order topology on be a linearly ordered set and Thenis compact iff has a glb and a lub. does not have a least upper bound denote the set of upperand let is an open cover ofthat has no finite subcover so is not compact. Similarly if then bounds ofIf does not have a greatest lower bound thenis not compact. For the converse letbe an open cover ofand letbe the set of such thatcan be covered by a finitely many elements ofIf every nonempty then compact else there exist Then is This is a contradiction.so

12. Supplement: Well Orderings Definition A well ordered set is a linearly ordered set such that every nonempty subset contains a minimal (or least) element is not the maximal element in Question Why is such an element unique? Definition If then its successor is given by Lemma Ifis an uncountable well ordered set then is uncountable. Proof. Otherwiseso let and observe thatsois uncountable.

13. Supplement: Well Orderings Theorem Let Then for every be a well ordered set. the closed interval is compact. Proof It suffices, by the preceding lemma, to show that every nonempty subsethas an glb and a lub. Since Let is well orderedcontains a minimal elementso then denote the set of upper bounds ofSince containsit is nonempty and therefore contains a minimal elementso then

14. Supplement: Countable Compactness Theorem Ifspace thenis a Proof Assume if and only if is countably compact has the Bolzano-Weierstrass property. If is countably compact and that is an infinite set thenis closed and there exist open such that Then is a countable open cover of that has no finite subcover, showing the only if part. Now assumehas the BW property and an open cover ofwith no finite subcover. Construct with is infinite andand observe that is This violates the BW property, showing the if part.

15. Supplement: Countable Compactness Exercise 3.3, Question15, page 181: (challenging) There are countably compact spaces that are not compact. on with the order topology is countably compact. It suffices be an uncountable set. The axiom of choiceProof Let Claim: by the theorem on the preceding page to show that implies that there exists a well order http://en.wikipedia.org/wiki/Well-order has the BW property. Let be countably infinite and let Sinceis countable LetIf (a basic open set) then hence

16. Assignment 14 Read pages 211-212, 213-220, 221-227, 231-232. Prepare to solve during Tutorial Thursday 25 March Exercise 7.3 problem 4, 6 Exercise 7.4 problems 6, 11 Exercise 7.5 problems 1, 8 Exercise 8.1 problems 6, 7