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 5.

2. Completeness Definition A metric space (X,d) is complete if every Be a CS and define sequences and that the sequence of Cauchy sequence in X converges to a point in X. Example 3.7.1 (a) R is complete. Proof Let Show that Verify ALL the examples on page 87 in Example 3.7.1 intervalsis nested and Cantor’s nested interval theorem implies that there existssuch that Show Corollary

3. Function Spaces Example 3.7.6 bounded, real-valued functions on a metric space is a complete metric space with metric Proof. Letbe a CS inSince for every is a CS inhence there exists such that Letand choose and set of continuous, for every such that sois continuous. Show that it is also bounded.

4. Proof of the Baire Category Theorem Lemma A subset A of a metric space X is nowhere Proof  we assume that A is nowhere dense in X dense in X iff each non-empty open set in X contains an open ball whose closure is disjoint from A. it follows that and thatis nonempty and open. Since and hence Sinceis open there exists such that Then  ifthen there exists an open ball with  a contradiction

5. Proof of the Baire Category Theorem Thm 3.18 Baire Category Theorem. Every complete metric space is of second category. Proof. Let X is a complete metric space. Assume to the contrary that X is not of the second category. Then X is of the first category. This means that there exists a sequenceof nowhere dense sets with such that Since X is open, the lemma implies there exists a open ball We may assume that the radius of isSince is open it contains an open ballwith andContinuing gives a nested sequence whose centres form a Cauchy sequence whose limitfor allcontradicting

6. Contractions Definition Let is a contraction (with respect to d) if be a metric space. A function there exists a number such that Theorem 3.18 (The Contraction Lemma) If X is complete and such that Proof. Chooseand define Then hence a unique is a contraction then there exists is Cauchy sequence. Since X is complete there existssuch that Corollary

7. Hyperspace Hausdorff distance between closed subsets a metric space, http://en.wikipedia.org/wiki/Hausdorff_distance

8. Iterated Function Systems See problem 6 in: http://www.math.leidenuniv.nl/~shille/FNA/Assignment_1_final_20090312.pdf >> S = [3 -2 1]; >> for k = 1:10 L = (1/3)*S; R = L + (2/3); S = [L R]; end >> plot(S,0*S,'b.')

9. Iterated Fractal Question What set is obtained as the limit of these iterations ?

10. http://www.math.byu.edu/~grant/courses/m634/f99/lec4.pdf Picard–Lindelöf Theorem http://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem http://math.byu.edu/mathwiki/index.php/Math_634 http://math.byu.edu/mathwiki/index.php/Math_635

11. Completion of Metric Spaces Theorem 3.19 Let (X,d) be a metric space. Then there exits a complete metric space (Y,d’) and an isometric embedding e : X  Y for which e(X) is a dense subset of Y. Furthermore, (Y,d’) is unique up to metric equivalence. Proof pages 91-92 (here is one part of the proof)

12. Assignment 5 Read pages 99-108 in Chapter 4 Study problems in Exercises on p.102 and p. 108-109