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
Orientability wikipedia , lookup
Michael Atiyah wikipedia , lookup
Sheaf (mathematics) wikipedia , lookup
Geometrization conjecture wikipedia , lookup
Covering space wikipedia , lookup
Surface (topology) wikipedia , lookup
Fundamental group wikipedia , lookup
Brouwer fixed-point theorem wikipedia , lookup
Continuous function wikipedia , lookup
MA4266 Topology Lecture 7. 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 1st Countable Spaces Theorem Every metric space is first countable Proof p X ,{B( p, 1n ) : n 1,2,3,...} is a local basis at p. Theorem A 2nd countable space 1st countable and separable. Proof Assume that X is 2nd countable with a countable basis B. Then for every p X the set B p {O B : p O} is a local basis at p. For every nonempty O B choose xO O and define the set D {xO : (O B) (O )}. Then D is countable and dense ( D X ) (why?). Bases Theorem 4.8 If X is a set and B P( X ) then B is a basis for a topology T on X iff B satisfies: (a) O X OB (b) O1 B, O2 B, x O3 B x O3 O1 O2 O1 O3 x O2 Proof Follows since T { OA O : A B} Bases Theorem 4.9 Bases B and B are equivalent iff (a) O B, x O, O B x O O and (b) O B, x O, O B x O O Remark Let T be the topology generated by B and let T be the topology generated by B. Then condition (a) is equivalent to T T (or T is weaker than T , or T is stronger than T , and condition (b) is equivalent to T T See 7.3 Comparison of Topologies on pages 211-213 Subbases Exercise 4.3 Problem 10. If X is a set and S P( X ) satisfies AS A X then the family B of finite intersections of members of S is a basis for a topology T on X . Then S is called a subbasis for T . 2 Example Consider ( R , T ) where T is the usual topology whose members consists of unions of open balls. Then the following set is a subbasis for T S {( a, b) R : a, b R, a b} {R (a, b) : a, b R, a b} Proof B {( a, b) (c, d ) : a, b, c, d R, a b, c d } is equivalent to the basis for the usual topology that consists of open balls as shown by Continuity Definition ( X , T ), (Y , T ) topological spaces, a X , f : X Y is continuous at a if V T [( f (a) V ) (U T [( a U ) ( f (U ) V )] )] f : X Y is continuous if a X [ f : X Y is continuous at a ] Alternative Definition f : X Y is continuous if V T [ U T [ f (U ) V ] ] Theorem 4.11 If f : X Y then following are equiv. 1 (2) (1) f is continuous, C closed f (C ) closed (3) A X , f ( A ) f ( A) (4) basis B for T O B, f 1 (O) T (5) subbasis S for T O S , f 1 (O) T Exotic Topologies for R Sorgenfrey Line = R with the half-open interval topology generated by B {[a, b) : a b }. Question Why is B a basis for some topology? Question Is Q dense in R Question Does this topology have a countable basis? Question Is [0,1] compact ? Is R connected ? R with the countable complement topology Question Does R have a countable dense subset ? Question Does this topology have a countable basis? Question Is [0,1] compact ? Is R connected ? Subspaces Definition Let ( X , T ) be a topological space and A X . Then T {O A : O T } is a topology A. It is called the relative or subspace topology. The pair ( A, T ) is a topological space and is called a subspace of ( X , T ). on Theorem 4.16 A subset D A is closed in ( A, T ) D C A for some C X closed in ( X , T ). Proof. If D A is closed in ( A, T ) then A \ D O A for some OT . Then C X \ O iff is closed in ( X , T ) and D A \ ( A \ D) A \ (O A) C A. If D C A for some C X closed in ( X , T ) then A \ D A \ (C A) ( X \ C ) A. Subspaces Definition A property of topological spaces that holds for all subspaces is called hereditary. Example 4.5.1 1st and 2nd countability. Example 4.5.2 Separability is not heriditary. Example 4.5.4 The Zariski Topology For n = 1 it is the finite compliment topology For n > 1 it is not the finite compliment topology It is not Hausdorff http://en.wikipedia.org/wiki/Oscar_Zariski http://en.wikipedia.org/wiki/Zariski_topology Assignment 7 Review Chapters 1- 4. Study all Exercises, be prepared to present solutions during the tutorial Thursday 4 Feb Be prepared for Test 1 on Friday 5 Feb.