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MA4266 Topology Lecture 17 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 Metrization Definition A topological space ( X , T ) s metrizable if there exists a metric d on X that gives the topology Theorem 8.15 The product of a countable number of metric spaces is metrizable. Proof n n 1 (Xn,d ) ~ dn , d n 1 d n : X n X n [0, ) are bounded metrics and ~ 2 d ( x, y) n1 (d n ( pn ( x), pn ( y)) / n) is a metric on 1/ 2 X n 1 X n that gives the product topology, see pages 254-255. T. Metrization Thm8.16 Every 2ndcountable regular space is metrizable Proof Let X be a 2nd countable regular space with basis n n 1 {B } . Lindelöff’s theorem 6.13 implies X is Lindelöf hence Theorem 8.9 implies X is normal. Therefore for each pair (i, j ) N N with Bi B j there exists f : X [0,1] such that f ( Bi ) 0 and f ( X \ B j ) 1. - why? These functions can be a function n n 1 F : X H 2 ( N ) 1 1 F ( x ) ( f ( x ), f ( x ),..., by 1 2 2 n f n ( x),...). Then F indexed { f } - why? Define is an embedding – what exactly does this mean ? Metrization Claims F is one-to-one – why ? F is continuous – why ? F : X F ( X ) H is an open mapping. Proof pages 257-258 in textbook. Since X is homeomorphic to a subspace of H it is metrizable – why ? Supplement: The Zarisky Topology X F where F is field, for example F R, Q, C , Z / pZ where p N is prime. Let P( X ) denote the ring of polynomials in n variables with coefficients in F . For A P ( X ) Define the variety of n A by V ( A) {v F : A(v) 0}. The Zarisky topology TZar on X is the topology generated by the basis BZar { X \ V ( A) : A P( F )}. See Ex. 4.5.4 125. n Definition Let Examples X R , X \ V (0) X \ X , X \ V (1) X \ X , 2 X \ V ( x1 x2 ) 4 open quadrants and X \ V ( x12 x22 1) X \ {circle} are open in TZar . Supplement: The Zarisky Topology ( X , TZar ) is a compact space. Proof Assume to the contrary that ( X , TZar ) is not Theorem compact (we will derive a contradiction) so that there O with no finite subcover. Then C { X \ O : O O } is a collection of closed sets exists an open cover having the finite intersection property (the intersection of every finite collection of sets in C is nonempty) and the intersection of all the sets in C is empty. From these two properties we can find a sequence C j in with C C j C j 1 (Prove it ! ) so K j i 1 Ci is a strictly decreasing sequence of closed sets. j Supplement: The Zarisky Topology j N define I j { A P( X ) : K j V ( A)} so I j is the set of all polynomials that vanish on K j . For each I j is an ideal in the ring P ( X ). This means A, B I j , Q P( X ) A B I j , QA I j . Claim 1. Each Claim 2. I1 I 2 I 3 strictly increases. Claim 3. I j 1 I j is an ideal. ( Prove these claims ! ) Hilbert's basis theorem every ideal in the ring of multivariate polynomials over a field is finitely generated. Proof http://en.wikipedia.org/wiki/Hilbert's_basis_theorem HBT there exists a basis B {b1 ,..., bL } for I meaning I b1P( X ) bL P( X ). Each bi I ki so B I m with m max {k1 ,..., kL }.So I m I a contradiction, why ?. Assignment 16 Read pages 253-260, 261-264 Prepare to solve during Tutorial Thursday 15 April Exercise 8.5 problems 3, 5, 6, 7 Exercise 8.6 problems 2, 7, 11 a, b, c, d