Topology: Urysohn Lemma Made Simple.

Urysohn Lemma Class 15: Urysohn Lemma Sebastian Vattamattam 23. November 2010 Sebastian Vattamattam TopologyUrysohn Lemma Denition X is normal if for disjoint closed sets A; B; subsets of X, there exist disjoint open sets U; V such thatA U; B V Lemma If X is normal, A is closed in X and open set U contains A, then there exists open set V such that A V V U Sebastian Vattamattam TopologyUrysohn Lemma Theorem Thm33:1 in [1] Urysohn Lemma X is normal, A; B disjoint, closed in X. [a; b] a closed interval in R Then there exists f : X ! [a; b] such that 1 f is continuous. 2 f 1(a) = A; f 1(b) = B Proof We take a = 0; b = 1 Sebastian Vattamattam TopologyUrysohn Lemma 1 Step 1: Let P := QT[0; 1] To show that, for p; q 2 P; p < q there exist open sets Up; Uq such that Up Uq Proof by induction Since P is countable, let P = (xn); an innite sequence, where x1 = 1; x2 = 0. For n 2 N; let Pn := fxk jk ng Sebastian Vattamattam TopologyUrysohn Lemma 1 P1 = f1g and the result is vacuously true. For P2 = f1; 0g ; 0 < 1, let U1 = X B U1 is an open neighborhood of A By lemma 1.2, there is an open set U0 such that A U0 U0 U1 Thus the result is true on P2 2 Suppose the result is true for p; q 2 Pn; p < q That is, there exist open sets Up;Uq such that Up Uq Let xn+1 = r and Pn+1 := Pn Sfrg Pn+1 is a nite subset of [0; 1] With the usual order relation on [0; 1]; Pn+1 is simply ordered. Since x1 = 1; x2 = 0 2 Pn+1; 0 = min Pn+1; 1 = max Pn+1 In Pn+1 every number, other than 0 and 1, has an immediate predecessor and an immediate successor. Sebastian Vattamattam TopologyUrysohn Lemma Since r =2 f0; 1g, let p be the immediate predecessor and q be the immediate successor of r in Pn+1 Since p < q in Pn, by hypothesis, there exist open sets Up;Uq such that Up Uq (1.1) Again by lemma 1.2, there exist open set Ur in X such that Up Ur Ur Uq (1.2) To show that the result holds for each pair of elements in Pn+1 By the hypothesis, it holds for every pair of elements in Pn Consider the pair (r ; s), where s 2 Pn If s p, then by 1.1 and 1.2, Us Up Up Ur If s q then r < q s and hence by 1.1 and 1.2, Ur Uq Us Thus for every pair of elements of Pn+1, the result holds. The conclusion follows by induction. 2 Step 2 Sebastian Vattamattam TopologyUrysohn Lemma To extend the result in Step 1 to all p 2 Q Let Up = if p < 0; X if p > 1 Now again, for p; q 2 Q; p < q Up Uq (1.3) 3 Step 3 For x 2 X; let Q(x) := fp 2 Qjx 2 Upg Sebastian Vattamattam TopologyUrysohn Lemma If p < 0; Up = and hence 0 p 8p 2 Q(x) Denef : X ! [0; 1]; f (x) = inf Q(x) = inf fpjx 2 Upg 4 Step 4 To show that 1 f 1(a) = A; f 1(b) = B 2 f is continuous. 1 Let a 2 A Since A U0 U0 Up; 8p 0 a 2 Up; 8p 0 Sebastian Vattamattam TopologyUrysohn Lemma ) f (a) = inf Q(a) = 0; 8a 2 A Let b 2 B ) b =2 X B = U1 Up; 8p 1 ) b =2 Up; 8p 1 f (b) = inf Q(b) = 1 2 To prove x 2 Ur ) f (x) r (1.4) x 2 Ur ) f (x) r (1.5) 1 Let x 2 Ur For r < s; x 2 Ur Us) x 2 Us ; 8s > r Therefore Q(x) fp 2 Qjp > rg ) f (x) = inf Q(x) r Sebastian Vattamattam TopologyUrysohn Lemma 2 Let x =2 Ur For s < r ;Us Us Ur So, x =2 Us ; 8s < r f (x) = inf Q(x) r 3 To prove f is continuous. Let x0 2 X; and c < f (x0) < d Let p; q 2 Q such that c < p < f (x0) < q < q Then Up Uq Let U := Uq Up; open f (x0) < q ) inf Q(x0) < q ) x0 2 Uq f (x0) > p ) inf Q(x0) > p ) x0 =2 Up Up Thus x0 2 U Sebastian Vattamattam TopologyUrysohn Lemma To show that f (U) (c; d) x 2 U ) x 2 Uq Uq By (1:5) f (x) q x =2 Up ) x =2 Up By (1:6) f (x) p Thus f (x) 2 [p; q] (c; d) f (U) (c; d) Therefore f is continuous. James R. Munkres,Topology, Second Edition, Prentice-Hall of India, New Delhi, 2002. Sebastian Vattamattam Topology