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Urysohn Lemma Class 15: Urysohn Lemma Sebastian Vattamattam 23. November 2010 Sebastian Vattamattam Topology

Urysohn Lemma Deﬁnition X is normal if for disjoint closed sets A, B, subsets of X , there exist disjoint open sets U, V such that A ⊂ 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 Topology

Urysohn 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 Topology

Urysohn Lemma 1 Step 1 : Let P := Q [0, 1] To show that, for p, q ∈ P, p < q there exist open sets Up , Uq such that ¯ Up ⊂ Uq Proof by induction Since P is countable, let P = (xn ), an inﬁnite sequence, where x1 = 1, x2 = 0. For n ∈ N, let Pn := {xk |k ≤ n} Sebastian Vattamattam Topology

Urysohn Lemma 1 P1 = {1} and the result is vacuously true. For P2 = {1, 0} , 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 ∈ Pn , p < q That is, there exist open sets Up , Uq such that ¯ Up ⊂ Uq Let xn+1 = r and Pn+1 := Pn {r } Pn+1 is a ﬁnite subset of [0, 1] With the usual order relation on [0, 1], Pn+1 is simply ordered. Since x1 = 1, x2 = 0 ∈ 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 Topology

Urysohn Lemma Since r ∈ {0, 1}, 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 ∈ 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 Topology

Urysohn Lemma To extend the result in Step 1 to all p∈Q Let φ if p < 0; Up = X if p > 1 Now again, for p, q ∈ Q, p < q ¯ Up ⊂ Uq (1.3) 3 Step 3 For x ∈ X , let Q(x) := {p ∈ Q|x ∈ Up } Sebastian Vattamattam Topology

Urysohn Lemma If p < 0, Up = φ and hence 0 ≤ p ∀p ∈ Q(x) Deﬁne f : X → [0, 1], f (x) = inf Q(x) = inf {p|x ∈ Up } 4 Step 4 To show that 1 f −1 (a) = A, f −1 (b) = B 2 f is continuous. 1 Let a ∈ A ¯ Since A ⊂ U0 ⊂ U0 ⊂ Up , ∀p ≥ 0 a ∈ Up , ∀p ≥ 0 Sebastian Vattamattam Topology

Urysohn Lemma ⇒ f (a) = inf Q(a) = 0, ∀a ∈ A Let b ∈ B ⇒ b ∈ X − B = U1 ⊃ Up , ∀p ≤ 1 / ⇒ b ∈ Up , ∀p ≤ 1 / f (b) = inf Q(b) = 1 2 To prove ¯ x ∈ Ur ⇒ f (x) ≤ r (1.4) x ∈ Ur ⇒ f (x) ≥ r (1.5) 1 ¯ Let x ∈ Ur ¯ For r < s, x ∈ Ur ⊂ Us ⇒ x ∈ Us , ∀s > r Therefore Q(x) ⊃ {p ∈ Q|p > r } ⇒ f (x) = inf Q(x) ≤ r Sebastian Vattamattam Topology

Urysohn Lemma 2 Let x ∈ Ur / ¯ For s < r , Us ⊂ Us ⊂ Ur So, x ∈ Us , ∀s < r / f (x) = inf Q(x) ≥ r 3 To prove f is continuous. Let x0 ∈ X , and c < f (x0 ) < d Let p, q ∈ 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 ∈ Uq f (x0 ) > p ⇒ inf Q(x0 ) > p ¯ ⇒ x0 ∈ Up ⊂ Up / Thus x0 ∈ U Sebastian Vattamattam Topology

Urysohn Lemma To show that f (U) ⊂ (c, d) ¯ x ∈ U ⇒ x ∈ Uq ⊂ Uq By (1.5) f (x) ≤ q / ¯ x ∈ Up ⇒ x ∈ Up / By (1.6) f (x) ≥ p Thus f (x) ∈ [p, q] ⊂ (c, d) f (U) ⊂ (c, d) Therefore f is continuous. vattamattam@gmail.com Sebastian Vattamattam Topology

Urysohn Lemma James R. Munkres,Topology, Second Edition, Prentice-Hall of India, New Delhi, 2002. Sebastian Vattamattam Topology

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