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URYSOHN's THEOREM and TIETZE EXTENSION THEOREM Definition URYSOHN'S THEOREM AND TIETZE EXTENSION THEOREM Tianlin Liu [email protected] Mathematics Department Jacobs University Bremen Campus Ring 6, 28759, Bremen, Germany Definition 0.1. Let x; y topological space X. We define the following properties of topological space X: ∈ T0: If x y, there is an open set containing x but not y or an open set containing y but not x. ≠ T1: If x y, there is an open set containing y but not x. T2: If x≠ y, there are disjoint open sets U; V with x U and y V . ≠ ∈ T3∈: X is a T1 space, and for any closed set A X and any x AC there are disjoint open sets U; V with x U and A V . ⊂ T4∈: X is a T1 space, and for any disjoint closed∈ sets A, ⊂B in X there are disjoint open sets U, V with A Uand B V . We say a T2 space is a Hausdorff space, a T⊂3 space is⊂ a regular space, a T4 space is a normal space. Definition 0.2. C X; a; b Space of all continuous a; b valued functions on X. ( [ ]) ∶= [ ] Date: February 11, 2016. 1 URYSOHN'S THEOREM AND TIETZE EXTENSION THEOREM 2 Theorem 0.3. (Urysohn's Lemma) Let X be a normal space. If A and B are disjoint closed sets in X, there exists f C X; 0; 1 such that f 0 on A and f 1 on B. Proof. ∈ ( [ ]) = = Step 1: Define a large collection of open sets in X (Lemma 4.14 in [1]) Let D be the set of dyadic rationals in 0; 1 , that is, D 1 1 3 1 3 7 1; 0; 2 ; 4 ; 4 ; 8 ; 8 ; 8 ::: . [ ] = { We want to construct} a sequence of open sets Uq in X, in- dexed by q D, such that: (1) For each q D, A Uq (2) B U1 and∈ for each q 1, B Uq (3) p; q D with∈ p ⫅q, we have Up Uq ⊂ < ∩ = ∅ To∀ this∈ end, we define< ⫅ U1 X B U0 U1 is closed open U0 A such that U0 U1 U0 = U 1 is closed open U 1 U0 such that U 1 U 1 ⫅ 2 ⇒ ∃ 4 ⫆ ⫅4 2 U 1 U1 is closed open U 3 U 1 such that U 3 U1 :::2 ⫅ ⇒ ∃ 4 ⫆ 2 4 ⫅ This⫅ construction⇒ ∃ satisfies (1),⫆ (2), and (3). ⫅ Step 2: Define the map f Define: f X 0; 1 via ∶ → [ ] x inf q x Uq By (1), f = 0 in A. By (2), f = 1 in B. ↦ { S ∈ } Step 3: Prove the map f is continuous It is sufficient to show that f −1 ; α and f −1 α; are open because we know half lines generate the topology on R. (−∞ ) ( ∞) f x α x Ur for some r α x r<α Ur −1 ( f) < ⇐⇒; α ∈ r<α Ur is open.< ⇐⇒ ∈ ⋃ f⇒x (−∞α ) =x⋃ Ur for some r α x Us for some C s α x s>α Us ( ) > ⇐⇒ ∉ > ⇐⇒ ∉ −1 C > f⇐⇒α; ∈ ⋃ s(>α )Us is open. ⇒ ( ∞) = ⋃ ( ) URYSOHN'S THEOREM AND TIETZE EXTENSION THEOREM 3 Theorem 0.4. (Tietze Extension Theorem) Let X be a normal space and A be a closed subset in X. If f A a; b is a continuous function, then f has a continuous extension F X a; b , i.e., F is continuous and F A f. ∶ → [ ] ∶ Proof. WLOG, assume that a; b 0; 1 . (just replace f by f−a . → [ ] S = b−a Step 1. [ ] = [ ] Let h A 0; k be a continuous function for some constant k. −1 k −1 2k Then B h 0; 3 and C h 3 ; k are closed in A, and ∶ → [ ] k therefore closed in X. By Urysohn Lemma, g X 0; 3 such = ([ ]) k = ([ ]) that g B 0 and g C 3 . k 2k ∃ ∶ → [ ] Note that g 3 , and h g 3 on A. ( ) = ( ) = Step 2. ≤ − ≤ Start with f A 0; 1 .(k=1 in Step 1.) Then g1 1 3. f g1 A 0∶ ; 2→3 [. ] Now apply≤ Step~ 1 to this function f g1 with k 2 3: 2 Then− f∶ →g1 [ g2 ~ ]2 3 and g2 1 3 2 3 2 9. In this way, we can obtain a sequence− gn with= the~ property that − − ≤ ( ~ ) ≤ ~ ⋅ ~ = ~ n (1)f g1 g2 gn 2 3 { } n−1 (2)gn 1 3 2 3 . − − ⋯ − ≤ ( ~ ) Step≤ 3. ~ ⋅ ( ~ ) Let sn g1 g2 gn. Then sn is a Cauchy sequence in X; R since sn sm= +gn+⋯1 + gm gn+1 gm 1 3( 2) 3 n 2 3 m−1 1 3C(2 3 n) 1 2 3 2 3 2 Y2 3− n Y = Y + ⋯ + Y ≤ Y Y + ⋯ + Y Y By≤ ( the~ )(( completeness~ ) +⋯+( ~ of) X;) <R( ,~ s)(n ~ )F(uniformly+ ~ +( ~ and) +⋯F) = ( ~X;)R . Now we claim that F is aC desired( ) extension→ of f: ∈ n StepC( 2(1)) f sn A 2 3 sn f uniformly on A F f on A. ⇒ Y − Y ≤ ( ~ ) ⇒ → ⇒ = References [1]Folland Real Analysis.
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