Lecture : Tietze Extension Theorem

Transcription

1 Proof of Lecture : Dr. Department of Mathematics Lovely Professional University Punjab, India November 26, 2014

2 Outline Introduction Proof of 1 Introduction 2 3 Proof of 4

3 Proof of The theorem is due to Heinrich Franz Friedrich Tietze. And it is know as the TietzeUrysohnBrouwer extension theorem also. The theorem generalizes Urysohn s lemma and is widely applicable, since all metric spaces and all compact Hausdorff spaces are normal. It can be generalized by replacing R with R ω for some indexing set ω, any retract of R ω, or any normal absolute retract whatsoever.

4 Proof of Mathematics could be described as an ultimate quest for generalization. Everyone knows that in a (3 4 5) right triangle, = 5 2. As Pythagoras knew, however, this is the case for any right triangle. That is, the sum of the squares of the legs equals the square on the hypotenuse. Or as an equation,a 2 + b 2 = c 2. This of course can handle a much broader range of triangles, but it is still not satisfactory enough. This formula gives no information about triangles that are not right, so this formula was later generalized to the so-called Law of Cosines a 2 = b 2 + c 2 2bc cos A. This formula can be applied to any triangle that you can draw. This quest towards generalization is a good way to think of mathematics, and a good way to lead into this short topological paper.

5 Proof of Although the last example is quite elementary, in a way, so is the question behind the. Take for instance the curve defined by f(x) = x on the closed interval [0, 1]. The graph of this curve is a line segment. Indeed, what we really have is a continuous function (hereafter called a map): f : [0, 1] R

6 Proof of It is now natural to ask if there exists a continuous function g defined over the whole space R such that f = g on [0, 1]. For this particular example, it is easy, just take g(x) = x. In general, however, it might not be possible to find such a function. For example, consider f(x) = 1/x on the open interval (0, ). This is certainly continuous, but it is not possible to extend this to a continuous function g over the whole real line. Thus, as we sought above, we wish to find the general case when we can extend a function from a subset to one on the whole space.

7 Proof of I now state the question that we seek to answer in this lecture. The answer will be the presented at here. Question Let S be a subset of a topological space X and let f be a map f : S R What conditions must be placed on S and X such that there exists a map g : X R and f = g on S?

8 Proof of Theorem Suppose (X, τ) is topological space. The space X is normal if and only if every continuous real function f defined on a closed subspace F of X into a closed interval [a, b] has a continuous extension f : X [a, b]

9 Proof of Proof: Part A As given 1 (X, τ) is space such that every continuous real valued function has a continuous extension 2 F is closed subspace of X. f : F [a, b] g : X [a, b] Our aim to show that (X, τ) is normal space.

10 Proof of Let F 1 and F 2 be two disjoint closed subset of X. Define a map f : F 1 F 2 [a, b] such that { a, x F 1 ; f(x) = b, x F 2. Here this continuous over the subspace F 1 F 2. By assumption f can be extended to a continuous map f : X [a, b] such that { a, x F 1 ; g(x) = b, x F 2. This map g satisfies Urysohn s lemma and hence (X, τ) is normal space.

11 Proof of 2 A map f : F [a, b] Proof: Part B As given a topological space (X, τ) 1 (X, τ) is normal space. be a continuous map, where F is closed subspace of X. Our aim is to show that that there exists a continuous extension of f over X.

12 Proof of Let a = 1 and b = 1. Now define a map f 0 : F [ 1, 1] such that f 0 (x) = f(x) x F Suppose A 0 and B 0 be two subset of F such that A 0 = {x: f 0 (x) 1 3 } (1) B 0 = {x: f 0 (x) 1 3 } (2) Here we can see that A 0 and B 0 are non-empty disjoint closed sets in F. This show that both are closed in X because F X. Now apply Urysohn lemma on A 0 and B 0, then there exists a continuous map g 0 : X [ 1 3, 1 3 ] such that g 0 (A 0 ) = 1 3 and g 0(B 0 ) = 1 3 (3)

13 Proof of Define a map f 1 on F such that f 1 = f 0 = g 0, then f 1 (x) = (f 0 g 0 )(x) = f 0 (x) g 0 (x) Finally, f 1 (x) 2 3 (4) Let A 1 = {x: f 1 (x) ( 1 3 )( 2 3 )} (5) B 1 = {x: f 1 (x) ( 1 3 )( 2 3 )} (6)

14 Proof of Here we can see that A 1 and B 1 are non-empty disjoint closed sets in X and hence by Urysohn lemma again there exists a continuous map g 1 : X [ ( 1 3 )( 2 3 ), ( 1 3 )( 2 3 )] such that g 1 (A 1 ) = ( 1 3 )( 2 3 ) and g 1(B 1 ) = ( 1 3 )( 2 3 ) (7) Again we define a map f 2 on F such that Then, f 2 = f 1 g 1 = f 0 g 0 g 1 = f 0 (g 0 + g 1 ) f 2 (x) = (f 0 (g 0 + g 1 ))(x) = f 0 (x) (g 0 + g 1 )(x)

15 Finally, Introduction Proof of f 2 (x) Continuing this process, we get a sequence of function defined on F such that And a sequence of function defined on X such that < f 0, f 1,..., f n,... > f n (x) < g 0, g 1,..., g n 1,... > g n (x) 1 3 ( ) 2 2 (8) 3 ( ) 2 n (9) 3 ( ) 2 n (10) 3

16 Proof of And Now write f n = f 0 (g 0 + g g n 1 ) n 1 s n = r=0 Now s n can be regarded as partial sums bounded by continuous map defined on X. Since the space of bounded continuous real valued function is complete and g n (x) 1 3 ( ) 2 n and 3 g r n=0 1 3 ( ) 2 n = 1 3

17 Proof of The sequence s n converge uniformly on X to g (say). Where g(x) 1. Now ( ) 2 n f n (x) < s n > converge uniformly on F to f 0 = f say 3 Hence g = f on F. Thus g is a continuous extension of F to X, which satisfies the given conditions.

18 Proof of Theorem (Urysohn Lemma) A topological space (X, τ) is normal if and only if given a pair of disjoint closed sets A, B X, there is a continuous function f : X R such that f(a) = {0} and f(b) = {1} Where set R is with usual topology, and I = [0, 1] = {x R : 0 x 1}.