Analysis with an introduction to proof 5th by steven lay ch03e

Chapter The Real Numbers Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.4, Slide Section 3.4 Topology of the Real Numbers Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.4, Slide Recall that the distance between two real numbers x and y is given by | x – y | Many of the central ideas in analysis are dependent on the notion of two points being “close” to each other If we are given some positive measure of closeness, say ε, we may be interested in all points y that are less than ε { y : | away from x:  x – y | < ε } We formalize this idea in the following definition Definition 3.4.1 Let x ∈ and let ε > A neighborhood of x is a set of the form   N (x ; ε) = { y ∈   The number ε     : | x – y | < ε }       is referred to as the radius of N (x ; ε) N  ε)  (x ; ( • ) x –ε x x +ε Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.4, Slide Sometimes want to consider points y that are close to x but different from x We can accomplish this by requiring | x – y | > Definition 3.4.2 Let x ∈ and let ε > A deleted neighborhood of x is a set of the form   N* (x ; ε) = { y ∈     : < | x – y | < ε }        ε) = (x – ε, x) ∪ (x, x + ε)   Clearly, N*(x ε) = N ε)\{x}  ;  (x ; N* )•( ( x x –ε If S ⊆ , then a point x in  (x ; ) x +ε can be thought of as being “inside” S, on the “edge” of S, Saying that x is “outside” S is the same as saying that x is “inside” the or “outside” S complement of S, \ S Using neighborhoods, we can make the intuitive ideas of “inside” and “edge” more precise Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.4, Slide Definition 3.4.3 Let S be a subset of neighborhood A point x in N of x such that N ⊆ S N ∩ S ≠ ∅ and N ∩ ( is an interior point of S if there exists a If for every neighborhood N of x, \ S) ≠ ∅, then x is called a boundary point of S The set of all interior points of S is denoted by int S, and the set of all boundary points of S is denoted by bd S Example 3.4.4(c) Let S be the interval [0, 5) ( [| ) ( •) ( | ) ( •)( ) | ) The point is an interior point of S because there exists a neighborhood of that is contained in S The same is true for any point x such that < x < So int S = (0, 5) But the point is different Every neighborhood of (no matter how small) will contain points of S and points that are not in S The same is true for So bd S = {0, 5} Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.4, Slide Open Sets and Closed Sets A set may contain all of its boundary, part of its boundary, or none of its boundary Those sets in either the first or last category are of particular interest Definition 3.4.6 Let S be a subset of If bd S ⊆ S, then S is said to be closed If bd S ⊆ \ S, then S is said to be open Theorem 3.4.7 (a) A set S is open iff S = int S Equivalently, S is open iff every point in S is an interior point of S (b) A set S is closed iff its complement \ S is open Proof: (a) If none of the points in S are boundary points of S, then all the points in S must be interior points and S = int S The converse also applies (b) is in We have bd S = bd ( \ S) So all the boundary is in S iff none of the boundary \ S ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.4, Slide How intersections and unions relate to open sets and closed sets? Theorem 3.4.10 (a) The union of any collection of open sets is an open set (b) The intersection of any finite collection of open sets is an open set Proof: (a) Let A be an arbitrary collection of open sets and let S If x ∈ S, then x ∈ A for some A ∈ A =  {A : A ∈ A } Since A is open, x is an interior point of A That is, there exists a neighborhood N of x such that N ⊆ A But A ⊆ S, so N ⊆ S and x is an interior point of S Hence, S is open (b) Let T = A , …, A be a finite collection of open sets and let n If T = ∅, we are done, since ∅ is open I n A i =1 i If T ≠ ∅, let x ∈ T Then x ∈ A for all i = 1, …, n i Since each set A is open, there exist neighborhoods N  (x ; εi) of x such that Ni (x ; εi) ⊆ Ai i i Let ε  = min {ε1, …, εn} Then N (x ; ε) ⊆ Ai for each i = 1, …, n, so N (x ; ε) ⊆ T Thus x is an interior point of T, and T is open ♦ Note: Part (b) of Theorem 3.4.10 does not necessarily hold for infinitely many open sets Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.4, Slide Example 3.4.12 For each n ∈ , let I But what is in A = (− n  1/n, ∞ Each of these sets is an open set (an open interval)  1/n) A ? n =1 n I ∞ A n =1 n = {0}, which is not open So the open sets have “shrunk down” (by intersecting) to a set that is not open Our next corollary follows easily from Theorem 3.4.10 and the fact that a set is open iff its complement is closed Corollary 3.4.11 (a) The intersection of any collection of closed sets is closed (b) The union of any finite collection of closed sets is closed Practice 3.4.13 To see that the union of an infinite collection of closed sets may not be closed, consider the closed sets An = [1/n, 2] for each n ∈ ∞ What is in Un =1 An ? ∞ Un =1 An = (0, 2], which is not closed So the closed sets have “built up” (by taking unions) to a set that is not closed Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.4, Slide Accumulation Points By using deleted neighborhoods we can define another important property of points and sets Definition 3.4.14 Let S be a subset of A point x in is an accumulation point of S if every deleted neighborhood of x contains a point of S That is, for every ε > 0, N*(x  ε) ∩ S ≠ ∅  ; The set of all accumulation points of S is denoted by S ′ If x ∈ S and x ∉ S ′, then x is called an isolated point of S Example 3.4.15 [0, 1] (a) If S is the interval (0, 1], then S ′ = (b) If S = {1/n : n ∈ }, then S ′ = {0} ∅ So (c) If S = , then S ′ = (d) If S is a finite set, then S ′ = Copyright © 2013, 2005, 2001 Pearson Education, Inc Note that ∉ S consists entirely of isolated points ∅ Section 3.4, Slide Definition 3.4.15 Let S be a subset of Then the closure of S, denoted cl S, is defined by cl S = S ∪ S ′, In terms of neighborhoods, a point x is in cl S where S ′ is the set of accumulation points of S iff every neighborhood of x intersects S To see this, let x ∈ cl S and let N be a neighborhood of x If x ∉ S, then x ∈ S ′ and every deleted neighborhood intersects S If x ∈ S, then N ∩ S contains x Thus in either case the neighborhood N must intersect S Conversely, suppose that every neighborhood of x intersects S If x ∉ S, then every neighborhood of x intersects S in a point other than x Thus x ∈ S ′, and so x ∈ cl S The basic relationships between accumulation points, closure, and closed sets are presented in the following theorem Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.4, Slide 10 Theorem 3.4.17 Let S be a subset of Then (a) S is closed iff S contains all of its accumulation points, (b) cl S is a closed set, (c) S is closed iff S = cl S, (d) cl S = S ∪ bd S Proof: (a) Suppose that S is closed and let x ∈ S is in the open set \ S If x ∉ S, then x  ′ We must show that x ∈ S Thus there exists a neighborhood N of x such that N ⊆ \ S But then N ∩ S = ∅, which contradicts x ∈ S ′ So we must have x ∈ S Conversely, suppose that S ′ ⊆ S let x ∈ \ S We shall show that \ S is open Then x ∉ S ′, so there exists a deleted neighborhood N*(x ; ε) that misses S Since x ∉ S, the whole neighborhood N (x ; ε) misses S; that is, N (x ; ε) ⊆ Thus To this end, \ S \ S is open and S is closed by Theorem 3.4.7(b) Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.4, Slide 11 Theorem 3.4.17 Let S be a subset of Then (a) S is closed iff S contains all of its accumulation points, N (b) cl S is a closed set, (c) S is closed iff S = cl S, ( )•( ( x (d) cl S = S ∪ bd S Proof: N*  ( y ; δ ) | | ) y z )  ε)  (x ; (b) By part (a) it suffices to show that if x ∈ (cl S )′, then x ∈ cl S So suppose that x is an accumulation point of cl S Then every deleted neighborhood N*(x ; ε) To this end, let intersects cl S We must show that N*(x ; ε) intersects S y ∈ N*(x ; ε) ∩ cl S Since N*(x ; ε) is an open set, there exists a neighborhood N ( y ; δ ) contained in N*(x ; ε) But y ∈ cl S, so every neighborhood of y intersects S That is, there exists a point z in N ( y ; δ ) ∩ S But then z ∈ N ( y ; δ ) ⊆ N*(x ; ε), so that x ∈ S ′ and x ∈ cl S The proofs of (c) and (d) are Exercise 18 ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.4, Slide 12 ... intersections and unions relate to open sets and closed sets? Theorem 3.4.10 (a) The union of any collection of open sets is an open set (b) The intersection of any finite collection of open sets is an. ..Section 3.4 Topology of the Real Numbers Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.4, Slide Recall that the distance between two real numbers x and y is given by | x – y | Many of... Practice 3.4.13 To see that the union of an infinite collection of closed sets may not be closed, consider the closed sets An = [1/n, 2] for each n ∈ ∞ What is in Un =1 An ? ∞ Un =1 An = (0, 2],