Analysis with an introduction to proof 5th by steven lay ch03f

Chapter The Real Numbers Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.6, Slide Section 3.6 Metric Spaces Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.6, Slide Definition 3.6.1 Let X be any nonempty set A function d  : X × X →  is called a metric on X if it satisfies the following conditions for all x, y, z ∈ X (1) d(x, y) ≥ (2) d(x, y) = if and only if x = y (3) d(x, y) = d( y, x) (4) d(x, y) ≤ d(x, z) + d(z, y) (triangle inequality) A set X together with a metric d is said to be a metric space Since a set may have more than one metric defined on it, we often identify both and denote the metric space by (X, d ) Example 3.6.2 (a) Let X = and define d :  × → by d (x, y) = | x – y | for all x, y ∈ The fact that d is a metric follows directly from the properties of absolute values In particular, condition (4) follows from the triangle inequality of Theorem 3.2.10(d) When we refer to as a metric space and not specify any particular metric, it is understood that we are using this absolute value metric Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.6, Slide Example 3.6.2 (b) Let X = × 2 = and define d d ( ( x1, y1 ),( x2 , y2 ) ) = for points (x1, y1) and ( x2, y2) in  :  × → by ( x2 − x1 ) + ( y2 − y1 ) 2 This metric is called the Euclidean metric on because it corresponds to our usual measure of distance between two points in the plane We can now see why condition (4) is called the triangle inequality If x, y, and z are the vertices of a triangle, then (4) states that the length of one side of the triangle must be less than or equal to the sum of the lengths of the other two sides z y x d Copyright © 2013, 2005, 2001 Pearson Education, Inc  (x, y) ≤d  (x, z) + d (z, y) Section 3.6, Slide Example 3.6.2 (c) Let X be a nonempty set and define the “discrete” metric d on X by 0, if x = y, d ( x, y ) =  1, if x ≠ y The first three conditions of a metric follow directly from the definition of d The triangle inequality can be established by considering the separate cases when the points x, y, and z are distinct or not If x ≠ z, then d (x, y) ≤ ≤ + d (z, y) = d (x, z) + d (z, y) If x = z = y, then d (x, y) = = + = d (x, z) + d (z, y) If x = z but x ≠ y, then z ≠ y so that d (x, y) = = + = d (x, z) + d (z, y) From this example we see that any nonempty set can be made into a metric space Definition 3.6.3 Let (X,  d ) be a metric space, let x ∈ X, and let ε is given by > The neighborhood of x of radius ε N (x; ε) = { y ∈ X : d  (x, y) < ε } Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.6, Slide Example 3.6.4 Let’s look again at the examples of metric spaces defined above and see what the neighborhoods look like geometrically (a) The metric d defined on The neighborhoods are just open intervals: N (x; ε)  (x,  y) = |  x – y | is the usual measure of distance in – ε, x + ε) 2 (b) The Euclidean metric on that are circular disks = (x by d produces neighborhoods N(0; 1) In particular, the neighborhood of radius centered at the origin = (0, 0) in is given by –2 2 N(0; 1) = {(x, y) : x + y < 1} –2 (c) The neighborhoods for the discrete metric defined in Example 3.6.2(c) depend on the size of the radius If ε ≤ 1, then the neighborhood contains only the center point itself If ε  > 1, then the neighborhood contains all of X In particular, if X = Copyright © 2013, 2005, 2001 Pearson Education, Inc , then N (0; 1) = {0} and N =  (0; 2) Section 3.6, Slide Example 3.6.5 For another example, let X= and define d1 by d1 ( ( x1, y1 ),( x2 , y2 ) ) = x1 − x2 + y1 − y2 It is clear that the first three conditions of a metric are satisfied by d1 To see that the triangle inequality also holds, let p1 = (x1, y1), p2 = (x2, y2), and p3 = (x3, y3) be arbitrary points in Then d1(p1, p2) = | x1 – x2 | + | y1 – y2 | = | x1 – x3 + x3 – x2 | + | y1 – y3 + y3 – y2 | N(0; 1) ≤ | x1 – x3 | + | x3 – x2 | + | y1 – y3 | + | y3 – y2 | = | x1 – x3 | + | y1 – y3 | + | x3 – x2 | + | y3 – y2 | –2 = d1(p1, p3) + d1(p3, p2) –2 Geometrically, the neighborhoods in this metric are diamond shaped If (X,  d ) is a metric space, then we can use Definition 3.6.3 for neighborhoods to characterize interior points, boundary points, open sets, and closed sets, just as we did Most of the theorems from Section 3.4 that relate to these concepts in Section 3.4 for also carry over to this more general setting with little or no change in their proofs Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.6, Slide One result that does require a new proof is the following theorem Theorem 3.6.6 Let (X,  d ) be a metric space Any neighborhood of a point in X is an open set Proof: Let x ∈ X and let ε > To see that N  ε) is an open set, we shall show that  (x; If y ∈ N(x any point y in N (x; ε) is an interior point of N (x; ε) δ We claim that N  δ ) ⊆ N (x; ε) If z ∈ N then d (z,  y)  ( y;  ( y; δ ), N  ε)  (x; It follows that d (z, x) ≤ d (z, y) + d ( y, x) x < δ + d ( y, x) = [ε – d (x, y)] + d ( y, x) = ε N  ( y; Thus z ∈ N  (x;  ε) It follows that N  ( y; δ ) ⊆ N (x; ε),  δ ) • y • z • and so y is an interior point of N (x; ε), and N (x; ε) is open ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.6, Slide The definition of a deleted neighborhood in a metric space (X, d N*(x; ε) = {  y ∈ X  ) is similar to that in : : < d (x, y) < ε } The definition of an accumulation point is also analogous: x is an accumulation point of a set S if for every ε > 0, N*(x; Once again, the closure of a set S,  ε) ∩ S ≠ ∅ denoted by cl S, is given by cl S ∪ S ′, where S ′ is the set of all accumulation points of S The properties of closure and closed sets given in Theorem 3.4.17 continue to apply in a general metric space Indeed, the proofs given there were all stated in terms of neighborhoods, so they carry over with no change at all But compact sets are different The definition is the same: a set S is compact iff every open cover of S contains a finite subcover But the Heine-Borel theorem no longer holds.To see this, we first need to define what it means for a set to be bounded in (X, d  ) Definition 3.6.10 A set S in a metric space (X, d Copyright © 2013, 2005, 2001 Pearson Education, Inc  ) is bounded if S ⊆ N (x; ε ) for some x ∈ X and some ε > Section 3.6, Slide Theorem 3.6.11 Let S be a compact subset of a metric space (X, d  ) Then (a) S is closed and bounded (b) Every infinite subset of S has an accumulation point in S Proof: (a) The first half of the proof of the Heine-Borel Theorem 3.5.5 was given entirely in terms of neighborhoods and open sets It applies without any changes in this more general setting (b) This proof is the same as the proof of the Bolzano-Weierstrass Theorem 3.5.6, except that the compactness of S is assumed (instead of using boundedness and the Heine-Borel theorem) We know that the accumulation point of S must be in S, since S is closed ♦ While a compact subset of (X, d not hold in general The proof of the converse in very dependent on the completeness of Copyright © 2013, 2005, 2001 Pearson Education, Inc  ) must be closed and bounded, the converse does (given in Theorem 3.5.5) was , a property not shared by all metric spaces Section 3.6, Slide 10 Example 3.6.12 Consider the set with the discrete metric of Example 3.6.2(c) N(p; 1) = {p} for every p ∈ We have seen that But neighborhoods are open sets by Theorem 3.6.6, so each singleton set consisting of one point is an open set Since every set is the union of the points in the set, this means that every set is open! Since the complement of an open set is closed, this means that every subset is also closed! = {(x, y): ≤ x ≤2 and ≤ y ≤ 1} Let S be the unit square: S and it is also bounded since S ⊆ N  (0;  2) = We claim that S is notTocompact see this, for each point p ∈ S, let Ap = {p} S⊆ Then S is a closed set, Then each Ap is an open set and U Ap p∈ S Thus F = {Ap  : p ∈ S} is an open cover for S But F contains no finite subcover of S, since each set in F covers only one point in S and there are infinitely many points in S We conclude that in this metric space the unit square S is closed and bounded, but not compact Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.6, Slide 11 ... metric space Any neighborhood of a point in X is an open set Proof: Let x ∈ X and let ε > To see that N  ε) is an open set, we shall show that  (x; If y ∈ N(x any point y in N (x; ε) is an interior... (triangle inequality) A set X together with a metric d is said to be a metric space Since a set may have more than one metric defined on it, we often identify both and denote the metric space by. .. our usual measure of distance between two points in the plane We can now see why condition (4) is called the triangle inequality If x, y, and z are the vertices of a triangle, then (4) states that