1 the Topology of Metric Spaces

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1 the Topology of Metric Spaces Introductory Analysis I Fall 2014 Notes on Metric Spaces These notes are an alternative to the textbook, from and including "Closed Sets and Open Sets" (page 58) to and excluding Cantor Sets (page 95) 1 The topology of metric spaces Assume M is a metric space with distance function d. I will depart a bit from the book's notation; instead of using Mr(p), I will denote by B(p; r) the open ball of center p 2 M and radius r > 0; i.e, B(p; r) = fq 2 M : d(q; p) < rg: The reason is that once we get to concrete metric spaces like Rn, it may be silly n to write (R )p(r). More importantly, I like my notation better. Incidentally a notation such as "r > 0" means: "r is a positive real number." The basic definition is Definition 1 A subset U of M is said to be open iff for every p 2 U there is r > 0 such that B(p; r) ⊂ U. In other words, the set U is open if and only if for every p 2 U there is r > 0 such that if q 2 U and d(q; p) < r, then q 2 U. Examples: 1. Let M be an arbitrary metric space. Then the empty set ; and M are open. 2. Let M be an arbitrary metric space. Let p 2 M and r > 0. Then B(p; r) is open. 3. Let M be a discrete metric space; i.e., d(p; q) = 1 whenever p =6 q. Then every subset of M is open. 4. Let M = Rm. The following sets are open: m U1 = fx = (x1; : : : ; xm) 2 R : x1 > 0g; m U2 = fx = (x1; : : : ; xm) 2 R : x1 > x2g; m U3 = fx = (x1; : : : ; xm) 2 R : jx1j + ··· + jxmj < 1g; 5. Let M = R. A subset U of R is open if and only if for each x 2 R there exist a; b 2 R such that x 2 (a; b) ⊂ U. In particular, all open intervals are open. You should be able to prove that the sets declared open in these examples are indeed open. 1 THE TOPOLOGY OF METRIC SPACES 2 Theorem 1 Let M be a metric space and let T be the family of all open subsets of M. This family has the following properties: 1. ;;M 2 T . S 2 T 2 2 T 2. If Uα for all α A; A some index set, then α2A Uα . In words, T is closed under arbitrary unions. An equivalent way this property is sometimes formulated is: Assume U is a collection ofS open subsets of M (finite, infinite, countable or non-countable). Then fU : U 2 Ug is open. 3. If U; V 2 T , then U \ V 2 T . The intersection of two open sets is open. By induction, it follows that the intersection of any finite number of open sets is open. Proof. 1. That ;;M are open was already mentioned. S 2 T 2 2 T 2. Let Uα for all α A; A some index set, let U = α2A Uα . Let p 2 U. Then there exists α 2 A such that p 2 Uα. Since Uα is open, there is r > 0 such that B(p; r) ⊂ Uα. Since Uα ⊂ U, B(p; r) ⊂ U. 3. Let U; V be open. If p 2 U \ V , then p 2 U and p 2 V . Since U is open, p 2 U implies that there exists r1 > 0 such that B(p; r1) ⊂ U. Similarly, because V is open, there exists r2 > 0 such that B(p; r2) ⊂ V . Let r = min(r1; r2). Then r > 0 and clearly B(p; r) ⊂ B(p; r1) \ B(p; r2) ⊂ U \ V . Exercise 1 Let M = R. 1. Prove that in R singleton sets are NOT open. 2 N f 2 R − g 2. For n let Un =T x : 1=n < x < 1=n . Prove that Un is open 2 N 1 for all n , but n=1 Un is not open. This shows that, in general, the intersection of an infinite family of open sets may not be open. Note: If M is any set, a family of subsets of M satisfying properties 1, 2, and 3 of Theorem 1 is called a topology for M. Definition 2 Let M be a metric space. A subset F of M is closed iff its complement MnF is open. Examples: 1 THE TOPOLOGY OF METRIC SPACES 3 1. Let M be an arbitrary metric space. Then the empty set ; and M are closed. 2. Let M be an arbitrary metric space. Let p 2 M and r ≥ 0. Then the closed ball of center p, radius r; that is, the set fq 2 M : d(q; p) ≤ rg is closed. 3. A particular case of the previous result, the case r = 0, is that in every metric space singleton sets are closed. 4. Let M be a discrete metric space; i.e., d(p; q) = 1 whenever p =6 q. Then every subset of M is closed. 5. Let M = Rm. The following sets are closed: m U1 = fx = (x1; : : : ; xm) 2 R : x1 ≥ 0g; m U2 = fx = (x1; : : : ; xm) 2 R : x1 ≥ x2g; m U3 = fx = (x1; : : : ; xm) 2 R : jx1j + ··· + jxmj ≤ 11g; 6. Let M = R. All closed intervals are closed. As mentioned in the examples, in R open intervals are open and closed intervals are closed. Intervals of the form (a; b]; [a; b) with a; b 2 R, a < b are neither open nor closed. By taking complements, the following theorem is an immediate consequence of Theorem 1 Theorem 2 Let M be a metric space. 1. ;;M 2 T are closed. 2 2. IfT Fα is a closed subset of M for all α A; A some index set, then α2A Uα is closed. In words, arbitrary intersections of closed subsets are closed. Equivalently: Assume F is a collection ofT closed subsets of M (finite, infinite, countable or non-countable). Then fF : F 2 Fg is closed. 3. If F; G, are closed then F [ G is closed. By induction, it follows that the Union of any finite number of closed sets is closed. Our next task will be to show that the basic metric properties can be de- scribed exclusively in terms of open sets. Another way of stating this is to say that one never has to mention the distance function again, except if one wants to. It will be convenient to first have another topological concept, that of neighborhood. Definition 3 Let M be a metric space. Let p 2 M. A subset V of M is said to be a neighborhood of p in M (or just a neighborhood of p) iff there exists an open set U such that p 2 U ⊂ V . 1 THE TOPOLOGY OF METRIC SPACES 4 Exercise 2 V is a neighborhood of p if and only if there exists r > 0 such that B(p; r) ⊂ V . We also have the following simple lemma Lemma 3 A subset U of a metric space is open if and only if it is a neighbor- hood of each of its points. Proof. Assume first U is open and let p 2 U. Then p 2 U ⊂ U, showing U is a neighborhood of p. Conversely, assume U is a neighborhood of eachS of its points. 2 2 ⊂ Then for each p U, there is Wp open, p Wp U. Let W = p2U Wp. By property 2 of Theorem 1, W is open. Since Wp ⊂ U for all p 2 U, we have W ⊂ U. On the other hand, if p 2 U, then p 2 Wp ⊂ W , thus U ⊂ W . It follows that U = W is open. Let us begin with sequences. We'll state it in the form of a theorem, but it is fairly immediate. Theorem 4 Let (pn) be a sequence in the metric space M; let p 2 M. The following statements are equivalent: 1. For each ϵ > 0 there exists N 2 N such that n ≥ N implies d(pn; p) < ϵ. 2. For each ϵ > 0 there exists N 2 R such that n > N implies d(pn; p) < ϵ. 3. For each open set U in M such that p 2 U, there exists N 2 R such that n > N implies pn 2 U. 4. For each neighborhood V of p, there exists N 2 R such that n > N implies pn 2 V . 5. If V is a neighborhood of p, then fn 2 N : pn 2= V g is finite. As mentioned before, most of the equivalences are obvious. In lieu of a proof I'll just make some comments about what could be the less obvious parts. Concerning the equivalence of 1 and 2, notice first that if the following statement "For each there exists N 2 N such that n ≥ N implies " (fill the blanks any way you wish), is equivalent to "For each there exists N 2 N such that n > N implies " In fact, if something holds for n ≥ N, it also holds for n > N. But the converse is also true; if we can find N 2 N so that a property holds for n > N, we can also find N 0 so it holds for n ≥ N 0; namely N 0 = N + 1. Maybe in words: if there is an element of N so the property holds when n is strictly larger than that element of N, then there is an element of N (namely the previous element plus 1) such that the property holds for n greater than or equal this element.
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