Introduction to Topology

Transcription

1 Chapter 2 Introduction to Topology In this chapter, we will use the tools we developed concerning sequences and series to study two other mathematical objects: sets and functions. For definitions of set notation, see the appendices. 2.1 The closure of a set. Closed and open sets. We will be studying the notions of closed and open subsets of R d. Notice that R 1 is just the real line (case d = 1), and R 2 the plane (case d = 2). You are welcome to just consider these two cases, although the definitions are valid for all dimensions d. The following theory is developed based on the notion of convergence of a sequence, which in turn is based on the distance formula. In d dimensions, the distance between two points A =(a 1,a 2,...,a d ) and B =(b 1,b 2,...,b d )is d(a, B) = p (b 1 a 1 ) 2 +(b 2 a 2 ) 2 + +(b d a d ) 2. A sequence of points (P k )inr d converges to a point Q if and only if the distance between P k and Q is eventually arbitrarily small. Just as for d = 2, this happens if and only if the sequences of coordinates of the P k converge to the corresponding coordinates of Q. Let S R d. In some cases there are points in R d that are infinitely close to S that are not in S. For example, zero is in some sense infinitely close to the interval (0, 1), but does not belong to it. There is a name for such points: Definition Given S R d and a 2 R d, we say that a is an adherent point of S if and only if there is a sequence (x n ) which is contained in S and converges to a. The set of all adherent points of S R d is denoted S and is called the closure of S. Lemma Every point of a set S R d is an adherent point of the set. 35

2 In set-theoretic notation, this lemma says that S S. Let a 2 S. We need to show that a is an adherent point. Define the sequence x n = a 8 n. This sequence is in S and converges to a, so we are done. Definition AsetS is closed if and only if it contains all of its adherent points. In set-theoretic notation, this says S is closed () S S. In fact, the set inclusion on the right hand side of the statement above can be replaced by equality, as demonstrated in the following theorem. *Theorem AsetS is closed if and only if S = S. 1 2 *Example For each of the following sets S find the set S. (Note: The first three are subsets of R, while the last two are subsets of R 2.) Explain your answer by refering to the definition, but you do not need to give a formal proof of your solution. Which sets are closed? 1. S =[2, 5). 2. S =[2, 5] c (the complement of [2, 5]). 3. S = Q (the set of all rational numbers). 1 Note: This theorem does not show that for all sets S, S is closed. The fact that S is closed will be proven below in Theorem An if and only if statement is really two statements that is, you need to show: 1. If S = S then S is closed, and 2. If S is closed then S = S. 36

3 4. S =? (the empty set). 5. S = {(x, y) 2 R 2 ; x>0 and y 0}. 6. S = {(x, y) 2 R 2 ; x 2 + y 2 < 4}. *Problem Explain why closed intervals are closed sets, as well as intervals of the form ( 1,b] and [a, 1). The following theorem justifies our use of the term closure for the process of adding in all adherent points of a set. 3 Theorem The closure S of a set S is closed. We need to show that S S. Idea: Write down what it means for a to be an adherent point of S (i.e. a 2 S), and use this to find a sequence of points in S that also converge to a (so that a is an adherent point of S, ora 2 S). Let a 2 S. This means that a is an adherent point of S; i.e.thereexistsasequence(a k ) such that 1. a k 2 S for all k, and 2. lim k!1 a k = a. Now consider a fixed a k 2 S. This means that a k is an adherent point of S; i.e. thereexistsa sequence (a k,n ) n such that 3 This theorem sounds like a tautology, but in fact is not obvious: if you have a non-closed set, and add in all of its adherent points, why shouldn t there be additional adherent points of the adherent points? 37

4 1. a k,n 2 S for all n, and 2. lim n!1 a k,n = a k. Since a k 2 S for all k, we get a 1,1 a 1,2 a 1,3 a 1,4...! a 1 a 2,1 a 2,2 a 2,3 a 2,4...! a 2 a 3,1 a 3,2 a 3,3 a 3,4...! a 3 a 4,1 a 4,2 a 4,3 a 4,4...! a # a such that a k,n 2 S for all k and n, a k 2 S for all k, and a 2 S. We will find a sequence x m in the table above with lim x m = a. Let = 1 m!1 2m. From the convergence statements above, we know that 1. There exists an k m such that a km a < 1 2m. 2. For the sequence a km,n (which converges to a km ) there exists an n m such that a km,n m a km < 1 2m. Let x m = a km,n m. Then by the triangle inequality, x m a apple x m a km + a km a < 1 2m + 1 2m = 1 m. We claim that x m is a sequence in S that converges to a. This is because for each m, x m 2 apple a 1 m,a+ 1 m, which is a nested sequence of closed intervals whose length goes to 0. Therefore a is an adherent point of S, i.e.a 2 S. *Theorem If S T R d then S T. *Theorem S is the smallest closed set containing S. 38

5 4 We can also define open sets: Definition AsetS R d is open if and only if its complement S c is closed. This is a non-standard definition of open sets, but it fits our set of tools well, and is equivalent to the more widely-used one given below. We state but will not prove the following theorem: Theorem Definition is equivalent to the following definition of an open set: AsetS R d is open if and only if for any element x 2 S, thereexists >0 such that the open ball B (x) :={y 2 R d d(x, y) < } S. *Example Use Definition to determine which of the sets in Example is open. *Problem Definition: The interior of a set S can be defined as: h i c. Int(S) = S c (I.e., take the complement of S, then the closure of S c, then the complement of S c.) Since Int(S) is the complement of a closed set, Int(S) is an open set. 1. Find Int([0, 5]). 2. Why is Int(S) S for any S? 3. Show that Int(S) is the largest open set contained in S. 2.2 Compact sets Definition AsetS R d is called bounded if and only if there exists a real number M>0 such that where O 2 R d is the origin and x = d(o, x). 5 x applem 8 x 2 S, 4 Lemma and Theorem show that S S and S is closed, so your job is to show that if F is a closed set and S F,thenS F. 5 Note that this gives defines the same property for sets S R as S is bounded if and only if there exists real numbers m apple M such that m apple x apple M for all x 2 S, which parallels the definition of a bounded sequence from Chapter 1. The reason to use notation as above is that it works well for higher dimensions d 2. 39

6 *Theorem AsetS R is not bounded if and only if there exists a monotone sequence a : N! S diverging to ±1. The following theorem is a version of the Axiom of Completeness that is better adapted to working directly with sets of real numbers. While this theorem is actually equivalent to the Axiom of Completeness, we will only prove it as a theorem that follows from the Axiom of Completeness. Theorem A nonempty set of real numbers that is bounded above has a least upper bound. Note: The least upper bound of a set S R is denoted lub(s). Likewise, a set S that is bounded below has a greatest lower bound, denoted glb(s). Let S R be a bounded set. Idea: We will construct a sequence of nested closed intervals whose lengths go to zero, and use the Axiom of Completeness to get a unique number g 2 R. Then we show that g is the least upper bound of S. Pick an element a 1 2 S, letb 1 be an upper bound of S, and let c 1 be the midpoint of the interval I 1 =[a 1,b 1 ]. Define a subinterval I 2 I 1 according to the following cases: 1. If c 1 is an upper bound of S, let I 2 =[a 1,c 1 ]. 2. If c 1 is not an upper bound of S, thenthereexistsa 2 2 S such that a 2 >c 1. In this case, let I 2 =[a 2,b 1 ] In either case, let us write I 2 =[a 2,b 2 ] and note that 1. a 2 2 S and b 2 is an upper bound of S, and 2. I 2 apple 1 2 I 1. 40

7 Repeating this procedure generates a sequence (I k ) of nested intervals whose lengths go to zero, so the Axiom of Completeness gives us a number g = lim k!1 a k = lim k!1 b k. Then g is the least upper bound of S, as shown by the following two claims: 1. The number g is an upper bound of S. 2. Moreover, if M is any upper bound of S, theng apple M. Subsets of R d that are both closed and bounded are of special interest: Definition AsetK R d is called compact if and only if it is closed and bounded. Compact sets show up a lot in mathematical arguments because of their nice properties. Here is one such property: *Theorem If K R d is compact and (s k ) is any sequence in K then there exists a convergent subsequence (s kn ) whose limit is a point in K. The real line R 1 is special among R d in that it has a natural ordering. Compact sets in R then have the following especially important property: *Theorem A nonempty compact set in R has a greatest and a least element. That is, if K is compact, there exist g 2 K and ` 2 K such that ` apple x apple g 8 x 2 K. 41