Chapter 3 The Real Numbers

Section 3.4 Topology of the Real Numbers

( )  Definition 3.4.1 N (x ; ) = { y  : | x – y | <  }. 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 , we may be interested in all points y that are less than  away from x: { y : | x – y | <  }. We formalize this idea in the following definition. Definition 3.4.1 Let x  and let  > 0. A neighborhood of x is a set of the form   N (x ; ) = { y   : | x – y | <  }. The number  is referred to as the radius of N (x ; ). N (x ; ) ( )  x x –  x + 

Sometimes want to consider points y that are close to x but different from x. We can accomplish this by requiring | x – y | > 0. Definition 3.4.2 Let x  and let  > 0. A deleted neighborhood of x is a set of the form   N* (x ; ) = { y   : 0 < | x – y | <  }. Clearly, N*(x ; ) = N (x ; )\{x}. N* (x ; ) = (x – , x)  (x, x + ) ( )  x ( ) x –  x +  If S  , then a point x in can be thought of as being “inside” S, on the “edge” of S, or “outside” S. Saying that x is “outside” S is the same as saying that x is “inside” the complement of S, \ S. Using neighborhoods, we can make the intuitive ideas of “inside” and “edge” more precise.

[ ) Definition 3.4.3 Example 3.4.4(c)  ( )  Let S be a subset of . A point x in is an interior point of S if there exists a neighborhood N of x such that N  S. If for every neighborhood N of x, N  S   and N  ( \ 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) | 5 3 [ ) ( ) ( )  ( ) ( )  ( ) Let S be the interval [0, 5). The point 3 is an interior point of S because there exists a neighborhood of 3 that is contained in S. The same is true for any point x such that 0  x  1. So int S = (0, 5). But the point 0 is different. Every neighborhood of 0 (no matter how small) will contain points of S and points that are not in S. The same is true for 5. So bd S = {0, 5}.

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. We have bd S = bd ( \ S). So all the boundary is in S iff none of the boundary is in \ S. 

How do 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 =  {A : A  A }. If x  S, then x  A for some 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 A1, …, An be a finite collection of open sets and let If T = , we are done, since  is open. If T  , let x  T. Then x  Ai for all i = 1, …, n. Since each set Ai is open, there exist neighborhoods Ni (x ; i) of x such that Ni (x ; i)  Ai. Let   = min {1, …, n}. Then N (x ; )  Ai for each i = 1, …, n, so N (x ; )  T. Thus x is an interior point of T, and T is open.  Note: Part (b) of Theorem 3.4.7 does not necessarily hold for infinitely many open sets.

Example 3.4.12 Corollary 3.4.11 Practice 3.4.13 For each n  , let An = ( 1/n, 1/n). Each of these sets is an open set (an open interval). But what is in 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 which is not closed. So the closed sets have “built up” (by taking unions) to a set that is not closed.

Accumulation Points Definition 3.4.14 Example 3.4.15 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  > 0, N*(x ; )  S  . The set of all accumulation points of S is denoted by S . Example 3.4.15 (a) If S is the interval (0, 1], then S  = [0, 1]. (b) If S = {1/n : n  }, then S  = {0}. Note that 0  S. (c) If S = , then S  = . So consists entirely of isolated points. (d) If S is a finite set, then S  = .

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 , where S  is the set of accumulation points of S. In terms of neighborhoods, a point x is in cl 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 N  S contains x. If x  S, then x  S  and every deleted neighborhood intersects S. 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.

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 . We must show that x  S. If x  S, then x is in the open set \ 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. To this end, let x  \ S. Conversely, suppose that S   S. We shall show that \ S is open. Then x  S , so there exists a deleted neighborhood N*(x ; ) that misses S. Since x  S, the whole neighborhood N (x ; ) misses S; that is, N (x ; )  \ S. Thus \ S is open and S is closed by Theorem 3.4.7(b).

 ( ) ( ) 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. N ( y ;  )  x ( ) ( ) ( ) | y | z Proof: N* (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 ; ) intersects cl S. We must show that N*(x ; ) intersects S. To this end, let y  N*(x ; )  cl S. Since N*(x ; ) is an open set, there exists a neighborhood N ( y ;  ) contained in N*(x ; ). But y  cl S, so every neighborhood of y intersects S. That is, there exists a point z in N ( y ;  )  S. But then z  N ( y ;  )  N*(x ; ), so that x  S  and x  cl S. The proofs of (c) and (d) are Exercise 18. 