Ifaandbare in the extended reals anda < b, then theopen interval.a; b/is defined by .a; b/D˚
xˇˇa < x < b :
The open intervals.a;1/and. 1; b/aresemi-infiniteifaandbare finite, and. 1;1/ is the entire real line.
Definition 1.3.2
Ifx0is a real number and > 0, then the open interval.x0 ; x0C/ is an -neighborhood ofx0. If a set S contains an -neighborhood ofx0, thenS is a neighborhoodofx0, andx0 is aninterior pointofS (Figure 1.3.2). The set of interior points ofS is theinteriorofS, denoted byS0. If every point ofSis an interior point (that is,S0DS), thenSisopen. A setSisclosedifSc is open.( ) x0 + x0 − x
0
x0 = interior point of S S = four line segments
Figure 1.3.2
The idea of neighborhood is fundamental and occurs in many other contexts, some of which we will see later in this book. Whatever the context, the idea is the same: some defi-nition of “closeness” is given (for example, two real numbers are “close” if their difference is “small”), and a neighborhood of a pointx0is a set that contains all points sufficiently close tox0.
Example 1.3.4
An open interval.a; b/ is an open set, because ifx0 2 .a; b/ and minfx0 a; b x0g, then.x0 ; x0C/.a; b/:
The entire lineRD . 1;1/is open, and therefore;.D Rc/is closed. However, ;is also open, for to deny this is to say that;contains a point that is not an interior point, which is absurd because;contains no points. Since;is open,R.D ;c/is closed. Thus,Rand; are both open and closed. They are the only subsets ofRwith this property (Exercise 18).
Adeleted neighborhoodof a pointx0is a set that contains every point of some neigh-borhood ofx0except forx0itself. For example,
S D˚
xˇˇ0 <jx x0j<
is a deleted neighborhood ofx0. We also say that it is adeleted-neighborhoodofx0.
Theorem 1.3.3
(a)
The union of open sets is open:(b)
The intersection of closed sets is closed:These statements apply to arbitrary collections, finite or infinite, of open and closed sets:
Proof (a)
LetGbe a collection of open sets and SD [˚GˇˇG2G :
Ifx0 2 S, thenx0 2 G0 for someG0 inG, and sinceG0 is open, it contains some -neighborhood ofx0. SinceG0S, this-neighborhood is inS, which is consequently a neighborhood ofx0. Thus,S is a neighborhood of each of its points, and therefore open, by definition.
(b)
Let F be a collection of closed sets and T D \˚FˇˇF 2F . Then Tc D [˚
FcˇˇF 2F (Exercise 7) and, since eachFcis open,Tcis open, from
(a)
. Therefore, T is closed, by definition.Example 1.3.5
If 1< a < b <1, the set Œa; bD˚xˇˇaxb
is closed, since its complement is the union of the open sets. 1; a/and.b;1/. We say thatŒa; bis aclosed interval. The set
Œa; b/D˚
xˇˇax < b
is ahalf-closedorhalf-open interval if 1< a < b <1, as is .a; bD˚
xˇˇa < xb I
however, neither of these sets is open or closed. (Why not?)Semi-infinite closed intervals are sets of the form
Œa;1/D˚
xˇˇax and . 1; aD˚
xˇˇxa ;
wherea is finite. They are closed sets, since their complements are the open intervals . 1; a/and.a;1/, respectively.
Example 1.3.4 shows that a set may be both open and closed, and Example 1.3.5 shows that a set may be neither. Thus, open and closed are not opposites in this context, as they are in everyday speech.
Example 1.3.6
From Theorem 1.3.3 and Example 1.3.4, the union of any collection of open intervals is an open set. (In fact, it can be shown that every nonempty open subset of Ris the union of open intervals.) From Theorem 1.3.3 and Example 1.3.5, the intersection of any collection of closed intervals is closed.It can be shown that the intersection of finitely many open sets is open, and that the union of finitely many closed sets is closed. However, the intersection of infinitely many open sets need not be open, and the union of infinitely many closed sets need not be closed (Exercises 8 and 9).
Definition 1.3.4
LetSbe a subset ofR. Then(a)
x0is alimit pointofSif every deleted neighborhood ofx0contains a point ofS.(b)
x0 is aboundary pointofS if every neighborhood ofx0contains at least one pointinSand one not inS. The set of boundary points ofS is theboundaryofS, denoted by@S. TheclosureofS, denoted byS, isS DS[@S.
(c)
x0is anisolated pointofSifx02Sand there is a neighborhood ofx0that contains no other point ofS.(d)
x0 isexteriortoSifx0is in the interior ofSc. The collection of such points is the exteriorofS.Example 1.3.7
LetS D. 1; 1[.1; 2/[ f3g. Then(a)
The set of limit points ofSis. 1; 1[Œ1; 2.(b)
@S D f 1; 1; 2; 3gandS D. 1; 1[Œ1; 2[ f3g.(c)
3is the only isolated point ofS.(d)
The exterior ofSis. 1; 1/[.2; 3/[.3;1/.Example 1.3.8
Forn1, let InD1
2nC1; 1 2n
and S D [1
nD1
In: Then
(a)
The set of limit points ofSisS[ f0g.(b)
@S D˚xˇˇxD0orx D1=n .n2/ andSDS[ f0g.
(c)
Shas no isolated points.(d)
The exterior ofSis . 1; 0/[" 1 [
nD1
1
2nC2; 1 2nC1
# [
1 2;1
:
Example 1.3.9
LetSbe the set of rational numbers. Since every interval contains a rational number (Theorem 1.1.6), every real number is a limit point ofS; thus,S D R. Since every interval also contains an irrational number (Theorem 1.1.7), every real number is a boundary point ofS; thus@S DR. The interior and exterior ofSare both empty, and Shas no isolated points.Sis neither open nor closed.The next theorem says thatSis closed if and only ifS DS(Exercise 14).
Theorem 1.3.5
A setSis closed if and only if no point ofSc is a limit point ofS:Proof
Suppose thatSis closed andx02Sc. SinceScis open, there is a neighborhood ofx0that is contained inScand therefore contains no points ofS. Hence,x0cannot be a limit point ofS. For the converse, if no point ofSc is a limit point ofSthen every point in Sc must have a neighborhood contained inSc. Therefore,Sc is open andSis closed.Theorem 1.3.5 is usually stated as follows.
Corollary 1.3.6
A set is closed if and only if it contains all its limit points:Theorem 1.3.5 and Corollary 1.3.6 are equivalent. However, we stated the theorem as we did because students sometimes incorrectly conclude from the corollary that a closed set must have limit points. The corollary does not say this. IfShas no limit points, then the set of limit points is empty and therefore contained inS. Hence, a set with no limit points is closed according to the corollary, in agreement with Theorem 1.3.5. For example, any finite set is closed and so is an infinite set comprised entirely of isolated points, such as the set of integers.