102 Chapter 3. Basic Topology ofR
3.4. Perfect Sets and Connected Sets 103 x1∈/K2,x2∈/K3,x3∈/ K4,. . . . Some care must be taken to ensure that each Kn is nonempty, for then we can use Theorem3.3.5to produce an
x∈ ∞
n=1
Kn ⊆P that cannot be on the list{x1, x2, x3, . . .}.
LetI1 be a closed interval that containsx1 in its interior (i.e.,x1 is not an endpoint ofI1). Now,x1is not isolated, so there exists some other pointy2∈P that is also in the interior ofI1. Construct a closed intervalI2, centered on y2, so thatI2⊆I1but x1∈/I2. More explicitly, ifI1= [a, b], let
ǫ= min{y2−a, b−y2,|x1−y2|}.
Then, the intervalI2= [y2−ǫ/2, y2+ǫ/2] has the desired properties.
$ %
x•1 [ y• ]
2
I1
I2
This process can be continued. Becausey2∈P is not isolated, there must exist another point y3 ∈ P in the interior of I2, and we may insist that y3 = x2. Now, constructI3 centered ony3and small enough so thatx2∈/ I3andI3⊆I2. Observe thatI3∩P =∅ because this intersection contains at leasty3.
If we carry out this construction inductively, the result is a sequence of closed intervalsIn satisfying
(i) In+1⊆In, (ii) xn∈In+1, and (iii) In∩P =∅.
To finish the proof, we let Kn =In∩P. For each n∈N, we have that Kn is closed because it is the intersection of closed sets, and bounded because it is contained in the bounded setIn. Hence, Kn is compact. By construction,Kn
is not empty andKn+1⊆Kn. Thus, we can employ the Nested Compact Set Property (Theorem3.3.5) to conclude that the intersection
∞
n=1
Kn=∅.
But eachKnis a subset ofP, and the fact thatxn∈In+1leads to the conclusion that ∞n=1Kn=∅, which is the sought-after contradiction.
104 Chapter 3. Basic Topology ofR
Connected Sets
Although the two open intervals (1,2) and (2,5) have the limit pointx= 2 in common, there is still some space between them in the sense that no limit point of one of these intervals is actually contained in the other. Said another way, the closure of (1,2) (see Definition3.2.11) is disjoint from (2,5), and the closure of (2,5) does not intersect (1,2). Notice that this same observation cannot be made about (1,2] and (2,5), even though these latter sets are disjoint.
Definition 3.4.4. Two nonempty setsA, B ⊆R are separated ifA∩B and A∩B are both empty. A set E ⊆ R is disconnected if it can be written as E=A∪B, whereAandB are nonempty separated sets.
A set that is not disconnected is called aconnected set.
Example 3.4.5. (i) If we letA= (1,2) andB= (2,5), then it is not difficult to verify that E = (1,2)∪(2,5) is disconnected. Notice that the sets C = (1,2] and D = (2,5) are not separated because C ∩D = {2} is not empty. This should be comforting. The union C∪D is equal to the interval (1,5), which better not qualify as a disconnected set. We will prove in a moment that every interval is a connected subset ofRand vice versa.
(ii) Let’s show that the set of rational numbers is disconnected. If we let A=Q∩(−∞,√
2) and B=Q∩(√ 2,∞), then we certainly have Q=A∪B. The fact thatA⊆(−∞,√
2) implies (by the Order Limit Theorem) that any limit point of Awill necessarily fall in (−∞,√
2]. Because this is disjoint from B, we get A∩B = ∅. We can similarly show that A∩B =∅, which implies that Aand B are separated.
The definition of connected is stated as the negation of disconnected, but a little care with the logical negation of the quantifiers in Definition3.4.4results in a positive characterization of connectedness. Essentially, a setEis connected if, no matter how it is partitioned into two nonempty disjoint sets, it is always possible to show that at least one of the sets contains a limit point of the other.
Theorem 3.4.6. A set E ⊆R is connected if and only if, for all nonempty disjoint sets A and B satisfying E = A∪B, there always exists a convergent sequence (xn)→xwith (xn) contained in one ofA orB, andxan element of the other.
Proof. Exercise3.4.6.
The concept of connectedness is more relevant when working with subsets of the plane and other higher-dimensional spaces. This is because, in R, the connected sets coincide precisely with the collection of intervals (with the un- derstanding that unbounded intervals such as (−∞,3) and [0,∞) are included).
3.4. Perfect Sets and Connected Sets 105 Theorem 3.4.7. A setE ⊆Ris connected if and only if whenever a < c < b with a, b∈E, it follows thatc∈E as well.
Proof. AssumeE is connected, and leta, b∈E anda < c < b. Set A= (−∞, c)∩E and B= (c,∞)∩E.
Because a∈ A andb ∈ B, neither set is empty and, just as in Example3.4.5 (ii), neither set contains a limit point of the other. IfE=A∪B, then we would have that E is disconnected, which it is not. It must then be that A∪B is missing some element ofE, andc is the only possibility. Thus,c∈E.
Conversely, assume thatEis an interval in the sense that whenevera, b∈E satisfya < c < bfor somec, thenc∈E. Our intent is to use the characterization of connected sets in Theorem 3.4.6, so let E = A∪B, where A and B are nonempty and disjoint. We need to show that one of these sets contains a limit point of the other. Picka0∈Aandb0∈B, and, for the sake of the argument, assume a0 < b0. Because E is itself an interval, the interval I0 = [a0, b0] is contained inE. Now, bisectI0 into two equal halves. The midpoint ofI0 must either be inA orB, and so chooseI1= [a1, b1] to be the half that allows us to havea1 ∈ A and b1 ∈ B. Continuing this process yields a sequence of nested intervals In = [an, bn], wherean ∈ A, bn ∈ B, and the length (bn−an) →0.
The remainder of this argument should feel familiar. By the Nested Interval Property, there exists an
x∈ ∞
n=0
In,
and it is straightforward to show that the sequences of endpoints each satisfy liman =xand limbn =x. But now x∈E must belong to eitherAorB, thus making it a limit point of the other. This completes the argument.
Exercises
Exercise 3.4.1. IfP is a perfect set andKis compact, is the intersectionP∩K always compact? Always perfect?
Exercise 3.4.2. Does there exist a perfect set consisting of only rational num- bers?
Exercise 3.4.3. Review the portion of the proof given in Example 3.4.2 and follow these steps to complete the argument.
(a) Because x ∈ C1, argue that there exists an x1 ∈ C∩C1 with x1 = x satisfying|x−x1| ≤1/3.
(b) Finish the proof by showing that for eachn∈N, there existsxn∈C∩Cn, different from x, satisfying|x−xn| ≤1/3n.
Exercise 3.4.4. Repeat the Cantor construction from Section3.1starting with the interval [0,1]. This time, however, remove the open middlefourth from each component.
106 Chapter 3. Basic Topology ofR (a) Is the resulting set compact? Perfect?
(b) Using the algorithms from Section3.1, compute the length and dimension of this Cantor-like set.
Exercise 3.4.5. LetA and B be nonempty subsets of R. Show that if there exist disjoint open sets U and V with A⊆ U and B ⊆V, then A and B are separated.
Exercise 3.4.6. Prove Theorem3.4.6.
Exercise 3.4.7. A setEistotally disconnected if, given any two distinct points x, y∈E, there exist separated setsAandB withx∈A,y∈B, andE=A∪B.
(a) Show thatQis totally disconnected.
(b) Is the set of irrational numbers totally disconnected?
Exercise 3.4.8. Follow these steps to show that the Cantor set is totally dis- connected in the sense described in Exercise3.4.7.
LetC= ∞n=0Cn, as defined in Section3.1.
(a) Given x, y ∈C, with x < y, set ǫ=y−x. For each n= 0,1,2, . . ., the set Cn consists of a finite number of closed intervals. Explain why there must exist anN large enough so that it is impossible forxandyboth to belong to the same closed interval of CN.
(b) Show thatC is totally disconnected.
Exercise 3.4.9. Let{r1, r2, r3, . . .}be an enumeration of the rational numbers, and for eachn∈Nset ǫn= 1/2n. DefineO=∞
n=1Vǫn(rn), and letF=Oc. (a) Argue that F is a closed, nonempty set consisting only of irrational
numbers.
(b) DoesF contain any nonempty open intervals? IsF totally disconnected?
(See Exercise3.4.7for the definition.)
(c) Is it possible to know whether F is perfect? If not, can we modify this construction to produce a nonempty perfect set of irrational numbers?