Problem Set Eight: Nested Intervals and Bisection
Definition: A point p is a accumulation point of a set A of reals iff  ε  0  x  A so that 0  x  p  ε .
Theorem: If p is an accumulation point of A then  ε  0 { x  A : 0  x  p  ε } is infinite.
Geometrically this means that every open interval containing p contains infinitely many points in A.
Theorem: (1) Each infinite set A contains a sequence ( x n ) of points with x n  x k whenever n  k.
(2) Each infinite set A has a countably infinite subset.
(3) (Bolzano-Weierstrass Theorem again): Every infinite bounded set of real numbers has an
accumulation point.
Remark: Roughlyly speaking part (2) says that countably infinite sets are the smallest sort of infinite sets.
Nested Intervals Theorem: If I n  [ a n , b n ] is a sequence of closed bounded intervals with I n  1  I n for
each n, then there is a point p in all the intervals I n .
Another Version of Nested Intervals: If I n  [ a n , b n ] is a sequence of closed bounded intervals with
I n  1  I n for each n and b n  a n  len (I n )  0 , then
(a) there is a unique point p in all the intervals I n , and
(b) if J is any open interval containing p then I n  J for some n.
Several interesting properties of sets can be derived by using the Nested Intervals Theorem and a method
loosely referred to as Bisection. This section ends with two examples.
Theorem Every closed bounded interval [a,b] is uncountable. Put another way, for any sequence ( a n )
there is a point c in [a,b] with c  a n for all n
Sketch of Proof: It is handy to use CBI to abbreviate “closed bounded interval.” Fix a sequence ( a n ) .
The main claim is that there a sequence ( I n ) of CBIs with I n  1  I n  [a, b] and a n  I n for all n. Once
that is established no point c   n  1 I n can be a term of the sequence. The construction of ( I n ) uses this
simple fact; for any CBI I and number p, there is a CBI J so that J  I and p  J .
The sequence ( I n ) is defined recursively. For I 1 choose any CBI so that I 1  [a, b] and a 1  I 1 . Next
choose any CBI I 2 so that I 2  I 1 and a 2  I 2 . Next choose any CBI I 3 so that I 3  I 2 and a 3  I 3 .
Continuing in this fashion produces the desired sequence ( I n ) .
Corollary: The set of irrational numbers is uncountable. The set of transcendental numbers is
uncountable.
Definition: Let A be a set of reals. A collection of sets C is a cover of A iff each element of A is in at
least one set from the collection C, i.e., A   { B : B  C } . A subcover of C is a subcollection of C that
also covers A. For instance C  { (x  2, x  2) : 0  x  2 } is a collection of open intervals that covers
[0,3], and C #  { (  2, 2), (  1, 3), ( 0, 4 ) } is a finite subcover of [0,3].
Heine-Borel Theorem: If C is a collection of open intervals that covers [a,b], then there are a finite
number of intervals in C which also cover [a,b]. Put another way every cover of [a,b] by open intervals
has a finite subcover.
Definition: The Jordan Content (or content) of a set A of reals is
con(A)  glb {

n
k 1
len (I k ) } ,
where the glb is taken over all covers C  { I k } kn  1 of A by a finite number of open intervals.
If A can’t be covered by a finite number of open intervals then con(A)   .
Example: Every finite set has content zero. Why? Suppose A is non-empty, has m elements and n is a
positive integer. Using the cover C  { (x  1/n, x  1/n) : x  A } to estimate con(A) gives
0  con(A)  2 m / n . Taking the limit as n   shows con(A) = 0. Similarly   (0, 1/n) for all n.
PROBLEMS
Problem 8-1. Find examples of the following.
(a) A nested sequence ( J n ) of bounded intervals with  n  1 J n   .
(b) A nested sequence ( K n ) of closed rays with  n  1 K n   .
(c) An infinite set with no accumulation points.
(d) A cover of (0,1] by open intervals having no finite subcover.
(e) A cover of [ 0 ,  ) by open intervals having no finite subcover.
Problem 8-2. Without giving formal proofs describe all accumulation points of these sets: [a,b]; (a,b); the
set of rational numbers in (0,1); the set of irrational numbers in (0,1); { (1/n)  (1/m) : m & n  1 } .
Problem 8-3. If A is an uncountable set of reals then A has at least one accumulation point.
Problem and Definitions 8.4. (a) The derived set of A, denoted by A  , is the set of all accumulation
points in A. Without giving formal proofs describe the derived set of (a,b) and { (1/n)  (1/m) : m & n  1} .
(b) The second derived set of A, denoted by A  , is the set of all accumulation points of A  . Without
giving formal proofs describe the second derived set of (a,b) and { (1/n)  (1/m) : m & n  1} .
(c) Prove that A   A  for any set A.
Problem 8-5. Prove these properties of content. (a) A  B implies con(A)  con(B) .
(b) con(  nk  1 A k )  
n
k 1
con(A
k
)
for any finite collection of sets.
(c) con ( [a, b] )  b  a . Hint, use problem 1-7.
(d) If A is bounded then con(A)  con( A  ) .
(e) If con(A)   then A is bounded and con( A  )  con(A) .
Problem 8-6. Prove that If D is dense in [a,b] then con ( D )  b  a . In particular  x  [a, b] : x rational 
and  x  [a, b] : x irrational  have content b  a .
Definition: There is a recursively defined sequence of sets { C n } with the following properties:
(i) C 0  [0, 1] ;
(ii) C n is the union of a finite number of disjoint CBIs, called the components of C n , and;
(iii) C n  1 is obtained from C n by removing the middle third of each component of C n .
For instance C 1  [ 0 , 1 / 3 ]  [ 2 / 3 , 1] , C 2  [ 0 , 1 / 9 ]  [ 2 / 9 , 1 / 3 ]  [ 2 / 3 , 7 / 9 ]  [ 8 / 9 , 1] , and so on.
Problem 8-7. The Cantor Set is C   n  1 C n . Prove it is uncountable, has content zero, and contain no
interval.