Class notes Aug. 30
The Principle of Mathematical Induction
Theorem 1. Let P ( n ) is a statement depending on n  S , where S  N and such that:
(i) P (1) is true;
(ii) for each n  S , if P ( n ) is true, then P ( n  1) is true.
Then P ( n ) is true for all n  S .
Proof:
We prove the theorem by contradiction. Assume that the conclusion is false, and
therefore P ( n ) is not true for all n  S . Define then A  {k  S : P(k ) is false} . By our
assumption A   , and since A  S  N , A is bounded below. Then by the Axiom of
Completeness A must have a greatest lower bound. Moreover, inf A  min A , that is A
contains its infimum (see proof of Lemma 1 below). Denote k0  min A  A . By
assumption (i), k0  1 , and since k0  min A , k0  1 A . Therefore P(k0  1) must be
true, and by (ii) it follows that P(k0 ) is also true. Therefore k0  A , contradiction!
QED
Lemma 1. Well Ordering Principle
Every nonempty subset of N has a smallest element.
Proof:
Let S  N , S   . Since any subset of N is bounded below by 1, S is bounded below.
Then by the Axiom of Completeness (which holds true for sets bounded below as well as
above), it follows that S has a greatest lower bound, b  inf S  R in the set of real
numbers. We want to prove that b  S .
Now, by the properties of infimum of a set, b  k for all k  S . Assume b  S . Since
k b
 0 . Again, by the properties
b  inf S , there exists k0  S , b  k0  b  1. Let   0
2
of infimum of a set, there exists k1  S , b  k1  b   . But by the choice of  , we actually
obtain:
b  k1  b    k0  b  1.
But then we found k0 , k1  S for which k0  k1  k0  b  b  1  b  1 impossible!
Therefore b  S .
QED
Let us now look at some examples of use of the Principle of Mathematical Induction.
Example 1.
Prove that for any n  N and any x  1 the following holds:
1  x n 1
1  x  x 2  ...  x n 
1 x
1  x2
Proof: Let P ( n ) be the above statement. If n  1 the left hand side is 1  x 
, so
1 x
assumption (i) in Theorem 1 is true.
Suppose now that P ( n ) is true for some n  N . For P ( n  1) we have:
1  x  x 2  ...  x n  x n 1 
1  x n 1
1  x n 1  x n 1  x n  2 1  x n  2
,
 x n 1 

1 x
1 x
1 x
so P ( n  1) is true.
Therefore, by Theorem 1, P ( n ) is true for all n  N .
Example 2. (which is Example 1.2.7. in text)
Define the sequence xn by the following recurrence equation: x1  1, xn 1 
1
xn  1, n  1 .
2
We want to prove that xn is increasing, that is, xn  xn1 , for all n  1 .
1
3
Proof: For n  1, x2  x1  1   x1 , so P (1) is true. Let now n  N such that P ( n ) is
2
2
true, that is, xn  xn1 . We have
1
1
xn  2  xn 1  1  xn  1  xn 1 ,
2
2
so P ( n  1) is true. Therefore, xn  xn1 , for all n  1 .
Example 3. (example 2 continued) We prove that xn  2, for all n  1 .
1
1
We have x1  1  2 , and if for some n  N , xn  2 , then xn 1  xn  1  2  1  2 .
2
2
Therefore, by the Principle of Mathematical Induction it follows that xn  2, for all n  1 .
Example 4. (example 2 continued) We give an alternate proof of xn  2, for all n  1 that
uses an explicit formula for xn . Let us write a few terms of the sequence:
1
3
1
3
7
x1  1, x2  x1  1  , x3  x2  1  2  1  2 .
2
2
2
2
2
1
2
3
2 1
2 1
2 1
, x3  2 , so we conjecture that
We observe a pattern: x1  0 , x2 
2
2
2
n
2 1
xn  n 1 for all n  N . We already checked for n  1 . Suppose that for some n  N ,
2
2n  1
1
1 2n  1
2n  1
2  2n  1 2n 1  1
xn  n 1 . Then xn 1  xn  1   n1  1  n  1 

, and by the
2
2
2 2
2
2n
2n
2n  1
Principle of Mathematical Induction it follows that xn  n 1 for all n  N .
2
Now we can get back to the proof of the statement in Example 3. We just proved that for
2n  1 2n
1
1
all n  N : xn  n 1  n 1  n 1  2  n 1  2 and we are done.
2
2
2
2
This proof provides a better insight into the sequence. From Example 3 we just found out
that 2 is an upper bound for the sequence. The above proof leads to stating that 2 is the
least upper bound. The proof of this statement is left to the reader.
We state without proof a modified version of the Principle of Mathematical Induction
which is used sometimes in proofs.
Theorem 2. Second Principle of Mathematical Induction.
Let P ( n ) is a statement depending on n  N . If
(i) P (1) is true;
(ii) For n  1 , if P(1), P(2),..., P(n  1) are all true, then P ( n ) is true.
Then P ( n ) is true for all n  N .
EXERCISES
1. Prove that n3  5n is divisible by 6 for all n  N .
n
n(n  1)
2. Prove that  k 
for all n  N .
2
k 1
n
n(n  1)(2n  1)
3. Prove that  k 2 
for all n  N .
6
k 1
4. Prove that 2n  n for all n  N .
5. Prove that n !  2 n for all n  4 .


2n  1
6. Prove that 2  sup  xn  n1 , n  1 .
2


Functions
Definition 1. Let A and B be two sets. A function f from A into B is a subset of
A  B  {( x, y ) : x  A, y  B} with the property that for every x  A there exists a single
y  B such that ( x, y )  f .
The set A is called the domain of the function f. The range of f, also referred to as the
image of the set A by the function f, is defined by { y  B :( x, y )  f , for some x  A} .
If f is a function from A into B and ( x, y )  f , then y is called the value of the function
at x and we write y  f ( x) .
If f : A  R , f is said to be a real-valued function on A.
Definition 2. Let f be a function from A into B, and let E  A . We define the image of
E under f by f ( E )  { f ( x) : x  E} .
Definition 3. Let f be a function from A into B, and let H  B . We define the inverse
image of H under f, f 1 ( H )  {x  A : f (x)  H } .
Example 5.
Let f : Z  Z , f ( x)  x 2 . Let E  {2, 1, 0} and H  {0,1, 4} . We have f ( E )  {0,1, 4} ,
and f 1 ( H )  {2, 1,0,1, 2}  E .
Notice that H  f ( E ) , so besides showing how to actually find the image and inverse
image of sets, the example also shows that in general f ( f 1 ( E ))  E .
Below are some well-known properties relating functions and sets. We state them as
Theorem 3.
Let f : A  B , and let A1 , A2  A, B1 , B2  B . Then:
f ( A1  A2 )  f ( A1 )  f ( A2 )
a)
f ( A1  A2 )  f ( A1 )  f ( A2 )
b)
f 1 ( B1  B2 )  f 1 ( B1 )  f 1 ( B2 )
f 1 ( B1  B2 )  f 1 ( B1 )  f 1 ( B2 )
Proof:
We will only prove parts 2 of a) and b).
Let y  f ( A1  A2 ) . By the definition of the image of a set by a function, there exists
Since
it
follows
that
x  A1  A2 : f ( x)  y .
x  A1  A2 , x  A1 and x  A2 ,
f ( x)  f ( A1 ) and f (x)  f ( A2 ) , and therefore y  f ( x)  f ( A1 )  f ( A2 ) and we are
done.
We now prove the second equality in part b). Let x  f 1 ( B1  B2 ) . Then
y  f ( x)  B1  B2 , so y  B1 and y  B2 , and therefore x  f 1 ( B1 )  f 1 ( B2 ) . So we
proved f 1 ( B1  B2 )  f 1 ( B1 )  f 1 ( B2 ) (1).
For the converse inclusion, let x  f 1 ( B1 )  f 1 ( B2 ) . Then there exist
y1  B1 , and y2  B2 such that f ( x)  y1 and f ( x)  y2 . But since f is a function
f ( x)  y1  y2 , and therefore f ( x)  B1  B2 , so x  f 1 ( B1  B2 ) . Thus we proved that
f 1 ( B1 )  f 1 ( B2 )  f 1 ( B1  B2 ) ,
which
together
with
(1)
proves
that
f 1 ( B1  B2 )  f 1 ( B1 )  f 1 ( B2 ) .
QED
Example 6. Here is the solution to Exercise 6 in Aug 28 notes, also exercise 8 below.
Let f , g : D  R, f ( D) and g ( D) are bounded subsets of R. We prove the following:
inf{ f ( x)  g ( x) : x  D}  inf{ f ( x) : x  D}  inf{g ( x) : x  D}
First notice that if both f and g are bounded then f  g is bounded. Indeed, suppose that
there exist m1 , m2 , M1 , M 2 such that m1  f ( x)  M1 , m2  g ( x)  M 2 , for all x  D . Then
m1  m2  f ( x)  g ( x)  M1  M 2 , for all x  D , so ( f  g )( D ) is a bounded set. Let
u1  inf{ f ( x) : x  D}, and u2  inf{g ( x) : x  D} . Since u1  f ( x), for all x  D , and
u2  g ( x), for all x  D , it follows that u1  u2  ( f  g )( x), for all x  D , so u1  u2 is a
lower bound for the set ( f  g )( D ) . But then from the definition of the greatest lower
bound, any lower bound has to be smaller than the greatest lower bound, so
inf{ f ( x)  g ( x) : x  D}  u1  u2 , and we are done.
Definition 4. A function f : A  B is one-to-one (or injective), if whenever
x1  x2 , f ( x1 )  f ( x2 ) .
Observation: In problems it is more useful to prove a statement equivalent to the
definition: if f ( x1 )  f ( x2 ), then x1  x2 .
Definition 5. A function f : A  B is onto (or surjective), if f ( A)  B .
A function f : A  B that is both one-to-one and onto has an inverse, denoted
f 1 : B  A , defined in the following way: f 1 ( y)  x the unique element in A, such that
f ( x)  y .
EXERCISES
7. Let f,g be real-valued functions defined on a nonempty set D, and such that
Range f  f ( D) and Range g  g ( D) are bounded subsets of R. Prove that
sup{ f ( x)  g ( x) : x  D}  sup{ f ( x) : x  D}  sup{g ( x) : x  D} ,
that is, sup( f  g )( D)  sup f ( D)  sup g ( D) .
8. Refer to Exercise 7. State and prove a statement regarding inf( f  g )( D) .
9. Define f : N  N , f (n)  2n  1 . Is f one-to-one? Is f onto? What is f ( N ) ?
Countable and Uncountable Sets. Cardinality
When discussing finite sets we all know what it means that two sets are of the same size,
it means they have the same number of elements. We also have an idea of what it means
for a set to be infinite, but what does it mean that two infinite sets are of the same size?
How do we measure the various degrees of infinity?
For example the sets A  {0,1, 2} and B  {4,5, 6} are both finite and have three elements.
But how do we compare N and S={2n  1: n  N } . Obviously they are both infinite, and
S is a proper subset of N, but does N have more elements than S?
Definition 6. Let A and B be two sets. We say that A and B have the same cardinality if
there exists a function f : A  B one-to-one and onto. In this case we write A B .
We will also refer to the sets A and B as being equivalent.
It can be easily shown using the properties of functions and compositions of functions
that the following properties hold:
Lemma 2.
Let A, B and C be sets. We have:
(i) A A .
(ii) If A B then B A .
(iii) If A B and B C , then A C .
Example 7.
Let us show that for the set S defined above, S N .
Define f : N  S , f (n)  2n  1 .
If for n, m  N , f (n)  f (m) , we have 2n  1  2m  1, so n  m , proving that f is one-toone. To prove that it is also onto, let m  S . We see from the definition of the function
that m has to be odd. Now if we show that we can find n  N : f (n)  m we are done,
since this would prove that S  f ( N ) , and as f ( N )  S it follows that f ( N )  S .
m 1
 N since m is odd, and we are done.
Now, f (n)  2n  1  m, n 
2
Definition 7. For each positive integer n, let Nn  {1, 2,..., n} . If A is a set we say:
(a) A is finite if A N n for some n, or A   , the empty set.
(b) A is infinite if A is not finite.
(c) A is countable if A N .