Linear Algebra, Theory And Applications, 2012a

1.8. WELL ORDERING AND ARCHIMEDEAN PROPERTY 21
Proposition 1.7.3 Let S be a nonempty set and suppose sup (S) exists. Then for every
δ>0,
S ∩ (sup (S) − δ, sup (S)] ≠ ∅.
If inf (S) exists, then for every δ>0,
S ∩ [inf (S) , inf (S)+δ) ≠ ∅.
Proof: Consider the first claim. If the indicated set equals ∅, then sup (S) − δ is an
upper bound for S which is smaller than sup (S) , contrary to the definition of sup (S) as
the least upper bound. In the second claim, if the indicated set equals ∅, then inf (S)+δ
would be a lower bound which is larger than inf (S) contrary to the definition of inf (S).
1.8 Well Ordering And Archimedean Property
Definition 1.8.1 A set is well ordered if every nonempty subset S, contains a smallest
element z having the property that z ≤ x for all x ∈ S.
Axiom 1.8.2 Any set of integers larger than a given number is well ordered.
In particular, the natural numbers defined as
N ≡{1, 2, ···}
is well ordered.
The above axiom implies the principle of mathematical induction.
Theorem 1.8.3 (Mathematical induction) A set S ⊆ Z, having the property that a ∈ S
and n +1∈ S whenever n ∈ S contains all integers x ∈ Z such that x ≥ a.
Proof: Let T ≡ ([a, ∞) ∩ Z) \ S. Thus T consists of all integers larger than or equal
to a which are not in S. The theorem will be proved if T = ∅. If T ≠ ∅ then by the well
ordering principle, there would have to exist a smallest element of T, denoted as b. It must
bethecasethatb>asince by definition, a/∈ T. Then the integer, b − 1 ≥ a and b − 1 /∈ S
because if b − 1 ∈ S, then b − 1+1 = b ∈ S by the assumed property of S. Therefore,
b − 1 ∈ ([a, ∞) ∩ Z) \ S = T which contradicts the choice of b as the smallest element of T.
(b − 1 is smaller.) Since a contradiction is obtained by assuming T ≠ ∅, it must be the case
that T = ∅ and this says that everything in [a, ∞) ∩ Z is also in S.
Example 1.8.4 Show that for all n ∈ N, 1 2 · 3 2n−1
4 ···
2n < √ 1
2n+1
.
If n = 1 this reduces to the statement that 1 2 <
then that the inequality holds for n. Then
1
2 · 3 ···2n − 1 2n +1
·
4 2n 2n +2
<
=
1 √
3
which is obviously true. Suppose
1 2n +1
√ 2n +1 2n +2
√ 2n +1
2n +2 .
1
The theorem will be proved if this last expression is less than √ 2n+3
. This happens if and
only if
( ) 2
1
1 2n +1
√ = > 2n +3 2n +3 (2n +2) 2
which occurs if and only if (2n +2) 2 > (2n +3)(2n + 1) and this is clearly true which may
be seen from expanding both sides. This proves the inequality.