Week 3

Question 1

Supremum

Answer 1

When M is the least upper bound for A, we call M the supremum of A and write M=sup(A)

Question 2

Infimum

Answer 2

When m is the greatest lower bound for A we call m the infimum of A and write m=inf(A)

Question 3

Let A ⊆ R and M ∈ R. Define M to be a least upper bound for A if and only if the following two conditions are satisfied:

Answer 3

a) M is an upper bound for A; and
b) for all upper bounds M’ for A, we have M ≤ M’

Question 4

Let A ⊆ R and let M ∈ R be an upper bound for A. Then M
is a least upper bound for A if and only if

Answer 4

∀ε > 0, ∃a ∈ A s.t. a > M − ε

Question 5

Let A ⊆ R and m ∈ R. Define m to be a greatest lower
bound for A if and only if:

Answer 5

a) m is an lower bound for A; and
b) for all lower bounds m′
for A, we have m′ ≤ m

Question 6

Theorem of uniqeness of least upper bounds and greatest lower bounds

Answer 6

Let A ⊆ R. Then A has at most one least upper bound and
at most one greatest lower bound

Question 7

Given a lower bound m for A, m is
a greatest lower bound for A if and only if

Answer 7

∀ε > 0, ∃a ∈ A s.t. a < m + ε

Question 8

The completeness axiom

Answer 8

Every non-empty subset of R
which is bounded above has a supremum.

Question 9

Theorem for necessary existence of infimum

Answer 9

Every non-empty subset of R which is bounded below has
an infimum.

Question 10

Theorem of ordered field and completeness axiom

Answer 10

There exists an ordered field R satisfying the completeness
axiom, which contains Q as a subfield. Any two such ordered fields are
isomorphic.

Question 11

Archimedes axiom

Answer 11

The natural numbers N ⊆ R are not bounded above. In particular, given any x ∈ R, there exists n ∈ N such that n > x.