Week 2

Question 1

Polynomial estimation lemma

Answer 1

Let n ∈ N, and suppose
we are given real numbers a_0, a_1, · · · , a_n with an > 0. Write
p(x) = a_nx^n + a_n−1x^n−1 + · · · + a_1x + a_0.
Then there exists N > 0 such that x ≥ N =⇒ 1/2a_nx^n ≤ p(x) ≤ 3/2 a_nx^n

Question 2

The ordered field axioms

Answer 2

The real numbers R are a set equipped with an addition +, and multiplication ·, and form a field under these operations, that
is, nine axioms are satisfied.
What this essentially means is that we can add, subtract, multiply and divide real numbers in the usual way.

Question 3

Order axiom for R

Answer 3

There is a relation > on R satisfying
j) For all a ∈ R, exactly one of the statements a = 0, a > 0 and 0 > a
is true;
k) For all a, b ∈ R, b > a ⇐⇒ b − a > 0;
l) For all a, b ∈ R with a > 0 and b > 0, we have a + b > 0 and
ab > 0.

Question 4

Upper bound

Answer 4

Let A ⊆ R. We say that M ∈ R is an upper bound for A if and only if for all a ∈ A, we have a ≤ M

Question 5

Bounded above

Answer 5

Define A to be bounded above if and only if there exists an upper bound for A

Question 6

Lower bound

Answer 6

m ∈ R is said to be a lower bound for A ⊆ R if and only if for all a ∈ A, we have m ≤ a

Question 7

Bounded below

Answer 7

Say that A is bounded below if and only if there exists a lower bound for A.

Question 8

Bounded above (symbolically)

Answer 8

Symbolically, the set A ⊆ R is bounded above if and only if
∃M ∈ R s.t. ∀a ∈ A, a ≤ M,

Question 9

Bounded below (symbolically)

Answer 9

A is bounded below if and only if
∃m ∈ R s.t. ∀a ∈ A, m ≤ a.

Question 10

Bounded

Answer 10

Let A ⊆ R. We say that A is bounded if and only if A is bounded above and bounded below.