Theory

Definition · The fundamental axiom of analysis#

If,

• $a_n \in \mathbb{R} \forall n \in \mathbb{N}$
• $A \in R$
• $a_{n} \leq a_{n+1} \forall n \in \mathbb{N}$
• $a_{n} < A \forall n \in \mathbb{N}$

Then, $\exists a \in \mathbb{R} : a_{n} \to n \text{ as } n \to \infty$.

Proof · Theorem 3.15; direct proof: Exercise 3.16#

See Theorem 3.15 and Exercise 3.16 of the book.

Proof · Exercise 1.54#

See Exercise 1.54 of the book.

Proof · Theorem 1.53#

See Theorem 1.53 of the book.

Theorem · 1.35 Intermediate Value Theorem#

If,

• $f: [a,b] \to \mathbb{R}$ is continuous
• $f(a) \geq 0 \geq f(b)$

Then, $\exists c \in [a,b] : f(c) = 0$

Proof · Theorem 1.35#

See Theorem 1.35 of the book.

Proof · Theorem 3.12#

See Theorem 3.12 of the book.

Theorem · 1.42 The mean value inequality#

Let,

• $U=\left( \alpha, \beta \right)$ be an open interval on $\mathbb{R}$.
• $a,b \in U$ and $a<b$
• $f: U \to \mathbb{R}$ be differentiable
• $f'(t) \leq K \ \forall t \in U$

Then, $f(b) - f(a) \leq K (b-a)$

Proof · Lemma 1.45; Theorem 1.42#

See Lemma 1.45 and Theorem 1.42 of the book.

Proof · Lemma 3.14; Theorem 1.42#

See Lemma 3.14 and Theorem 1.42 of the book.

Definition · Supremum#

Let $A \subseteq \mathbb{R}$ be non empty.

$\alpha$ is the least upper bound (supremum) of A if

1. $\alpha \geq a \ \forall a \in A$.
2. $\beta \geq a \ \forall a \in A \implies \beta \geq \alpha$

Theorem 3.7 Supremum Principle#

Let,

• $A \subseteq \mathbb{R}$ be non empty.
• $A$ is bounded above, $\exists K \in \mathbb{R} : a \leq K\ \forall a \in A$

Then, $A$ has a supremum.

Proof · Theorem 3.7; Exercises 3.9 or 3.10#

See Theorem 3.7 and Exercises 3.9 or 3.10 of the book. Exercise 3.9 is worked in the chapter 3 notes.