MT2025 Analysis I lecture notes

Remaining TODOs: 1

1. The real numbers

1.1. Field axioms

Definition 1.1.1: A field $(𝐹, +, \times)$ is a set $𝐹$ with two binary operations “sum” $+ : 𝐹^{2} \to 𝐹$ and “product” $\times : 𝐹^{2} \to 𝐹$ , such that the 10 field axioms A1-A4, M1-M4, D, Z are satisified.

A1-A4 govern addition:

A1: $\forall 𝑎, 𝑏 \in 𝐹, 𝑎 + 𝑏 = 𝑏 + 𝑎$ (commutativity)
A2: $\forall 𝑎, 𝑏, 𝑐 \in 𝐹, (𝑎 + 𝑏) + 𝑐 = 𝑎 + (𝑏 + 𝑐)$ (associativity)
A3: $\exists 0 \in 𝐹$ s.t. $𝑎 + 0 = 𝑎 \forall 𝑎 \in 𝐹$ (additive identity)
A4: $\forall 𝑎 \in 𝐹, \exists - 𝑎 \in 𝐹$ s.t. $𝑎 + (- 𝑎) = 0$ (additive inverse)

M1-M4 deal with multiplication:

M1: $\forall 𝑎, 𝑏 \in 𝐹, 𝑎 \times 𝑏 = 𝑏 \times 𝑎$ (commutativity)
M2: $\forall 𝑎, 𝑏, 𝑐 \in 𝐹, (𝑎 \times 𝑏) \times 𝑐 = 𝑎 \times (𝑏 \times 𝑐)$ (associativity)
M3: $\exists 1 \in 𝐹$ s.t. $\forall 𝑥 \in 𝐹, 𝑥 \times 1 = 𝑥$ (multiplicative identity)
M4: $\forall 𝑥 \in 𝐹 \ {0}, \exists 𝑥^{- 1} \in 𝐹$ s.t. $𝑥 \times 𝑥^{- 1} = 1$ . (multiplicative inverse)
D: $\forall 𝑎, 𝑏, 𝑐 \in 𝐹, 𝑎 \times (𝑏 + 𝑐) = (𝑎 \times 𝑏) + (𝑎 \times 𝑐)$ (distributivity)
Z: $0 \neq 1$

Remark: Without axiom Z,

{0}

would be a field, which is undesirable.

Proposition 1.1.2: If $𝑎 + 𝑥 = 𝑎 \forall 𝑎 \in ℝ$ , then $𝑥 = 0$ , i.e. $0$ is the unique additive identity.

Proof: As $𝑎 + 𝑥 = 𝑎$ , then

\begin{aligned} (- 𝑎) + 𝑎 + 𝑥 & = (- 𝑎) + 𝑎 \\ = 0 (by A1 and A4) \end{aligned}

By A2,

\begin{aligned} LHS & = ((- 𝑎) + 𝑎 + 𝑥) \\ = 0 + 𝑥 (by A1 and A4) \\ = 𝑥 (by A1 and A3) . \end{aligned}

Hence

𝑥 = 0

Remark: Associativity means that unbracketed finite sums and products are unambiguous.

This isn’t true for infinite sums and products in general.

Proposition 1.1.3: $𝑎 \times 0 = 0$ for all $𝑎 \in ℝ$ .

Proof:

\begin{aligned} 𝑎 \times 0 & = 𝑎 \times (0 + 0) \\ = (𝑎 \times 0) + (𝑎 \times 0) \end{aligned}

so $𝑎 \times 0 = 0$ by additive inverse.

Proposition 1.1.4: If $𝑎, 𝑏 \neq 0$ then $𝑎 \times 𝑏 \neq 0$ .

Proof: By contraposition.

Suppose $𝑎 \times 𝑏 = 0$ and $𝑎 \neq 0$ .

Then

𝑎^{- 1} \times (𝑎 \times 𝑏) = 𝑎^{- 1} \times 0 = 0 .

By M2, M1, M4, LHS $= 1 \times 𝑏 = 0$ , so by M3, $𝑏 = 0$ .

Remark:We write

𝑎 𝑏

to mean

𝑎 \times 𝑏

Remark:We write

𝑎 - 𝑏

to mean

𝑎 + (- 𝑏)

, and

𝑎 / 𝑏 = 𝑎 𝑏^{- 1}

Remark:For

𝑎 \in 𝑅

and positive integer

𝑘

, we define

𝑎^{1} = 𝑎

, and

𝑎^{𝑘 + 1} = 𝑎^{𝑘} 𝑎

Remark: For

𝑎 \neq 0

, we define

𝑎^{0} = 1

Remark: For

𝑙 > 0

and

𝑎 \neq 0

, we define

𝑎^{- 𝑙} = {(𝑎^{𝑙})}^{- 1}

1.2. Order axioms

Definition 1.2.1: The order axioms say that there is a subset $ℙ \subseteq ℝ$ such that:

P1: If $𝑎 \in ℙ, 𝑏 \in ℙ$ then $𝑎 + 𝑏 \in ℙ$
P2: If $𝑎 \in ℙ, 𝑏 \in ℙ$ then $𝑎 𝑏 \in ℝ$
P3: For all $𝑎 \in ℝ$ , precisely one of $𝑎 \in ℙ, 𝑎 = 0, - 𝑎 \in ℙ$ holds (trichotomy)

We say “ $𝑎$ is positive” when $𝑎 \in ℙ$ .

As an interval, $ℙ = (0, \infty)$ .

Definition 1.2.2: For

𝑎, 𝑏 \in ℝ

, we write

𝑎 \leq 𝑏 \Leftrightarrow 𝑏 - 𝑎 \in ℙ \cup {0}

and

𝑎 < 𝑏 \Leftrightarrow 𝑏 - 𝑎 \in ℙ

Proposition 1.2.3: $\leq$ is a total order on $ℝ$ .

Proof:

$\leq$ is reflexive: $𝑎 - 𝑎 = 0 \in ℙ \cup {0}$ , so $𝑎 \leq 𝑎$ for all $𝑎 \in ℝ$ .

$\leq$ is antisymmetric: suppose $𝑎 \leq 𝑏$ and $𝑏 \leq 𝑎$ . Then $𝑏 - 𝑎 \in ℙ \cup {0}$ and $- (𝑏 - 𝑎) = 𝑎 - 𝑏 \in ℙ \cup {0}$ . These contradict P3 unless $𝑏 - 𝑎 = 0 ⟹ 𝑎 = 𝑏$ .

$\leq$ is transitive: suppose $𝑎 \leq 𝑏$ and $𝑏 \leq 𝑐$ . Then $𝑏 - 𝑎, 𝑐 - 𝑏 \in ℙ \cup {0}$ . If a pair of $𝑎, 𝑏, 𝑐$ are equal, clearly $𝑎 \leq 𝑐$ . Otherwise, $𝑏 - 𝑎, 𝑐 - 𝑏 \in ℙ$ so by P1, $𝑐 - 𝑎 = (𝑐 - 𝑏) + (𝑏 - 𝑎) \in ℙ$ so $𝑎 \leq 𝑐$ .

Every pair $𝑎, 𝑏 \in ℝ$ is comparable:

If $𝑎 = 𝑏$ , done
If $𝑎 < 𝑏$ , done
By trichotomy, the only other option is $𝑎 - 𝑏 = - (𝑏 - 𝑎) \in ℙ$ , so $𝑏 < 𝑎$ (so $𝑏 \leq 𝑎$ ).

Proposition 1.2.4: Let $𝑎, 𝑏, 𝑐 \in ℝ$ .

If $𝑎 \leq 𝑏$ then $𝑎 + 𝑐 \leq 𝑏 + 𝑐$
If $𝑎 \leq 𝑏$ and $𝑐 \geq 0$ then $𝑎 𝑐 \leq 𝑏 𝑐$

Proof (i):

(𝑏 + 𝑐) - (𝑎 + 𝑐) = 𝑏 - 𝑎 \in ℙ \cup {0}

by assumption.

Proof (ii):

$𝑏 𝑐 - 𝑐 𝑎 = 𝑐 (𝑏 - 𝑎)$ .

If $𝑐 = 0$ or $𝑏 = 𝑎$ , then $𝑐 (𝑏 - 𝑎) \in ℙ \cup {0}$ .

If $𝑐 > 0$ and $𝑏 > 𝑎$ , then $𝑐 (𝑏 - 𝑎) \in ℙ$ by P2.

Proposition 1.2.5: $ℂ$ is not an ordered field, i.e. there is no $ℙ \subseteq ℂ$ satisfying axioms P1-3.

Proof: Suppose for the sake of contradiction that there is such a $ℙ$ .

By P3, one of $𝑖 \in ℙ, 𝑖 = 0, - 𝑖 \in ℙ$ .

If $𝑖 = 0$ , then $1 = 𝑖^{4} = 0^{4} = 0$ , contradicting axiom Z.

If $𝑖 \in ℙ$ , then by P2, $- 𝑖 = 𝑖^{3} \in ℙ$ . This contradicts P3.

If $- 𝑖 \in ℙ$ , we get the same contradiction.

Therefore $ℂ$ is not an ordered field.

Definition 1.2.6: We define the maximum function to be

max : ℝ^{2} \to ℝ; max (𝑎, 𝑏) = {\begin{cases} 𝑏 & if 𝑏 \leq 𝑎 \\ 𝑎 & if 𝑎 < 𝑏 . \end{cases}

This is well-defined by the trichotomy axiom P3.

We define the minimum similarly:

min : ℝ^{2} \to ℝ = {\begin{cases} 𝑎 & if 𝑎 \leq 𝑏 \\ 𝑏 & if 𝑏 < 𝑎 . \end{cases}

We define the modulus function to be

| \cdot | : ℝ \to ℝ; | 𝑥 | = {\begin{cases} 𝑥 & if 𝑥 \geq 0 \\ - 𝑥 & if 𝑥 < 0 . \end{cases}

Lemma 1.2.7: The function $𝑥 \mapsto 𝑥^{2}$ is increasing on $[0, \infty)$ .

Proof: For

𝑥, 𝑦 \in [0, \infty)

s.t.

𝑥 \leq 𝑦

𝑦^{2} - 𝑥^{2} = (𝑦 - 𝑥) (𝑦 + 𝑥) \in [0, \infty]

by P2. This satisfies the definition of

𝑥^{2} \leq 𝑦^{2}

, so

𝑥 \mapsto 𝑥^{2}

is increasing on this interval.

Proposition 1.2.8 (triangle inequality): Given $𝑎, 𝑏 \in ℝ$ , $| 𝑎 + 𝑏 | \leq | 𝑎 | + | 𝑏 |$ .

Proof:

\begin{aligned} {(| 𝑎 + 𝑏 |)}^{2} & = {(𝑎 + 𝑏)}^{2} \\ = 𝑎^{2} + 2 𝑎 𝑏 + 𝑏^{2} \\ = {| 𝑎 |}^{2} + 2 𝑎 𝑏 + {| 𝑏 |}^{2} \\ \leq {| 𝑎 |}^{2} + 2 | 𝑎 | | 𝑏 | + {| 𝑏 |}^{2} \\ = {(| 𝑎 | + | 𝑏 |)}^{2} . \end{aligned}

By Lemma 1.2.7, $| 𝑎 + 𝑏 | \leq | 𝑎 | + | 𝑏 |$ .

Theorem 1.2.9 (Bernoulli's inequality): Let $𝑛 \in ℕ$ and $𝑥 > - 1$ , then ${(1 + 𝑥)}^{𝑛} \geq 1 + 𝑛 𝑥$ .

Proof: By induction.

When $𝑛 = 0$ , LHS $= 1 =$ RHS.

Suppose true for $𝑛$ . Then

\begin{aligned} {(1 + 𝑥)}^{𝑛 + 1} & = (1 + 𝑥) {(1 + 𝑥)}^{𝑛} \\ \geq (1 + 𝑥) (1 + 𝑛 𝑥) by ih \\ = 1 + (𝑛 + 1) 𝑥 + 𝑛 𝑥^{2} \\ \geq 1 + (𝑛 + 1) 𝑥 . \end{aligned}

Hence true by induction.

1.3. The completeness axiom

Remark: Many fields also satisfy the order axioms (e.g.

ℚ, ℝ

). One way that

ℚ

is different from

ℝ

is that the rationals have some “gaps”, e.g.

∄ 𝑞 \in ℚ

s.t.

𝑞^{2} = 2

Definition 1.3.1: Let $𝑆 \subseteq ℝ$ .

$𝛼 ≔ max 𝑆$ is the maximum of $𝑆$ if $𝛼 \in 𝑆$ and $𝛼 \geq 𝑠 \forall 𝑠 \in 𝑆$ . The minimum is similarly defined.

$𝑢$ is an upper bound for $𝑆$ if $𝑢 \geq 𝑠 \forall 𝑠 \in 𝑆$ ; a lower bound is defined similarly.

$𝑆$ is bounded if $𝑆$ has a lower and an upper bound.

Definition 1.3.2: The supremum of a set

𝑆

sup 𝑆

, is the least upper bound of

𝑆

Proposition 1.3.3: For $𝑆 \subseteq ℝ$ :

If $max 𝑆$ exists, it is unique.
If $sup 𝑆$ (“least upper bound”) exists, it is unique.
If $max 𝑆$ exists, $max 𝑆 = sup 𝑆$ .

Proof:

Suppose $𝑎, 𝑏$ are maxima of $𝑆$ . Then $𝑎 \geq 𝑏$ as $𝑎$ is a max and $𝑏 \in 𝑆$ . Also $𝑏 \geq 𝑎$ by the same argument. Hence $𝑎 = 𝑏$ .
A least upper bound is an upper bound which is less than or equal to any upper bound. For $𝑎, 𝑏$ two least upper bounds, $𝑎 \leq 𝑏$ and $𝑏 \leq 𝑎$ so $𝑎 = 𝑏$ .
$max 𝑆 \leq sup 𝑆$ because $max 𝑆 \in 𝑆$ . But $max 𝑆$ is an upper bound itself, so $sup 𝑆 \leq max 𝑆$ . Therefore $max 𝑆 = sup 𝑆$ .

Definition 1.3.4:The completeness axiom is:

C: Every nonempty bounded above $𝑆 \subseteq ℝ$ has a supremum.

Remark: To verify that $𝛼$ is the supremum of a set $𝑆$ involved showing that:

$𝛼 \geq 𝑠 \forall 𝑠 \in 𝑆$
if $𝛽 \geq 𝑠 \forall 𝑠 \in 𝑆$ then $𝛽 \geq 𝛼$

Alternatively, the second step can be replaced with showing that

\forall 𝜀 > 0, \exists 𝑠 \in 𝑆 s.t. 𝛼 - 𝜀 \leq 𝑠 \leq 𝛼 .

TODO approximation property

Remark: There are many equivalent statements of the completeness axioms.

Definition 1.3.5: The infimum of a set $𝑆$ is the greatest lower bound of $𝑆$ . We don’t need any addition axioms to show theexistance of an infimum equivalent to the supremum, because for $𝑆 \neq \emptyset$ we can define

inf 𝑆 ≔ - sup {- 𝑠 | 𝑠 \in 𝑆} .

Remark: The reals are essentially unique in satisfying the field, order and completeness axioms.

Theorem 1.3.6: There is a unique $𝛼 > 0$ s.t. $𝛼^{2} = 2$ .

Proof: Let $𝑆 = {𝑥 \in ℝ | 𝑥^{2} \leq 2}$ .

Note that $1 \in 𝑆$ and that 2 bounds $𝑆$ from above. We set $𝛼 = sup 𝑆$ , which exists by the completeness axiom.

By trichotomy, $𝛼^{2} = 2$ or $𝛼^{2} > 2$ or $𝛼^{2} < 2$ .

Suppose for the sake of contradiction that $𝛼^{2} > 2$ . Pick $0 < ℎ < \frac{𝛼^{2} - 2}{2 𝛼}$ , then ${(𝛼 - ℎ)}^{2} > 2$ . Then no element of $𝑆$ lies in $(𝛼 - ℎ, 𝛼)$ . This contradicts the approximation property.

Suppose for the sake of contradiction that $𝛼^{2} < 2$ . Pick $ℎ < 1$ (so that $ℎ^{2} < ℎ$ ), then

0 < ℎ < min {1, \frac{2 - 𝛼^{2}}{2 𝛼 + 1}},

giving ${(𝛼 + ℎ)}^{2} < 2$ . Therefore $𝛼 + ℎ \in 𝑆$ , but $𝛼 + ℎ > 𝛼 = sup 𝑆$ , contradiction.

By trichotomy, $𝛼^{2} = 2$ .

To show uniqueness, suppose that $𝛽 > 0$ and $𝛽^{2} = 2$ . Then $𝛼^{2} = 𝛽^{2} = 2$ and

0 = 𝛼^{2} - 𝛽^{2} = (𝛼 - 𝛽) (𝛼 + 𝛽) .

Since $𝛼 + 𝛽 > 0$ , it must be that $𝛼 - 𝛽 = 0$ so $𝛼 = 𝛽$ .

Remark: We can generalise these arguments to show the uniqueness of

𝑛

th roots for positive reals.