# 3 Pure

Number theory and Analysis, topics apparently elevated from immediate application. Pure mathematics. Rigour, maturity and the small print (*well behaved*, *subject to conditions*) not urgently required for basic understanding.

Principle reference texts:

A Pythagorean Introduction to Number Theory* Takloo-Bighash (2018)

interestingly motivated

Cambridge (2022/23) courses covered:

- analysis I

## 3.1 Real Analysis

Following Tao.

### 3.1.1 Peano Axioms

Defining the set of natural numbers \(\mathbb{N}:={0,1,2,\dots}\).

- \(0\) is a natural number.
- if \(n\) is a natural number, \(n++\) is also a natural number (closure under ++)

- can assign the labels as define \(1:=0++\), \(2:=1++\), …

- \(n++\neq 0\) for any \(n \in \mathbb{N}\).
- if \(n++=m++\) then \(n=m\) (different natural numbers have different successors).
- (induction) Let \(P(n)\) be any property pertaining to a natural number \(n\). Suppose that \(P(0)\) is true, and suppose that whenever \(P(n)\) is true, \(P(n++)\) is also true. Then \(P(n)\) is true for every natural number \(n\).

Can define addition and multiplication recursively:

- define \(0+m:=m\). then define \((n++)+m:=(n+m)++\)
- define \(1 \times m := m\) then define \((n++)\times m := (n \times m) + m\)

As a corollary of addition we have cancellation \(a+b=a+c \implies b=c\), prove by induction, base case \(a=0\).

**strong induction**: Let \(P(n)\) be any property pertaining to a natural number \(n\). Suppose that whenever \(P(a_0), P(a_0++), \dots P(n)\) are true, \(P(n++)\) is also true. Then \(P(n)\) is true for every natural number \(n>a_0\)

**example** If all \(c\) cities are each connected by a one-way road, show that there is a route through all cities.

Take a city \(c_0\). Divide other cities into set \(A\): cities to \(c_0\) and set \(B\): cities from \(c_0\). Then there is a route through \(A\) (strong induction), and a route through \(B\). Since all cities are connected, the last city of the \(A\) route connects to \(c_0\) which also connects to the first city of the \(B\) route by construction. \(\qquad\blacksquare\)

### 3.1.2 Sets

Modern analysis, like most of modern mathematics, is concerned with numbers, sets, and geometry (Tao)

Set axioms (includes redundancy):

We can define equality of sets: \(A=B\) iff \(x \in A \implies x \in B\) and \(y \in B \implies y \in A\).

Then we have (includes redundancy) a set of set axioms:

- If \(A\) is a set, then \(A\) is also an object. It is meaningful to ask if set \(A\) \(\in\) set \(B\).
- (empty set). \(\exists\) a set \(\varnothing\), such that \(\forall x, x \notin \varnothing\).

- single-choice: we can choose an element \(x\) from a non-empty set \(A\)
- finite-choice: we can choose one element \(x_1,\dots,x_n\) from each of non-empty sets \(A_1, \dots , A_n\)
- axiom of choice: choose from infinitely many sets \(A_1,A_2, \dots\)

- (Singleton sets and pair sets). If \(a\) is an object, then there exists a set \({a}\) whose only element is \(a\), i.e., for every object \(y\), \(y \in {a}\) if and only if \(y = a\). Furthermore, if \(a\) and \(b\) are objects, then there exists a set \({a, b}\) whose only elements are \(a\) and \(b\).
- (Pairwise union). Given any two sets \(A\), \(B\), there exists a set \(A \cup B\) such that \(x \in A \cup B \iff (x \in A \, \textrm{or} \,x\in B)\)

Note associative and commutative, \(A \cup (B \cup C)=(A \cup B) \cup C\). - (Axiom of specification). Let \(A\) be a set, and for each \(x \in A\), let \(P(x)\) be a property pertaining to \(x\). Then there exists a set, \(\{x \in A : P(x)\,\, \textrm{is true}\,\,\}\) (or simply \(\{x ∈ A : P(x)\}\), i.e. \(y \in \{ x \in A:P(x) \,\,\textrm{is true}\,\,\} \iff (y \in A \,\,\textrm{and} \,\, P(y) \,\,\textrm{is true}\,\,)\).

Can now define:

- define subset: \(A \subset B := x \in A \implies x \in B\). \(A\) is a proper subset if \(A \neq B\).
- define intersection: \(A \cap B := \{x : x\in A\,\, \textrm{and}\,\, x \in B\}\)
- define difference: \(A - B\) or \(A \backslash B := \{x \in A \,\, \textrm{and}\,\,x \notin B\}\)

With Intersection and Union, sets form **a Boolean algebra**. Let \(A,B,C\) be subsets of set \(X\):

- (minimal element) \(A \cup \varnothing = A\) and \(A \cap \varnothing = \varnothing\)
- (maximal element) \(A \cup X = X\) and \(A \cap X = A\)
- (identity) \(A \cup A = A\) and \(A \cap A = A\)
- (associativity) \(A \cup (B \cup C)=(A \cup B) \cup C\) and \(A \cap (B \cap C)=(A \cap B) \cap C\)
- (commutativity) \(A \cup B = B \cup A\) and \(A \cap B = B \cap A\)
- (distributivity) \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\) and \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\)
- (partition) \(A \cup X \backslash A = X\) and \(A \cap X \backslash A = \varnothing\)
- (De Morgan Laws) \(X\backslash (A \cup B) = (X \backslash A) \cap (X \backslash B)\) and \(X\backslash (A \cap B) = (X \backslash A) \cup (X \backslash B)\)