$$\def\R{\mathbb{R}} \def\N{\mathbb{N}} \def\Z{\mathbb{Z}} \def\Q{\mathbb{Q}} \def\C{\mathbb{C}} \def\1{\textbf{1}} \def\eps{\varepsilon} \def\epsilon{\varepsilon} \renewcommand{\geq}{\geqslant} \renewcommand{\leq}{\leqslant}$$

# Chapter 1 - Set Theory

## 1.3 Ordinal Sets and Ordinal Numbers

 Main: Index Previous: 1.2 Equivalence of Sets. The Power of a Set Next: 1.4 Systems of Sets

### Results

Isomorphism
Two ordered sets $M$ and $M'$ are said to be isomorphic if there exists an isomorphism between them, i.e a bijective function $f:M\rightarrow M'$ such that $a\leq b$ for $a,b\in M$ iff $f(a)\leq f(b)$ for $f(a), f(b)\in M'$.

Definition 1

An ordered set $M$ is said to be well-ordered if every nonempty subset $A$ of $M$ has a smallest (or 'first') element, i.e. an element $\mu$ such that $\mu < a$ for all $a\in A$.

Definition 2

The order type of a well-ordered set is called an ordinal number or simply an ordinal. If the set is infinite, the ordinal is said to be transfinite.

Theorem 1

The ordered sum of a finite number of well-ordered sets $M_1, M_2, \ldots, M_n$ is itself a well-ordered set.

Corollary

The sum of a finite number of ordinal numbes is itself an ordinal number.

Theorem 2

The ordered product of two well-ordered sets $M_1$ and $M_2$ is itself a well-ordered set.

Corollary 1

The ordered product of a finite number of well-ordered sets is itself a well-ordered set.

Corollary 2

The ordered product of a finite number of ordinal numbers is itself an ordinal number.

Theorem 3

Two given ordinal numbers $\alpha$ and $\beta$ satisfy one and only one of the relations $$\alpha < \beta,\qquad \alpha = \beta,\qquad \alpha > \beta.$$

Theorem 4

Let $A$ and $B$ be well-ordered sets. Then either $A$ is equivalent to $B$ or one of the sets is of greater power than the other, i.e. the powers of $A$ and $B$ are comparable.

Well-ordering Theorem

Every set can be well-ordered.

Axiom of choice

Given any set $M$, there is a "choice function" $f$ such that $f(A)$ is an element of $A$ for every nonempty subset $A\subset M$.

Note: Skipping these exercises for now.