# Real Analysis

Ordinals

# Ordinal Numbers

Suppose there are sets \(X\), \(Y\), and orders on them, namely \((X, <)\) and \((Y, <)\). Say they have the same *order-type* if there exists a bijection \(f : X \rightarrow Y\) such that \(x < y\) if and only if \(f(x) < f(y)\). \(f\) is called an *order-isomorphism*.

Recall \(X\) is *well-ordered* if every non-empty subset of \(X\) has a least element.

Construct \(\emptyset\), \(\{\emptyset\}\), \(\{\emptyset, \{\emptyset\}\}\), \(\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}\), etc. Ordering on them is defined as set containment.

An *ordinal* is a set that is

- transitive/complete (every member is a subset), and
- strictly well-ordered by membership.

## Observations

- If \(\alpha\) is an ordinal then \(S(\alpha) = \alpha \cup \{\alpha\}\) is an ordinal, called the
*successor ordinal*. - If \(\alpha\) is an ordinal and \(\beta \in \alpha\) then \(\beta\) is an ordinal.
- If \(A\) is a set of ordinals then \(\sup A = \bigcup A\) is an ordinal.

## Theorems

**Theorem.** Any well-ordered set is order-isomorphic to some ordinal \(\alpha\).

\(\omega\), called the *limit ordinal* is the first infinite ordinal. It is not the successor of any ordinal (i.e. \(\omega \neq S(\alpha)\) for any ordinal \(\alpha\)). \(\epsilon_0\) is the first ordinal such that \(\epsilon_0 = \omega^{\epsilon_0}\). \(\omega_1\) is the first uncountable ordinal.

\[ 0, 1, 2, 3, \dots, \omega, \omega + 1, \omega + 2, \dots, \\ \underbrace{\omega + \omega}_{\omega \cdot 2}, \omega \cdot 2 + 1, \dots, \omega \cdot 3, \dots, \\ \underbrace{\omega\omega}_{\omega^2}, \omega^2 + 1, \dots, \omega^3, \dots, \\ \omega^\omega, \omega^\omega + 1, \dots, \omega^{\omega^\omega}, \dots, \\ \epsilon_0, \epsilon_0 + 1, \dots, \epsilon_1, \dots, \\ \omega_1, \omega_1 + 1 \]

# Transfinite Induction

Recall strong induction: Let \(S_n = \{i \in \mathbb{N} : i < n\}\), called a *section*. \(A \subset \mathcal{N}\) is *inductive* if for all \(n \in \mathcal{N}\), \(S_n \subset A\) implies \(n \in A\).

Recall the principle of strong induction: If \(A \subset \mathbb{N}\) is inductive then \(A = \mathcal{N}\).

*Proof (by contradiction).* If \(A \neq \mathbb{N}\) then \(\mathbb{N} \setminus A\) has a smallest element \(n\). But then \(S_n \subset A\) which implies \(n \in A\), a contradiction. QED.

**Theorem.** Every set can be well-ordered.

The proof depends on the *axiom of choice*.

Well-order the index set of statements \(J\). Let \(S_\alpha = \{\gamma \in J : \gamma < \alpha\}\), called a section. A set \(A \subset J\) is *inductive* if for all \(\alpha \in J\), \(S_\alpha \subset A\) implies \(\alpha \in A\).

The principle of transfinite induction: Suppose \(J\) is well-ordered. A set \(A \subset J\) is inductive implies \(A = J\).

## Applications

**Theorem.** There exists a set \(K\) in \(\mathbb{R}^2\) that intersect every line in the plane twice.

*Proof (depends on the axiom of choice).* Let \(L\) be the set of all lines in \(\mathbb{R}^2\) (so \(|L| = |\mathbb{R}|\)). Well-order \(L\) with the type \(J\) of the first ordinal with the same cardinality as \(\mathbb{R}\) (all elements of \(J\) have cardinality less than that of \(\mathbb{R}\)). Write \(L = \{L_\alpha\}_{\alpha \in J}\). Let \[\begin{align}
A = \{\alpha \in J : \exists \text{ set } K_\alpha \text{ such that } & (1) |K_\alpha| < |\mathbb{R}| \text{ and}\\
& (2) \text{ no three points are colinear} \text{ and} \\
& (3) |K_\alpha \cap L_\beta| = 2 \text{ if } \alpha \leq \beta \text{ and} \\
& (4) K_\beta \subset K_\alpha \text{ if } \beta < \alpha\}
\end{align}\] We need to show that \(A\) is inductive, so \(A = J\).

Base case: Clearly, \(1 \in A\) (let \(K_1\) be the set of two points on \(L_1\)).

Inductive step: If \(S_\alpha \subset A\), let \(K = \bigcup\limits_{\beta < \alpha} K_\beta\). \(K\) has the same cardinality as \(\mathbb{R}\) (by (1)) and has no three point colinear (by (2) and (3)). The set of all lines through \(K_\alpha\) have cardinality less than that of \(\mathbb{R}\) so it can’t hit all of \(L_\alpha\). Pick 1 or 2 points to form \(K_\alpha = K \cup \{\text{1 or 2 extra points}\}\). So \(\alpha \in A\). QED.

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