This talk on Ordinals was held on Friday February 10, 2017 in MC 4045. The talk was given by Eddie Onochie.

## Abstract

In this talk we will talk about ordinals and the philosophy of infinity. We will define what ordinals are and how to construct them. We will also define transfinite recursion and use the axiom of choice to give meaning to the "cardinality" of a set.

## Summary

Notes by Akshay Tiwary.

We use natural numbers for two things. For ordering things and for measuring the size of a set. Ordinals generalize natural numbers but also work for infinite sets.

Let's talk about well-ordered sets first.

**Definition**. Let $<$ be a binary relation on set $A$. We say that $(A, <)$ is a strict total-ordering if the following two conditions hold.

For every $x,y \in A$ we have exactly one of $x < y$ or $x = y$ or $y < x$.

For all $x,y,z \in A$ we have that if $x < y$ and $y < x$ then $x < z$.

A totally ordered set $(A, <)$ is well ordered if every non-empty subset of $A$ has a smallest element with respect to $<$. \end{defn}

We define the natural numbers as follows. Let $0 := \varnothing$, and ``n := (n -

\cup {n - 1}`` for non-zero natural numbers. But this is circular because

we don't have induction yet. So for this, we need the axiom of induction. This says the following.

**Axiom of Infinity** There is a set $I$ such that $\varnothing \in I$ and for every $x \in I$ we have $S(x) \in I$ where $S(x) = x \cup \{x\}$.

Define $\omega$ to be the smallest subset of $I$ that satisfies this property (call any set that satisfies this property *inductive*). Formally this is the following:

$\omega$ is the formal definition of the natural numbers.

Here is a theorem that is our first result.

**Theorem**. $(\omega, \in)$ is a strict well order and for every $n \in \omega$, $(n, \in)$ is also a strict well order.

Here is a lemma that we will need.

**Lemma**. Let $n \in \omega$. Then either $0 = n$ or $0 \in n$.

Proof: Observe that $J = \{0\} \cup \{n \in \omega : 0 in n\} \subseteq \omega$ and is inductive.

Now let's prove the first part of our theorem.

Suppose that $X \subseteq \omega$ has no least element and $X \subseteq \omega$. Let

Since $X$ has no least element we cannot have $0 \in X$. Hence $S(0) \cap X = \varnothing$ and this means that $0 \in J$. Now suppose that $n \in J$ which means $S(n) \cap X = \varnothing$. Now $S(n) \not in X$ because it would be least. Hence $S(S(n)) \cap X = \varnothing$ so $S(n) \in J$.

Now here's an important theorem we will not prove.

**Proposition**. Every element of $\omega$ is a subset of $\omega$.

Using this we can finally define an ordinal.

An **ordinal** is a set satisfying the following two properties.

$(\alpha, \in)$ is a strict well-ordering

Every element of $\alpha$ is a subset of $\alpha$.

Observe that subsets of ordinals are not always ordinals as $\{\{0\}\} \subseteq 2$ but $\{\{0\}\} \not\in 2$. However we have the following result.

**Proposition**. If $\alpha, \beta$ are ordinals with $\alpha \neq \beta$ and $\alpha \subseteq \beta$ then $\alpha \in \beta$.

We have a very cool result for ordinals that we will not prove here.

**Proposition**. Let $\alpha, \beta$ be ordinals. Then either $\alpha \in \beta$, $\alpha = \beta$ or $\beta \in \alpha$.

Let $E$ be a set of ordinals. Let $X = \bigcup E$. Then $X$ is an ordinal. [Exercise] You can check that $\sup E$ is the least upper bound of $E$ where $<$ is membership.

We can state the principle of transfinite induction as follows.

**Proposition: Transfinite Induction**. Suppose that $P(x)$ is a well-formed statement about an ordinal $x$. Suppose

P(0) is true.

If $P(\alpha)$ is true for some ordinal $\alpha$ then $P(S(\alpha))$ is true.

If $\alpha$ is a limit ordinal and $\alpha > 0$ then $P(\beta)$ is true for all $\beta < \alpha$ then $P(\alpha)$ is true.

Since we don't have time we will explain transfinite recursion by using it to define ordinal addition.

**Definition**. Let $\beta$ be an ordinal. First define $\beta + 0 = \beta$. Then for an ordinal $\alpha$ define $\beta + S(\alpha) = S(\beta + \alpha)$. Lastly if $\alpha$ is a limit ordinal (an ordinal that is not a successor ordinal i.e not of the form $S(x)$ for any ordinal $x$) then define

Here is a huge theorem.

**Proposition**. Let $(A, <)$ be any strict well ordering. Then there is a unique ordinal $\alpha$ such that $(A, <)$ is isomorphic to $(\alpha, \in)$ as an ordering (please google/ask what an isomorphism of orderings is).

Hence if we have the axiom of choice, then every set is well-orderable. Hence we can define the notion of a *cardinal* for any set.

**Definition**. An ordinal $\alpha$ is a cardinal if there is no ordinal $\beta < \alpha$ such that there is a bijection $f : \beta \to \alpha$.

So the axiom of choice guarantees that there is a cardinal for every set.

Okay that's enough math. Let's talk about the Philosophy of infinity.

A finitist will deny infinite mathematics. But an ultrafinitist denies the mathematics of very large numbers. Read Harvey M Friedman's account on meeting an ultrafinitist "Philosophical Problems in Logic".

I have seen some ultrafinitists go so far as to challenge the existence of

$2^{100}$ as a natural number, in the sense of there being a series of “points” of that length. There is the obvious “draw the line” objection, asking where in $2^1$, $2^2$, $2^3$, ... , $2^{100}$ do we stop having “Platonistic reality”? Here this … is totally innocent, in that it can be easily be replaced by $100$ items (names) separated by commas. I raised just this objection with the (extreme) ultrafinitist Yessenin-Volpin during a lecture of his. He asked me to be more specific. I then proceeded to start with $2^1$ and asked him whether this is “real” or something to that effect. He virtually immediately said yes. Then I asked about $2^2$, and he again said yes, but with a perceptible delay. Then $2^3$, and yes, but with more delay. This continued for a couple of more times, till it was obvious how he was handling this objection. Sure, he was prepared to always answer yes, but he was going to take $2^{100}$ times as long to answer yes to $2^{100}$ then he would to answering $2^1$. There is no way that I could get very far with this.