# Understanding the differences between $\omega + 1$ and $1 + \omega$ using Kenneth Kunen's definition of Ordinal Addition

In Kenneth Kunen's The Foundations of Mathematics, the following statement is proven:

If $$R$$ well-orders $$A$$, then there is a unique $$\alpha \in ON$$ such that $$(A;R) \cong (\alpha;\in)$$.

In case there is any confusion (because I've seen that sometimes there are slight differences in convention), Kunen establishes the following definitions:

• $$R$$ well-orders $$A$$ iff $$R$$ totally orders $$A$$ strictly and $$R$$ is well-founded on $$A$$

• $$R$$ is well-founded on $$A$$ iff for all non-empty sets $$X \subseteq A$$, there is a $$y \in X$$ that is $$R$$-minimal in $$X$$

• $$y \in X$$ is $$R$$-minimal in $$X$$ iff $$\neg \exists z (z \in X \land zRy)$$. Alternatively, $$y \in X$$ is $$R$$-maximal in $$X$$ iff $$\neg \exists z (z \in X \land yRz))$$

• $$R$$ totally orders $$A$$ strictly iff $$R$$ is transitive and irreflexive on $$A$$ and satisfies trichotomy on $$A$$

• Informally we define $$ON = \{x: x \text{ is an ordinal} \}$$, and the author has proven that $$ON$$ is well-ordered by $$\in$$

• If $$<$$ and$$\prec$$ are relations, then their lexicographic product on $$S \times T$$ is the relation $$\triangleleft$$ on $$S \times T$$ defined by:

$$\langle s,t \rangle \triangleleft \langle s',t' \rangle \leftrightarrow [s < s' \lor [s=s' \land t \prec t']]$$

Finally:

• $$F$$ is an isomorphism from $$(A;<)$$ onto $$(B; \triangleleft)$$ iff $$F$$ is a bijection $$(F: A \to B$$) and $$\forall x,y \in A [ x < y \leftrightarrow F(x) \triangleleft F(y)]$$. Then, $$(A;<)$$ and $$(B;\triangleleft)$$ are isomorphic (in symbols, $$(A;<) \cong (B;\triangleleft)$$ ) iff there exists an isomorphism from $$(A;<)$$ onto $$(B;\triangleleft)$$.

After proving the above statement, Kunen provides the following two definitions (which are what my question relates to):

Definition 1: If $$R$$ well-orders $$A$$ then $$\text{type}(A;R)$$ is the unique $$\alpha \in ON$$ such that $$(A;R) \cong (\alpha;\in)$$. We also write $$\text{type}(A)$$ (when $$R$$ is clear from context). or $$\text{type}(R)$$ (when $$A$$ is clear from context).

Definition 2: $$\alpha \cdot \beta = \text{ type}(\beta \times \alpha)$$ and $$\alpha + \beta = \text{ type}(\{0\} \times \alpha \ \ \cup \ \ \{1\} \times \beta)$$.

Kunen adds: "In both cases, we're using lexicographic order to compare ordered pairs of ordinals"

Using Kunen's definitions of ordinal multiplication and addition, I want to make sure that I am correctly understanding the major features. The two examples I would like to consider are:

1. $$\omega + 1$$ versus $$1 + \omega$$

2. $$\omega \cdot 2$$ versus $$2 \cdot \omega$$ (Edit: Because this post is running long, I'll just focus on the addition)

Because of Kunen's definitions of ordinal arithmetic, a useful lemma is the following:

If $$<$$ and $$\prec$$ are well-orders of $$S,T$$, respectively, then their lexicographic product $$\triangleleft$$ on $$S \times T$$ is a well-order of $$S \times T$$.

$$\color{blue}{\text{ Also, although I only did an informal proof, I am fairly certain that } \{0\} \times \alpha \ \ \cup \ \ \{1\} \times \beta}$$ $$\color{blue}{\text{is well-ordered, too}}.$$

Consider $$\omega + 1$$

$$\omega$$ is an ordinal and $$1$$ is an ordinal, which we will denote as $$\{0\}$$. Presumably, the implicit relation associated with these sets is $$\in$$. Thus:

$$\omega + \{0\} = \text{type}(\{0\} \times \omega \ \cup \ \{1\} \times \{0\})=\text{type}(\{ \langle 0,0 \rangle, \langle 0,1 \rangle, \langle 0,2 \rangle,...,\langle 1,0 \rangle\})$$

Importantly, note that $$\langle 1,0 \rangle$$ is a maximal element. Let $$A$$ be the set $$\{ \langle 0,0 \rangle, \langle 0,1 \rangle, \langle 0,2 \rangle,...,\langle 1,0 \rangle\}$$.

By the initial statement this post began with: there is a unique $$\alpha \in ON$$ such that $$(A;\in) \cong (\alpha;\in)$$

Consider $$1 + \omega$$

$$\{0\} + \omega = \text{type}(\{0\} \times \{0\} \ \cup \ \{1\} \times \omega)=\text{type}(\{ \langle 0,0 \rangle, \langle 1,0 \rangle, \langle 1,1 \rangle \langle 1,2 \rangle, ...\})$$

Importantly, note that there is no maximal element. Let $$A'$$ be the set $$\{ \langle 0,0 \rangle, \langle 1,0 \rangle, \langle 1,1 \rangle \langle 1,2 \rangle, ...\}$$

By the initial statement this post began with: there is a unique $$\alpha' \in ON$$ such that $$(A';\in) \cong (\alpha';\in)$$

Now, the trouble I am having in both cases is determining which function I should use to describe the isomorphisms...and then, using these functions, demonstrate that $$\alpha = S(\omega)$$ and $$\alpha'=\omega$$

For $$\alpha \in ON$$ such that $$(A;\in) \cong (\alpha;\in)$$, a good candidate seems to be something along the lines of:

$$f: \begin{cases} \langle 0,n \rangle \mapsto n \\ \langle 1, 0 \rangle \mapsto \omega \end{cases}$$

However, the fact that I am mapping $$\langle 1, 0 \rangle$$ to $$\omega$$ seems arbitrary. Couldn't I have mapped it to any ordinal that lays "above" the natural numbers?

For $$\alpha' \in ON$$ such that $$(A';\in) \cong (\alpha';\in)$$, a good candidate seems to be:

$$f': \begin{cases} \langle 0,0 \rangle \mapsto 0 \\ \langle 1, n \rangle \mapsto n+1 \end{cases}$$

Once again, however, I'm not quite sure I understand why my mappings are justified. I feel like I am creating these functions rather blindly, and without the proper motivation/justification.

I apologize if this question does not make much sense (still getting used to working with ordinals).

• In the case of your bijection for $\omega+1$, if you map that last element to something bigger than $\omega$, then you haven’t given a bijection to an ordinal, which is what you need. Dec 15, 2020 at 13:26

Ordinal addition is "stacking up orders". So $$\alpha+\beta$$ means that we put a copy of $$\beta$$ on top of a copy of $$\alpha$$. Now, what does it mean for $$1$$ and $$\omega$$? Well, putting a copy of $$\omega$$ on top of $$1$$ is just $$\{-1\}\cup\Bbb N$$, really. And quite easily, this is again isomorphic to $$\Bbb N$$ by the obvious function. On the other hand, $$\omega+1$$ means adding a copy of $$1$$, i.e. a point, on top of $$\omega$$. So $$\omega+1$$ is $$\Bbb N$$ with a "cap", a maximum. Clearly, not the same order as $$\omega$$.
Since you bring up order multiplication, the idea is that $$\alpha\cdot\beta$$ is replacing each point in $$\beta$$ with a copy of $$\alpha$$. So $$\omega\cdot 2$$ means consider $$\{0,1\}$$ and then replacing each of those with a copy of $$\omega$$: so we really get $$\omega+\omega$$ (in fact, the very same set, by definition!). On the other hand, $$2\cdot\omega$$ means replacing each natural number with a copy of $$\{0,1\}$$. But this is the same as considering $$\{0,1\}$$ and then $$\{2,3\}$$, and then $$\{4,5\}$$, etc. So $$\{2n,2n+1\}$$ is the $$n$$th copy. Easily, again, this is the same as $$\omega$$.
• Draw it on a piece of paper. Put two forks and then three knives (left-to-right, in this case). Play with it. Things can be simple, too. And yes, a point is exactly a single element, or the ordinal $1$. Dec 16, 2020 at 2:05