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:

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


  • $F$ 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).

  • 1
    $\begingroup$ 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. $\endgroup$ Dec 15, 2020 at 13:26

1 Answer 1


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 $\omega$.

  • $\begingroup$ I appreciate the attempt at "dumbing it down" for me to offer intuition, but I fear that I do not know enough about ordinals to fully appreciate/comprehend the analogy you provided. I don't really understand the notion of putting a copy of an ordinal "on top" of another ordinal. I cannot create an appropriate image in my head that is faithfully capturing your idea. Also, what exactly do you mean by "point"? A set with a singular element? $\endgroup$
    – S.Cramer
    Dec 16, 2020 at 2:04
  • $\begingroup$ 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 $1$. $\endgroup$
    – Asaf Karagila
    Dec 16, 2020 at 2:05

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