I have recently started studying Set Theory in a self-thaught way, for that purpose I have been following Kunen's book: Set Theory: An Introduction to Independence Proofs. I'm in Chapter I section 7 and it has been defined the ordinals addition but I don't quite understand that definition. I have seen that in other books authors defines the addition using transfinite induction and it seems easier but now I want to understand Kunen's one.

$$\alpha + \beta=type(\alpha \times \{ 0 \} \cup \beta \times \{1\}, R) \:\text{where } $$ $$R=\{ \langle \langle \xi,0 \rangle, \langle \eta , 0\rangle \rangle : \xi<\eta<\alpha\} \; \cup \{\langle \langle \xi,1 \rangle, \langle \eta , 1\rangle \rangle : \xi<\eta<\beta\} \; \cup [(\alpha\times\{0\})\times(\beta\times\{1\})]. $$

With $type(A,R)$ is the unique ordinal $C$ such that $\langle A, R\rangle \cong C$ when $\langle A,R\rangle$ is a well-ordering set.

What I think I understood so far is that this definition tries to order two non-disjoint sets having that $\alpha<\beta$ and keeping the order inside $\alpha$ and $\beta$. What I can't understand is how I can get that cardinal (in for example a finite case), maybe I'm being stubborn and I should "ignore" this definition and trying to understand the more simplified one given in the later results.

Thank you for your time.

  • $\begingroup$ In my opinion this is the most intuitive definition: it actually gives a picture of the sum order. Are you familiar with lexicographic ordering? $\endgroup$ Apr 26, 2020 at 18:59
  • $\begingroup$ @BrianM.Scott Yes, i'm more less familiar for having studiet it in Naive Set Theory (althought not used it much) what I can't see is what it has to do, i mean, the relation that is defined is not the lexicographic order, right? Thank you for your answer! $\endgroup$
    – Partizanki
    Apr 26, 2020 at 19:03

3 Answers 3


As mentioned by Brian, it is essentially the lexicographic ordering.

For example, say $2=\{0_2,1_2\}$ and $3=\{0_3,1_3,2_3\}$.

According to the definition, we first extend $2$ and $3$ to ordered pairs:

$$2\times\{0\}=\{(0_2,0),(1_2,0)\}\quad{\rm and}\quad 3\times\{1\}=\{(0_3,1),(1_3,1),(2_3,1)\}.$$

Then what is $R$? Although it is written as the union of sets, we can write it in a chain like this:


The set $2\times\{0\}\cup 3\times\{1\}$ is linearly ordered and isomorphic to $5=\{0_5,1_5,2_5,3_5,4_5\}$ in which


Thus, $2+3=5$.

  • $\begingroup$ Amazing thanks! Now that I understand the finite case I will try to work on my own on the infinite case. $\endgroup$
    – Partizanki
    Apr 26, 2020 at 20:04

The definition that Ken is using amounts to placing a copy of the ordinal $\beta$ after the ordinal $\alpha$. Since the sets $\alpha$ and $\beta$ are not actually disjoint (unless one of them is $0$), we first use a small trick to make disjoint copies of them, replacing $\alpha$ by $\alpha\times\{0\}$ and $\beta$ by $\beta\times\{1\}$. We give these the obvious orders, which I’ll call $\le_\alpha$ and $\le_\beta$: for $\xi,\eta\in\alpha$ we set $\langle\xi,0\rangle\le_\alpha\langle\eta,0\rangle$ iff $\xi\le\eta$, and we define $\le_\beta$ similarly. As sets these relations are




Now we have disjoint copies of $\alpha$ and $\beta$ — copies in the sense that they are order-isomorphic to $\alpha$ and $\beta$, respectively — and we define an order that places the copy of $\beta$ after the copy of $\alpha$. We do this by imposing the reverse lexicographic order on $(\alpha\times\{0\})\cup(\beta\times\{1\})$. That is, we order first on the second coordinate and then on the first: we define

$$\langle\xi,i\rangle\,R\,\langle\eta,j\rangle\text{ iff }i<j,\text{ or }i=j\text{ and }\xi\le\eta\;.$$

If you check the various possibilities, you’ll see that this makes

$$\langle\xi,0\rangle\,R\,\langle\eta,0\rangle\text{ iff }\langle\xi,0\rangle\le_\alpha\langle\eta,0\rangle\text{ iff }\xi\le\eta$$

for $\xi,\eta\in\alpha$,

$$\langle\xi,1\rangle\,R\,\langle\eta,1\rangle\text{ iff }\langle\xi,1\rangle\le_\beta\langle\eta,1\rangle\text{ iff }\xi\le\eta$$

for $\xi,\eta\in\beta$, and $\langle\xi,0\rangle\,R\,\langle\eta,1\rangle$ whenever $\xi\in\alpha$ and $\eta\in\beta$. In short, $R$ orders $\alpha\times\{0\}$ just like $\le_\alpha$ and $\beta\times\{1\}$ just like $\le_\beta$, and it places all of $\alpha\times\{0\}$ before all of $\beta\times\{1\}$.

  • $\begingroup$ Ok now I understand, my problem was that I didn't know how that definition of $R$ would be an order but now it's much clearer, amazing answer! Now I think i'm stucking with the concept of isomorphism. $\endgroup$
    – Partizanki
    Apr 26, 2020 at 19:59
  • $\begingroup$ @Partizanki: Fortunately, in this case the order-isomorphisms that are involved are pretty straightforward: for instance, the map $f:\alpha\to\alpha\times\{0\}$ defined by $f(\xi)=\langle\xi,0\rangle$ is easily seen to be an order-isomorphism from the order $\langle\alpha,\le\rangle$ to the order $\langle\alpha\times\{0\},\le_\alpha\rangle$. For any $\xi,\eta\in\alpha$, $\xi\le\eta$ iff $\langle\xi,0\rangle\le_\alpha\langle\eta,1\rangle$. $\endgroup$ Apr 26, 2020 at 20:02
  • $\begingroup$ Yes, I more less understand the order-isomorphsim thanks to your explanation, I meant the isomorphsim that gives you the $type$ as previously defined. $\endgroup$
    – Partizanki
    Apr 26, 2020 at 20:07
  • $\begingroup$ @Partizanki: Ah, yes, that one. The proof from the axioms in Theorem 7.6 makes it look worse than it is. Intuitively you just map the least element of $A$ with respect to $R$ to $0$, the least element of the rest of $A$ to $1$, and so on. If you’ve mapped all of the $R$-predecessors of some $a\in A$ in this fashion, their images turn out to be an initial segment of the ordinals, and you just map $a$ to the smallest ordinal that hasn’t yet been used. $\endgroup$ Apr 26, 2020 at 20:17
  • $\begingroup$ Ok I had the intuition but your comment has just done it, now everything is clearer, thank you very much! Amazing answers. $\endgroup$
    – Partizanki
    Apr 26, 2020 at 20:54

As Kunen explains it you are putting a copy of $\beta$ after $\alpha$ and looking at the resulting order type. Whether $\alpha \lt \beta$ is not important. If they are both finite it is just regular addition. If you add $\omega+2$ and $\omega+1$ the order type is $\omega, 2, \omega, 1$. The $2$ gets absorbed into the start of the following $\omega$ and the result is $\omega + \omega + 1$

  • $\begingroup$ Ok I think I get it but not using the definition but the 7.18 (5) Lemma that states: If $\beta$ is a limit ordinal, $\alpha + \beta= sup\{ \alpha + \xi : \xi < \beta\}.$ Anyway I have to think more about your answer and thank you very much for answering! $\endgroup$
    – Partizanki
    Apr 26, 2020 at 20:11

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