# Problem with understanding the ordinals addition definition

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.

• In my opinion this is the most intuitive definition: it actually gives a picture of the sum order. Are you familiar with lexicographic ordering? Apr 26, 2020 at 18:59
• @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! Apr 26, 2020 at 19:03

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:

$$(0_2,0)<(1_2,0)<(0_3,1)<(1_3,1)<(2_3,1).$$

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

$$0_5<1_5<2_5<3_5<4_5.$$

Thus, $$2+3=5$$.

• Amazing thanks! Now that I understand the finite case I will try to work on my own on the infinite case. 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

$$\le_\alpha=\{\langle\langle\xi,0\rangle,\langle\eta,0\rangle\rangle:\xi\le\eta<\alpha\}$$

and

$$\le_\beta=\{\langle\langle\xi,1\rangle,\langle\eta,1\rangle\rangle:\xi\le\eta<\beta\}\;.$$

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

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 $$\beta\times\{1\}$$.

• 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. Apr 26, 2020 at 19:59
• @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 $\langle\xi,0\rangle\le_\alpha\langle\eta,1\rangle$. Apr 26, 2020 at 20:02
• Yes, I more less understand the order-isomorphsim thanks to your explanation, I meant the isomorphsim that gives you the $type$ as previously defined. Apr 26, 2020 at 20:07
• @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. Apr 26, 2020 at 20:17
• Ok I had the intuition but your comment has just done it, now everything is clearer, thank you very much! Amazing answers. 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 $$\omega + \omega + 1$$

• 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! Apr 26, 2020 at 20:11