Problem with understanding the ordinals addition definition

Question

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.

Answer 1

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, $2+3=5$.

Answer 2

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

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

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

Answer 3

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 $\omega + \omega + 1$