0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

1
$\begingroup$

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, $2+3=5$.

$\endgroup$
1
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
6
  • $\begingroup$ 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
0
$\begingroup$

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 $\omega + \omega + 1$

$\endgroup$
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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.