The Basics of Ordinal Numbers (long)

[Achernar]

In order to help me understand my infinites, I started reading up on Ordinal Numbers on MathWorld. I’ve run into three snags so far. If you know the answers to these, feel free to be terse; I don’t need an in-depth explanation.

1. Ordinal Addition

Let (A, <=) and (B, <=) be disjoint totally ordered sets with order types alpha and beta. Then the ordinal sum is defined at set (C = A union B, <=) where, if c[sub]1[/sub] and c[sub]2[/sub] are both from the same subset, the order is the same as in the subset, but if c[sub]1[/sub] is from A and c[sub]2[/sub] is from B, then c[sub]1[/sub] < c[sub]2[/sub] has order type alpha + beta.

My snag has to do with A and B being disjoint. If 2 = {0, 1} and 3 = {0, 1, 2} then 2 and 3 clearly aren’t disjoint. Is 2 + 3 undefined?

2. Ordinal Multiplication

[Quote]
Let (A, <=) and (B, <=) be totally ordered sets. Let C = A × B be the Cartesian Product and define order as follows. For any a[sub]1[/sub], a[sub]2[/sub] in A and b[sub]1[/sub], b[sub]2[/sub] in B,[ol][li]If a[sub]1[/sub] < a[sub]2[/sub], then (a[sub]1[/sub], b[sub]1[/sub]) < (a[sub]2[/sub], b[sub]2[/sub]),If a[sub]1[/sub] = a[sub]2[/sub], then (a[sub]1[/sub], b[sub]1[/sub]) and (a[sub]2[/sub], b[sub]2[/sub]) compare the same way as b[sub]1[/sub], b[sub]2[/sub] (i.e., lexicographical order)[/ol][/li][/Quote]
This seems straightforward enough, but I get the opposite result as I expect from transfinite multiplication. That is:

omega × 2 = {(0, 0), (0, 1), (1, 0), (1, 1), (2, 0), (2, 1), (3, 0), (3, 1), ··· } = omega
2 × omega = {(0, 0), (0, 1), (0, 2), (0, 3), ···, (1, 0), (1, 1), (1, 2), ···} = omega + omega

What’s going on here?

3. Ordinal Exponentiation. This is the one that’s really bugging me.

Let alpha and beta be any ordinal numbers, then ordinal exponentiation is defined so that if beta = 0, then alpha[sup]beta[/sup] = 1. If beta is not a limit ordinal, then choose gamma such that gamma + 1 = beta,

alpha[sup](successor of beta)/sup * alpha.

I don’t even understand what this is saying. Is * supposed to denote ordinal multiplication? What does gamma have to do with anything?

[ultrafilter]

Not sure, but I want the answers too.

[Cabbage]

1. Given any two ordinals, at least one will be a subset of the other, so when adding two ordinals, you always run into this problem. The easy way around it, when adding 2+3, for example, is to define 2+3 as:

[2 x {0}] + [3 x {1}]

(x denoting cross product). This “makes” the two ordinals disjoint, so you can add them as defined above.

2. You’ve got the right idea. The simple answer is that ordinal multiplication is not commutative. Note that ordinal addition isn’t commutative, either: 1+omega=omega, which is distinct from omega+1.

3. This seems a little muddled, I’m quite sure there’s a typo there. I believe they intended to write:

alpha[sup]beta[/sup] = alpha[sup]gamma+1[/sup] = alpha[sup]gamma[/sup] * alpha

where * means ordinal multiplication.

Note that ordinal exponentiation is completely different from cardinal exponentiation (for that matter, addition and multiplication are different for the two, as well). As ordinals:

2[sup]omega[/sup] = sup{2[sup]n[/sup]:n < omega} = omega

while as cardinals 2[sup]omega[/sup] = c = cardinality of the continuum > omega.

[Achernar]

1. Excellent. That makes a lot of sense. I figured it was something like that, and I “understood” how addition worked, but seeing it in set notation really helps.

2. I understand that multiplication is not commutative. My problem is that on the same page (Ordinal Multiplication) it says this:

2 * omega = omega != omega * 2

But, I get that 2 * omega != omega = omega * 2.

3. Wow, that’s actually a really bad typo. I think I’ll e-mail them about it. Exponentiation is pretty difficult, so I’m trying to do a few exercises, but I need to make sure I understand multiplication first.

[Cabbage]

About ordinal multiplication, you’re right; the website is misleading, at best. The catch is this quote:

However, Dauben (1990, p. 104) and Moore (1982, p. 40) define multiplication in the reverse order.

Some texts define alphabeta as the order type of alpha x beta with the lexicographic order (as done on the website); other texts define alphabeta the same way, only with beta x alpha instead of alpha x beta.

I can’t see any benefit of doing it one way or the other, but the website defines it the former way, but then their comment about 2omega vs. omega2 assumes defining it the latter way. Your evalutation of 2omega and omega2 is actually the correct one, given the definition of multiplication on the website.

[Achernar]

Thank you very much for your help so far Cabbage. One more question I think before I understand. There are two ways to define multiplication, and exponentiation is defined in terms of multiplication. Will the choice of multiplication affect how exponentiation works out?

I think I prefer the way that’s not on the website, because I like saying that omega + omega = 2 * omega, as opposed to omega * 2.

[Cabbage]

Yeah, how you define multiplication should influence how you define exponentiation:

1. These are the definitions I’m used to seeing: alpha*beta = the order type of beta x alpha. Then you can define exponentiation for a successor ordinal as:

alpha[sup]beta+1[/sup] = alpha[sup]beta[/sup] * alpha.

2. If, on the other hand, you define multiplication as the website does: alpha*beta = the order type of alpha x beta, I would think, for consistency, you should define exponentiation as:

alpha[sup]beta+1[/sup] = alpha * alpha[sup]beta[/sup]

Either way, you should get 2[sup]omega+1[/sup] = omega + omega, for example.

[John_Kentzel-Griffin]

While you could define multiplication either way, there is a motivation for defining it the opposite way from the website. If you define exponentation the other way you have:

1. a[sup]b+c[/sup] = a[sup]b[/sup]a[sup]c[/sup]
   If you define exponentation as the website defines it, you have:
2. a[sup]b+c[/sup] = a[sup]c[/sup]a[sup]b[/sup]

As long as you’re consistent, either way would work. But, as you can see by the smilies, the first way is less confusing than the way the website defines it.

[ultrafilter]

*Originally posted by DrMatrix *
**As long as you’re consistent, either way would work. But, as you can see by the smilies, the first way is less confusing than the way the website defines it. **

Where the hell were those smilies when I was in college? I needed them!