# Distributivity Concern With Ordinals

1. Sep 25, 2006

### Oxymoron

1. Why does left-distributivity work for ordinals $\alpha$, $\beta$, and $\gamma$ but not right-distributivity?

2. Suppose I have the ordinal $\omega$. Then why does the second equality hold?

$$(\omega + 1) \cdot \omega = \omega \cdot \omega + 1 \cdot \omega = \omega \cdot \omega$$

Why is it not $\omega \cdot \omega + \omega$?

Does 2. have anything to do with the reason behind 1.?

2. Sep 25, 2006

### matt grime

The answers, presumably, are because the definitions allow us to

1. find a counter example
2. show that w.w+1.w = w.w

As I would be guessing what the precise axioms of ordinal arithmetic are I can't say anymore than that.

3. Sep 25, 2006

### HallsofIvy

Staff Emeritus
For any ordinal, w, w+ 1= w.

4. Sep 25, 2006

### matt grime

For any infinite ordinal 1+w=w, not w+1=w, surely?

5. Sep 25, 2006

### Oxymoron

For any ordinal $\omega$, does the following calculation say that $1+\omega = \omega$?

$$1+\omega = \sup\{1+n\,:\,n\in\omega\} = \sup\{m \,:\,0 < m \in \omega\} = \omega$$

So would the following be correct?

$$(1+1)\cdot\omega = 1\cdot\omega + 1\cdot\omega = \omega + \omega = 2\cdot\omega$$

$$\omega\cdot(1+1) = \omega\cdot 1 + \omega\cdot 1 = \omega\cdot(1+\omega)\cdot 1 = \omega\cdot\omega\cdot 1 = \omega\cdot\omega$$

Which shows that for an ordinal w,

$$(1+1)\cdot\omega \neq \omega\cdot(1+1)$$

So basically, this is my counter-example that Matt proposed. Where I took $\alpha = \omega$, $\beta = \gamma = 1$.

Last edited: Sep 25, 2006
6. Sep 25, 2006

### Oxymoron

Yeah, I think he meant to write 1+w=w because w+1 does not equal w because w+1 has a maximal element and w does not.

7. Sep 27, 2006

### Oxymoron

Ok, so I can show that for the ordinal $\omega$ that

$$(1+1)\cdot\omega \neq \omega\cdot(1+1)$$

but this does not help me with three ordinals. Besides, Im meant to show that

$$\alpha\cdot(\beta + \gamma) = \alpha\beta + \alpha\gamma$$

and I don't see that there is anything to prove! We all know that ordinals are left-distributive so there is nothing to show!!! What am I meant to do?

8. Sep 27, 2006

### matt grime

When you say 'we all know that' have you actually proved it?

It is common to assume some result ina course, and ask for its proof as an exercise.

And I don't understand your concern abuot the other issue. You *have* found 3 ordinals for which it fails, so what if two of them are the same? (At least, assuming your argument is correct; I don't think it is). Or do you think you have to prove it for all triples of ordinals?

Last edited: Sep 27, 2006
9. Sep 27, 2006

### Oxymoron

My first question was to determine why left-distributivity and right-distributivity did not coincide with ordinals. You told me to find a counterexample and I did. Now, my second quest is to discover exactly why

$$\alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma$$

but Im stumped.

10. Sep 27, 2006

### matt grime

No, you first question was to determine why right distributivity does not hold. No one mentioned anything about showing it agrees with left distibrutivity, though I think it is a small step from there assuming that ordinal multiplication is commutative.

Just write out the definitions and see what you can conclude.

11. Sep 27, 2006

### Oxymoron

Yes, my mistake.

Ok. Well I know that I can define the addition of two ordinals $\beta$ and $\gamma$ to be the unique ordinal for which there is an order-preserving bijection:

$$((\{1\}\times \beta)\cup(\{2\}\times\gamma),\leq) \rightarrow \beta + \alpha$$

Therefore I can assume in my calculations that

$$\alpha+\beta$$ is isomorphic to $$(\alpha\times\{1\})\cup(\beta\times\{2\})$$

and

$$\alpha\cdot\beta$$ is isomorphic to $$\alpha\times\beta$$

$$\alpha\cdot(\beta + \gamma) \cong \alpha\times(\beta + \gamma)$$
$${}\quad\quad\quad\quad\cong \alpha \times ((\beta \times\{1\})\cup(\gamma\times\{2\}))$$
$${}\quad\quad\quad\quad\cong(\alpha \times \beta \times \{1\})\cup(\alpha\times\beta\times\{2\})$$
$${}\quad\quad\quad\quad\cong(\alpha \times \beta) + (\alpha \times \beta)$$
$${}\quad\quad\quad\quad\cong\alpha\cdot\beta + \alpha\cdot\beta$$

The problem with doing it this way is that I need to check whether at each step the obvious bijection is in fact an order-isomorphism. Otherwise I could probably do this using induction on $\gamma$. What do you think?

Last edited: Sep 27, 2006
12. Sep 27, 2006

### matt grime

presumably you mean a.b is isomorphic to axb with the lexicographic ordering. Ordinals are ordered sets. You appear to be ignoring that fact.

13. Sep 27, 2006

### Oxymoron

yes, I believe the ordering needs to be lexicographic.

Does this mean that my attempt method is flawed? Completely incorrect? Close? Or is this a hint?

Last edited: Sep 27, 2006
14. Sep 27, 2006

### matt grime

It's an observation. You say you're having trouble proving these are whatever, but you're not paying attention to the ordering, so it's not surprising. If you just paid attention to the ordering it will probably be obvious why these are order-isomorphisms.

15. Sep 27, 2006

### Oxymoron

Ah! Of course. The proposition that says

For every well-ordered set there exists a unique ordinal number such that there is an order-preserving bijection from the well-ordered set to the ordinal.

Therefore, since ordinals are ordered sets the bijections must be order-preserving. Thanks Matt, I had to read through the proposition again to understand why it works, but I wouldn't have figured it out as quickly as I did without your guidance.