Distributivity Concern With Ordinals

[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.?

 

[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.

 

[HallsofIvy]

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

 

[matt grime]

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

 

[Oxymoron]

Posted by 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.

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.

 

[Oxymoron]

Posted by Matt Grime:

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

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.

 

[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, I am 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?

 

[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?

 

[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 I am stumped.

 

[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.

 

[Oxymoron]

Posted by 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.

Yes, my mistake.

Posted by Matt Grime:

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

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?

 

[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.

 

[Oxymoron]

yes, I believe the ordering needs to be lexicographic.

Ordinals are ordered sets. You appear to be ignoring that fact.

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

 

[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.

 

[Oxymoron]

Posted by 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.

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.