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Topic: A canonical form for small ordinals
Replies: 12   Last Post: Jan 19, 2014 2:35 PM

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 quasi Posts: 12,012 Registered: 7/15/05
Re: A canonical form for small ordinals
Posted: Jan 17, 2014 6:04 AM

William Elliot wrote:
>Paul wrote:
>

>>I saw it asserted without proof that all ordinals, alpha,
>>less than epsilon_0, can be uniquely expressed as
>>
>> (omega^beta)*(gamma + 1)
>>
>>for some gamma and some beta such that beta < alpha,
>>where omega is the smallest infinite ordinal. This isn't
>>clear to me. Could anyone help or give a reference?

Note that alpha = 0 must be excluded since the ordinal 0 has
no such representation.

>Cantor's normal form.

Assuming 0 < alpha < epsilon_0, then as suggested by
William Elliot, Cantor's normal form clinches it.

Proposition:

For all ordinals alpha > 0 there are exist unique ordinals
beta and gamma such that (omega^beta)*(gamma + 1).

Existence:

Let beta be the least exponent in the Cantor normal form for
alpha. Then the normal form factors as

(omega^beta)*(gamma + 1)

for some gamma.

Uniqueness:

Suppose alpha has two representations:

alpha = (omega^beta1)*(gamma1 + 1)

alpha = (omega^beta2)*(gamma2 + 1)

From the uniqueness of the normal form for alpha, together with
the uniqueness of the normal form for gamma1 + 1 and gamma2 + 1,
it follows that beta1 = beta2, and based on that, the uniqueness
of the normal form for alpha then yields gamma1 = gamma2.

This completes the proof of the proposition.

Now suppose 0 < alpha < epsilon_0.

Applying the proposition, there are unique ordinals beta and
gamma such that alpha = (omega^beta)*(gamma + 1).

Then

alpha = (omega^beta)*(gamma + 1)

=> omega^beta <= alpha

But the above, together with alpha < epsilon_0, yields

beta < alpha

as required.

Bottom line:

Sometimes William Elliot is right on target, and this was
one of those times.

quasi

Date Subject Author
1/16/14 Paul
1/16/14 ross.finlayson@gmail.com
1/16/14 fom
1/17/14 Paul
1/17/14 William Elliot
1/17/14 quasi
1/17/14 quasi
1/17/14 quasi
1/17/14 Paul
1/18/14 ross.finlayson@gmail.com
1/18/14 ross.finlayson@gmail.com
1/19/14 Paul
1/19/14 ross.finlayson@gmail.com