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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: 10,450
Registered: 7/15/05
Re: A canonical form for small ordinals
Posted: Jan 17, 2014 6:10 AM
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quasi wrote:
>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).


I meant: such that alpha = (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



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