quasi
Posts:
11,254
Registered:
7/15/05


Re: A canonical form for small ordinals
Posted:
Jan 17, 2014 6:10 AM


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

