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.
For all ordinals alpha > 0 there are exist unique ordinals beta and gamma such that (omega^beta)*(gamma + 1).
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.
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).
alpha = (omega^beta)*(gamma + 1)
=> omega^beta <= alpha
But the above, together with alpha < epsilon_0, yields
beta < alpha
Sometimes William Elliot is right on target, and this was one of those times.