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Ordinals-proof help

  1. Aug 30, 2006 #1
    prove that [tex]\epsilon_0[/tex] is an [tex]\epsilon[/tex] number and that it's the smallest number.
    [tex]\phi(n)=\omega^{\omega^{\omega^{...^{\omega}}}}[/tex] where [tex]\omega[/tex] appears n times.
    an epsilon number is a number which satisfies the equation [tex]\omega^{\epsilon}=\epsilon[/tex].
    for the first part of proving that it's an epsilon number i used the fact that [tex]1+\omega=\omega[/tex], for the second part im not sure i understand how to prove it:
    i mean if we assume there's a number smaller than [tex]\epsilon_0[/tex] that satisfy that it's an epsilon number, then [tex]\omega^{\epsilon^{'}}=\epsilon^{'}<\epsilon_0=\omega^{\epsilon_0}[/tex]
    i know that there exists a unique ordinal such that [tex]\epsilon_0=\epsilon^{'}+\alpha[/tex]
    if [tex]\alpha[/tex] is a finite ordinal then [tex]\epsilon_0=\epsilon^{'}[/tex] and it's a contradiction, but how to prove it when alpha isnt a finite ordinal?
    Last edited: Aug 30, 2006
  2. jcsd
  3. Aug 30, 2006 #2

    matt grime

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    I don't know if this is at all valid in the theory of ordinals, but surely lims and exponentials commute, hence e_0 is an epsilon number.

    Secondly, if e is an epsilon number then e=w^e (=>w) = w^w^e (=>w^w) =... hence e must be greater than w, w^w, w^w, w^w^w,.. and therefore e must be greater thanor equal to the smallest ordinal larger than all of w, w^w, w^w^w, etc which is precisely e_0.

    (This is exactly the same as showing that if t is larger than 0.9, 0.99, 0.999,... then t=>1.)
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