prove that [tex]\epsilon_0[/tex] is an [tex]\epsilon[/tex] number and that it's the smallest number.(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\epsilon_0=\lim_{n<\omega}\phi(n)[/tex]

[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?

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# Ordinals-proof help

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