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MathematicalPhysicist
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prove that [tex]\epsilon_0[/tex] is an [tex]\epsilon[/tex] number and that it's the smallest number.
[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 I am 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 isn't a finite ordinal?
[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 I am 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 isn't a finite ordinal?
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