On proving $omega^{epsilon} = epsilon$












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$begingroup$


How can I prove that
$$omega^{epsilon} = epsilon$$
where $omega ={0, 1, 2, 3, 4, ...}$ and $epsilon = {omega, omega^{omega}, omega^{omega^{omega}}, omega^{omega^{omega^omega}}, ...}$






elementary-set-theory ordinals







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  • 3




    $begingroup$
    $epsilon_0$ is actually defined to be the supremum of ${ omega, omega^{omega}, dots }$. I think this will clarify your problem.
    $endgroup$
    – Shervin Sorouri
    Jan 19 at 13:55


















-1












$begingroup$


How can I prove that
$$omega^{epsilon} = epsilon$$
where $omega ={0, 1, 2, 3, 4, ...}$ and $epsilon = {omega, omega^{omega}, omega^{omega^{omega}}, omega^{omega^{omega^omega}}, ...}$






elementary-set-theory ordinals







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    $epsilon_0$ is actually defined to be the supremum of ${ omega, omega^{omega}, dots }$. I think this will clarify your problem.
    $endgroup$
    – Shervin Sorouri
    Jan 19 at 13:55
















-1












-1








-1





$begingroup$


How can I prove that
$$omega^{epsilon} = epsilon$$
where $omega ={0, 1, 2, 3, 4, ...}$ and $epsilon = {omega, omega^{omega}, omega^{omega^{omega}}, omega^{omega^{omega^omega}}, ...}$






elementary-set-theory ordinals







share|cite|improve this question











$endgroup$




How can I prove that
$$omega^{epsilon} = epsilon$$
where $omega ={0, 1, 2, 3, 4, ...}$ and $epsilon = {omega, omega^{omega}, omega^{omega^{omega}}, omega^{omega^{omega^omega}}, ...}$






elementary-set-theory ordinals




elementary-set-theory ordinals






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edited Jan 19 at 14:25









Henno Brandsma

110k347116




110k347116










asked Jan 19 at 13:39









FreeMindFreeMind

9231133




9231133








  • 3




    $begingroup$
    $epsilon_0$ is actually defined to be the supremum of ${ omega, omega^{omega}, dots }$. I think this will clarify your problem.
    $endgroup$
    – Shervin Sorouri
    Jan 19 at 13:55
















  • 3




    $begingroup$
    $epsilon_0$ is actually defined to be the supremum of ${ omega, omega^{omega}, dots }$. I think this will clarify your problem.
    $endgroup$
    – Shervin Sorouri
    Jan 19 at 13:55










3




3




$begingroup$
$epsilon_0$ is actually defined to be the supremum of ${ omega, omega^{omega}, dots }$. I think this will clarify your problem.
$endgroup$
– Shervin Sorouri
Jan 19 at 13:55






$begingroup$
$epsilon_0$ is actually defined to be the supremum of ${ omega, omega^{omega}, dots }$. I think this will clarify your problem.
$endgroup$
– Shervin Sorouri
Jan 19 at 13:55












1 Answer
1






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$begingroup$

As $varepsilon_0 = sup {omega, omega^omega, omega^{(omega^omega)}, dots }$ we know (by the recursive definition of ordinal exponentiation) that $omega^{varepsilon_0} = sup {omega^omega, omega^{(omega^omega)}, dots,}$, which is just the supremum of the tail of the same set, so that it also equals $varepsilon_0$.






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    $begingroup$

    As $varepsilon_0 = sup {omega, omega^omega, omega^{(omega^omega)}, dots }$ we know (by the recursive definition of ordinal exponentiation) that $omega^{varepsilon_0} = sup {omega^omega, omega^{(omega^omega)}, dots,}$, which is just the supremum of the tail of the same set, so that it also equals $varepsilon_0$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      As $varepsilon_0 = sup {omega, omega^omega, omega^{(omega^omega)}, dots }$ we know (by the recursive definition of ordinal exponentiation) that $omega^{varepsilon_0} = sup {omega^omega, omega^{(omega^omega)}, dots,}$, which is just the supremum of the tail of the same set, so that it also equals $varepsilon_0$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        As $varepsilon_0 = sup {omega, omega^omega, omega^{(omega^omega)}, dots }$ we know (by the recursive definition of ordinal exponentiation) that $omega^{varepsilon_0} = sup {omega^omega, omega^{(omega^omega)}, dots,}$, which is just the supremum of the tail of the same set, so that it also equals $varepsilon_0$.






        share|cite|improve this answer











        $endgroup$



        As $varepsilon_0 = sup {omega, omega^omega, omega^{(omega^omega)}, dots }$ we know (by the recursive definition of ordinal exponentiation) that $omega^{varepsilon_0} = sup {omega^omega, omega^{(omega^omega)}, dots,}$, which is just the supremum of the tail of the same set, so that it also equals $varepsilon_0$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 19 at 14:25

























        answered Jan 19 at 14:16









        Henno BrandsmaHenno Brandsma

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