Function on the set of limit countable ordinalsA question about $dotS^Q$-semiproperness and revised countable support iterated forcing of length a limit ordinalCofinality of countable ordinals in ZF, and in toposesOn a characterization of the recursively inaccessible ordinalsMemorable ordinalsProblem with definability in the constructible hierarchySplitting of ordinals of oscillation ranks of a Baire $1$ functionDo these ordinals exist?Connection between countable ordinals and Turing degreesFormalizations of The Matchstick Diagram Representation of OrdinalsIs the smallest $L_alpha$ with undefinable ordinals always countable?

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Function on the set of limit countable ordinals


A question about $dotS^Q$-semiproperness and revised countable support iterated forcing of length a limit ordinalCofinality of countable ordinals in ZF, and in toposesOn a characterization of the recursively inaccessible ordinalsMemorable ordinalsProblem with definability in the constructible hierarchySplitting of ordinals of oscillation ranks of a Baire $1$ functionDo these ordinals exist?Connection between countable ordinals and Turing degreesFormalizations of The Matchstick Diagram Representation of OrdinalsIs the smallest $L_alpha$ with undefinable ordinals always countable?













4












$begingroup$


Let $Lambda$ be the set of all countable limit ordinals. Does there exist an injective function $f:Lambdatoomega_1$ with the properties:



  1. $forall lambdainLambda:~f(lambda)<lambda$


  2. $forallalpha<omega_1~~existsbeta<omega_1~~foralllambda>beta:~f(lambda)>alpha$ ?









share|cite|improve this question









$endgroup$







  • 5




    $begingroup$
    Though the answer below answers your question, you might be interested to know that an injective function $f:Lambdatoomega_1$ automatically satisfies $2$: for any $gammaleqalpha$ there is at most one ordinal $a_gamma$ such that $f(a_gamma)=gamma$. Then $beta=sup_gammaleqalphaa_gamma$ will work.
    $endgroup$
    – Wojowu
    Apr 24 at 8:25















4












$begingroup$


Let $Lambda$ be the set of all countable limit ordinals. Does there exist an injective function $f:Lambdatoomega_1$ with the properties:



  1. $forall lambdainLambda:~f(lambda)<lambda$


  2. $forallalpha<omega_1~~existsbeta<omega_1~~foralllambda>beta:~f(lambda)>alpha$ ?









share|cite|improve this question









$endgroup$







  • 5




    $begingroup$
    Though the answer below answers your question, you might be interested to know that an injective function $f:Lambdatoomega_1$ automatically satisfies $2$: for any $gammaleqalpha$ there is at most one ordinal $a_gamma$ such that $f(a_gamma)=gamma$. Then $beta=sup_gammaleqalphaa_gamma$ will work.
    $endgroup$
    – Wojowu
    Apr 24 at 8:25













4












4








4





$begingroup$


Let $Lambda$ be the set of all countable limit ordinals. Does there exist an injective function $f:Lambdatoomega_1$ with the properties:



  1. $forall lambdainLambda:~f(lambda)<lambda$


  2. $forallalpha<omega_1~~existsbeta<omega_1~~foralllambda>beta:~f(lambda)>alpha$ ?









share|cite|improve this question









$endgroup$




Let $Lambda$ be the set of all countable limit ordinals. Does there exist an injective function $f:Lambdatoomega_1$ with the properties:



  1. $forall lambdainLambda:~f(lambda)<lambda$


  2. $forallalpha<omega_1~~existsbeta<omega_1~~foralllambda>beta:~f(lambda)>alpha$ ?






set-theory ordinal-numbers






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Apr 24 at 6:13









ar.grigar.grig

1




1







  • 5




    $begingroup$
    Though the answer below answers your question, you might be interested to know that an injective function $f:Lambdatoomega_1$ automatically satisfies $2$: for any $gammaleqalpha$ there is at most one ordinal $a_gamma$ such that $f(a_gamma)=gamma$. Then $beta=sup_gammaleqalphaa_gamma$ will work.
    $endgroup$
    – Wojowu
    Apr 24 at 8:25












  • 5




    $begingroup$
    Though the answer below answers your question, you might be interested to know that an injective function $f:Lambdatoomega_1$ automatically satisfies $2$: for any $gammaleqalpha$ there is at most one ordinal $a_gamma$ such that $f(a_gamma)=gamma$. Then $beta=sup_gammaleqalphaa_gamma$ will work.
    $endgroup$
    – Wojowu
    Apr 24 at 8:25







5




5




$begingroup$
Though the answer below answers your question, you might be interested to know that an injective function $f:Lambdatoomega_1$ automatically satisfies $2$: for any $gammaleqalpha$ there is at most one ordinal $a_gamma$ such that $f(a_gamma)=gamma$. Then $beta=sup_gammaleqalphaa_gamma$ will work.
$endgroup$
– Wojowu
Apr 24 at 8:25




$begingroup$
Though the answer below answers your question, you might be interested to know that an injective function $f:Lambdatoomega_1$ automatically satisfies $2$: for any $gammaleqalpha$ there is at most one ordinal $a_gamma$ such that $f(a_gamma)=gamma$. Then $beta=sup_gammaleqalphaa_gamma$ will work.
$endgroup$
– Wojowu
Apr 24 at 8:25










1 Answer
1






active

oldest

votes


















10












$begingroup$

No. The first property is known as $f$ being regressive. Fodor’s Lemma says that any regressive function on a stationary set is constant on a stationary subset. In particular, because $Lambda$ is club (and thus stationary), such $f$ cannot be injective.






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    active

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    10












    $begingroup$

    No. The first property is known as $f$ being regressive. Fodor’s Lemma says that any regressive function on a stationary set is constant on a stationary subset. In particular, because $Lambda$ is club (and thus stationary), such $f$ cannot be injective.






    share|cite|improve this answer









    $endgroup$

















      10












      $begingroup$

      No. The first property is known as $f$ being regressive. Fodor’s Lemma says that any regressive function on a stationary set is constant on a stationary subset. In particular, because $Lambda$ is club (and thus stationary), such $f$ cannot be injective.






      share|cite|improve this answer









      $endgroup$















        10












        10








        10





        $begingroup$

        No. The first property is known as $f$ being regressive. Fodor’s Lemma says that any regressive function on a stationary set is constant on a stationary subset. In particular, because $Lambda$ is club (and thus stationary), such $f$ cannot be injective.






        share|cite|improve this answer









        $endgroup$



        No. The first property is known as $f$ being regressive. Fodor’s Lemma says that any regressive function on a stationary set is constant on a stationary subset. In particular, because $Lambda$ is club (and thus stationary), such $f$ cannot be injective.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 24 at 6:54









        Monroe EskewMonroe Eskew

        8,03922565




        8,03922565



























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