How can we determine $2^kappa$ for singular $kappa$ assuming that $2^lambda=lambda^+$ whenever...












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I got an exercise in set theory and can't seem to solve it:
If we assume that $2^lambda = lambda^+ $ holds for every singular cardinal with $2^{operatorname{cof}(lambda)}<lambda$, then how can we determine the value of $2^kappa$ from the value of $2^{<kappa}$ for all singular cardinals $kappa$?
I tried using Silver's Theorem, but couldn't seem to get it right...






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    $begingroup$


    I got an exercise in set theory and can't seem to solve it:
    If we assume that $2^lambda = lambda^+ $ holds for every singular cardinal with $2^{operatorname{cof}(lambda)}<lambda$, then how can we determine the value of $2^kappa$ from the value of $2^{<kappa}$ for all singular cardinals $kappa$?
    I tried using Silver's Theorem, but couldn't seem to get it right...






    set-theory cardinals







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I got an exercise in set theory and can't seem to solve it:
      If we assume that $2^lambda = lambda^+ $ holds for every singular cardinal with $2^{operatorname{cof}(lambda)}<lambda$, then how can we determine the value of $2^kappa$ from the value of $2^{<kappa}$ for all singular cardinals $kappa$?
      I tried using Silver's Theorem, but couldn't seem to get it right...






      set-theory cardinals







      share|cite|improve this question











      $endgroup$




      I got an exercise in set theory and can't seem to solve it:
      If we assume that $2^lambda = lambda^+ $ holds for every singular cardinal with $2^{operatorname{cof}(lambda)}<lambda$, then how can we determine the value of $2^kappa$ from the value of $2^{<kappa}$ for all singular cardinals $kappa$?
      I tried using Silver's Theorem, but couldn't seem to get it right...






      set-theory cardinals




      set-theory cardinals






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      edited Dec 26 '18 at 22:43









      Andrés E. Caicedo

      65.7k8160250




      65.7k8160250










      asked Dec 26 '18 at 18:36









      Jan T.Jan T.

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          Let me consider separately two cases, and afterward indicate how to tell the cases apart.



          Case 1: $2^mu$ has the same value for all sufficiently large $mu<kappa$. In this case, we can invoke the Bukovsky-Hechler theorem (see for example Corollary 5.17 in http://fa.its.tudelft.nl/~hart/onderwijs/set_theory/Jech/05-AC_and_cardinal_arithmetic.pdf ) to conclude that $2^kappa=2^{<kappa}$.



          Case 2: As $mu$ ranges over infinite cardinals $<kappa$, the exponential $2^mu$ increases cofinally often. So the supremum $2^{<kappa}$ of these exponentials is a cardinal $lambda$ with cf$(lambda)=,$cf$(kappa)<kappa$. So $2^{text{cf}(lambda)}$ is one of the cardinals whose supremum defines $lambda$ and is, by the case hypothesis, $<lambda$. Now the hypothesis of your exercise gives us that $2^lambda=lambda^+$. So we have
          $$
          lambda=2^{<kappa}leq2^kappaleq2^lambda=lambda^+,
          $$
          which means $2^kappa$ has one of just two possible values, namely $lambda$ and $lambda^+$. But the first of these isn't really possible, by König's theorem, because it has cofinality $<kappa$. So $2^kappa=lambda^+=(2^{<kappa})^+$.



          To complete the answer, I still need to tell you how to distinguish the two cases, given only $2^{<kappa}$. Fortunately, I already pointed out that in Case 2, $2^{<kappa}$ (there called $lambda$) has cofinality $<kappa$. In Case 1, on the other hand, $2^{<kappa}$ is equal to $2^mu$ for arbitrarily large $mu<kappa$ and therefore, by König's theorem again, cannot have cofinality $<kappa$.



          So the final result is (under the hypothesis of the exercise) that $2^kappa$ equals $2^{<kappa}$ if this has cofinality $geqkappa$, and equals $(2^{<kappa})^+$ otherwise. (Less rigorous but easy to remember summary: $2^kappa$ has the smallest value it could possibly have, given $2^{<kappa}$ and given König's theorem.)






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            $begingroup$

            Let me consider separately two cases, and afterward indicate how to tell the cases apart.



            Case 1: $2^mu$ has the same value for all sufficiently large $mu<kappa$. In this case, we can invoke the Bukovsky-Hechler theorem (see for example Corollary 5.17 in http://fa.its.tudelft.nl/~hart/onderwijs/set_theory/Jech/05-AC_and_cardinal_arithmetic.pdf ) to conclude that $2^kappa=2^{<kappa}$.



            Case 2: As $mu$ ranges over infinite cardinals $<kappa$, the exponential $2^mu$ increases cofinally often. So the supremum $2^{<kappa}$ of these exponentials is a cardinal $lambda$ with cf$(lambda)=,$cf$(kappa)<kappa$. So $2^{text{cf}(lambda)}$ is one of the cardinals whose supremum defines $lambda$ and is, by the case hypothesis, $<lambda$. Now the hypothesis of your exercise gives us that $2^lambda=lambda^+$. So we have
            $$
            lambda=2^{<kappa}leq2^kappaleq2^lambda=lambda^+,
            $$
            which means $2^kappa$ has one of just two possible values, namely $lambda$ and $lambda^+$. But the first of these isn't really possible, by König's theorem, because it has cofinality $<kappa$. So $2^kappa=lambda^+=(2^{<kappa})^+$.



            To complete the answer, I still need to tell you how to distinguish the two cases, given only $2^{<kappa}$. Fortunately, I already pointed out that in Case 2, $2^{<kappa}$ (there called $lambda$) has cofinality $<kappa$. In Case 1, on the other hand, $2^{<kappa}$ is equal to $2^mu$ for arbitrarily large $mu<kappa$ and therefore, by König's theorem again, cannot have cofinality $<kappa$.



            So the final result is (under the hypothesis of the exercise) that $2^kappa$ equals $2^{<kappa}$ if this has cofinality $geqkappa$, and equals $(2^{<kappa})^+$ otherwise. (Less rigorous but easy to remember summary: $2^kappa$ has the smallest value it could possibly have, given $2^{<kappa}$ and given König's theorem.)






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Let me consider separately two cases, and afterward indicate how to tell the cases apart.



              Case 1: $2^mu$ has the same value for all sufficiently large $mu<kappa$. In this case, we can invoke the Bukovsky-Hechler theorem (see for example Corollary 5.17 in http://fa.its.tudelft.nl/~hart/onderwijs/set_theory/Jech/05-AC_and_cardinal_arithmetic.pdf ) to conclude that $2^kappa=2^{<kappa}$.



              Case 2: As $mu$ ranges over infinite cardinals $<kappa$, the exponential $2^mu$ increases cofinally often. So the supremum $2^{<kappa}$ of these exponentials is a cardinal $lambda$ with cf$(lambda)=,$cf$(kappa)<kappa$. So $2^{text{cf}(lambda)}$ is one of the cardinals whose supremum defines $lambda$ and is, by the case hypothesis, $<lambda$. Now the hypothesis of your exercise gives us that $2^lambda=lambda^+$. So we have
              $$
              lambda=2^{<kappa}leq2^kappaleq2^lambda=lambda^+,
              $$
              which means $2^kappa$ has one of just two possible values, namely $lambda$ and $lambda^+$. But the first of these isn't really possible, by König's theorem, because it has cofinality $<kappa$. So $2^kappa=lambda^+=(2^{<kappa})^+$.



              To complete the answer, I still need to tell you how to distinguish the two cases, given only $2^{<kappa}$. Fortunately, I already pointed out that in Case 2, $2^{<kappa}$ (there called $lambda$) has cofinality $<kappa$. In Case 1, on the other hand, $2^{<kappa}$ is equal to $2^mu$ for arbitrarily large $mu<kappa$ and therefore, by König's theorem again, cannot have cofinality $<kappa$.



              So the final result is (under the hypothesis of the exercise) that $2^kappa$ equals $2^{<kappa}$ if this has cofinality $geqkappa$, and equals $(2^{<kappa})^+$ otherwise. (Less rigorous but easy to remember summary: $2^kappa$ has the smallest value it could possibly have, given $2^{<kappa}$ and given König's theorem.)






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Let me consider separately two cases, and afterward indicate how to tell the cases apart.



                Case 1: $2^mu$ has the same value for all sufficiently large $mu<kappa$. In this case, we can invoke the Bukovsky-Hechler theorem (see for example Corollary 5.17 in http://fa.its.tudelft.nl/~hart/onderwijs/set_theory/Jech/05-AC_and_cardinal_arithmetic.pdf ) to conclude that $2^kappa=2^{<kappa}$.



                Case 2: As $mu$ ranges over infinite cardinals $<kappa$, the exponential $2^mu$ increases cofinally often. So the supremum $2^{<kappa}$ of these exponentials is a cardinal $lambda$ with cf$(lambda)=,$cf$(kappa)<kappa$. So $2^{text{cf}(lambda)}$ is one of the cardinals whose supremum defines $lambda$ and is, by the case hypothesis, $<lambda$. Now the hypothesis of your exercise gives us that $2^lambda=lambda^+$. So we have
                $$
                lambda=2^{<kappa}leq2^kappaleq2^lambda=lambda^+,
                $$
                which means $2^kappa$ has one of just two possible values, namely $lambda$ and $lambda^+$. But the first of these isn't really possible, by König's theorem, because it has cofinality $<kappa$. So $2^kappa=lambda^+=(2^{<kappa})^+$.



                To complete the answer, I still need to tell you how to distinguish the two cases, given only $2^{<kappa}$. Fortunately, I already pointed out that in Case 2, $2^{<kappa}$ (there called $lambda$) has cofinality $<kappa$. In Case 1, on the other hand, $2^{<kappa}$ is equal to $2^mu$ for arbitrarily large $mu<kappa$ and therefore, by König's theorem again, cannot have cofinality $<kappa$.



                So the final result is (under the hypothesis of the exercise) that $2^kappa$ equals $2^{<kappa}$ if this has cofinality $geqkappa$, and equals $(2^{<kappa})^+$ otherwise. (Less rigorous but easy to remember summary: $2^kappa$ has the smallest value it could possibly have, given $2^{<kappa}$ and given König's theorem.)






                share|cite|improve this answer









                $endgroup$



                Let me consider separately two cases, and afterward indicate how to tell the cases apart.



                Case 1: $2^mu$ has the same value for all sufficiently large $mu<kappa$. In this case, we can invoke the Bukovsky-Hechler theorem (see for example Corollary 5.17 in http://fa.its.tudelft.nl/~hart/onderwijs/set_theory/Jech/05-AC_and_cardinal_arithmetic.pdf ) to conclude that $2^kappa=2^{<kappa}$.



                Case 2: As $mu$ ranges over infinite cardinals $<kappa$, the exponential $2^mu$ increases cofinally often. So the supremum $2^{<kappa}$ of these exponentials is a cardinal $lambda$ with cf$(lambda)=,$cf$(kappa)<kappa$. So $2^{text{cf}(lambda)}$ is one of the cardinals whose supremum defines $lambda$ and is, by the case hypothesis, $<lambda$. Now the hypothesis of your exercise gives us that $2^lambda=lambda^+$. So we have
                $$
                lambda=2^{<kappa}leq2^kappaleq2^lambda=lambda^+,
                $$
                which means $2^kappa$ has one of just two possible values, namely $lambda$ and $lambda^+$. But the first of these isn't really possible, by König's theorem, because it has cofinality $<kappa$. So $2^kappa=lambda^+=(2^{<kappa})^+$.



                To complete the answer, I still need to tell you how to distinguish the two cases, given only $2^{<kappa}$. Fortunately, I already pointed out that in Case 2, $2^{<kappa}$ (there called $lambda$) has cofinality $<kappa$. In Case 1, on the other hand, $2^{<kappa}$ is equal to $2^mu$ for arbitrarily large $mu<kappa$ and therefore, by König's theorem again, cannot have cofinality $<kappa$.



                So the final result is (under the hypothesis of the exercise) that $2^kappa$ equals $2^{<kappa}$ if this has cofinality $geqkappa$, and equals $(2^{<kappa})^+$ otherwise. (Less rigorous but easy to remember summary: $2^kappa$ has the smallest value it could possibly have, given $2^{<kappa}$ and given König's theorem.)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 27 '18 at 2:31









                Andreas BlassAndreas Blass

                50.3k452109




                50.3k452109






























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