Assuming GCH: if $mathrm{cf}(kappa) leq lambda < kappa$, then $kappa^lambda = kappa^+$ (Jech Theorem 5.15)












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I am trying to fill in the details for part (ii) of Theorem 5.15 in Jech's Set Theory:




Theorem 5.15 If GCH holds and $kappa$ and $lambda$ are infinite cardinals, then



(i) If $kappa leq lambda$, then $kappa^lambda = lambda^+$.



(ii) If $mathrm{cf}(kappa) leq lambda < kappa$, then $kappa^lambda = kappa^+$.



(iii) If $lambda < mathrm{cf}(kappa)$, then $kappa^lambda = kappa$.




In the proof of (ii) he just states that it follows from these two lemmas:




Lemma 5.7 If $|A| = kappa geq lambda$, then the set $[A]^lambda$ has cardinality $kappa^lambda.$




(Here $[A]^lambda = {X subset A : |X| = lambda}.$)




Lemma 5.8 If $lambda$ is an infinite cardinal and $kappa_i > 0$ for each $i < lambda$, then



$$sum_{i<lambda}{kappa_i} = lambdacdotsup_{i<lambda}{kappa_i}.$$




I am struggling to construct an explicit proof using these results and a few facts about cardinal arithmetic such as absorption for infinite cardinals and cardinalities of power sets.



Since, assuming GCH, $kappa^+ = 2^kappa = |mathcal{P}(A)|$ for some $A$ with cardinality $kappa$, I thought I could come up with some sequence ${mu_i : i < mathrm{cf}(kappa)}$ such that



$$kappa^+ = 2^kappa = |mathcal{P}(A)| = sum_{i < mathrm{cf}(kappa)} |[A]^{mu_i}| = sum_{i < mathrm{cf}(kappa)} kappa^{mu_i} = mathrm{cf}(kappa)cdotsup_{i<mathrm{cf}(kappa)}kappa^{mu_i} = mathrm{cf}(kappa)cdotkappa^lambda = kappa^lambda$$



Is this approach a good idea? How would I go about finding the appropriate $mu_i$? I suppose it will have to make use of the assumption that $mathrm{cf}(kappa) leq lambda < kappa$, but I don't see how.



I'd much appreciate any hints on how to proceed (either with my suggestion or another way).






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    $begingroup$


    I am trying to fill in the details for part (ii) of Theorem 5.15 in Jech's Set Theory:




    Theorem 5.15 If GCH holds and $kappa$ and $lambda$ are infinite cardinals, then



    (i) If $kappa leq lambda$, then $kappa^lambda = lambda^+$.



    (ii) If $mathrm{cf}(kappa) leq lambda < kappa$, then $kappa^lambda = kappa^+$.



    (iii) If $lambda < mathrm{cf}(kappa)$, then $kappa^lambda = kappa$.




    In the proof of (ii) he just states that it follows from these two lemmas:




    Lemma 5.7 If $|A| = kappa geq lambda$, then the set $[A]^lambda$ has cardinality $kappa^lambda.$




    (Here $[A]^lambda = {X subset A : |X| = lambda}.$)




    Lemma 5.8 If $lambda$ is an infinite cardinal and $kappa_i > 0$ for each $i < lambda$, then



    $$sum_{i<lambda}{kappa_i} = lambdacdotsup_{i<lambda}{kappa_i}.$$




    I am struggling to construct an explicit proof using these results and a few facts about cardinal arithmetic such as absorption for infinite cardinals and cardinalities of power sets.



    Since, assuming GCH, $kappa^+ = 2^kappa = |mathcal{P}(A)|$ for some $A$ with cardinality $kappa$, I thought I could come up with some sequence ${mu_i : i < mathrm{cf}(kappa)}$ such that



    $$kappa^+ = 2^kappa = |mathcal{P}(A)| = sum_{i < mathrm{cf}(kappa)} |[A]^{mu_i}| = sum_{i < mathrm{cf}(kappa)} kappa^{mu_i} = mathrm{cf}(kappa)cdotsup_{i<mathrm{cf}(kappa)}kappa^{mu_i} = mathrm{cf}(kappa)cdotkappa^lambda = kappa^lambda$$



    Is this approach a good idea? How would I go about finding the appropriate $mu_i$? I suppose it will have to make use of the assumption that $mathrm{cf}(kappa) leq lambda < kappa$, but I don't see how.



    I'd much appreciate any hints on how to proceed (either with my suggestion or another way).






    set-theory cardinals







    share|cite|improve this question











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      $begingroup$


      I am trying to fill in the details for part (ii) of Theorem 5.15 in Jech's Set Theory:




      Theorem 5.15 If GCH holds and $kappa$ and $lambda$ are infinite cardinals, then



      (i) If $kappa leq lambda$, then $kappa^lambda = lambda^+$.



      (ii) If $mathrm{cf}(kappa) leq lambda < kappa$, then $kappa^lambda = kappa^+$.



      (iii) If $lambda < mathrm{cf}(kappa)$, then $kappa^lambda = kappa$.




      In the proof of (ii) he just states that it follows from these two lemmas:




      Lemma 5.7 If $|A| = kappa geq lambda$, then the set $[A]^lambda$ has cardinality $kappa^lambda.$




      (Here $[A]^lambda = {X subset A : |X| = lambda}.$)




      Lemma 5.8 If $lambda$ is an infinite cardinal and $kappa_i > 0$ for each $i < lambda$, then



      $$sum_{i<lambda}{kappa_i} = lambdacdotsup_{i<lambda}{kappa_i}.$$




      I am struggling to construct an explicit proof using these results and a few facts about cardinal arithmetic such as absorption for infinite cardinals and cardinalities of power sets.



      Since, assuming GCH, $kappa^+ = 2^kappa = |mathcal{P}(A)|$ for some $A$ with cardinality $kappa$, I thought I could come up with some sequence ${mu_i : i < mathrm{cf}(kappa)}$ such that



      $$kappa^+ = 2^kappa = |mathcal{P}(A)| = sum_{i < mathrm{cf}(kappa)} |[A]^{mu_i}| = sum_{i < mathrm{cf}(kappa)} kappa^{mu_i} = mathrm{cf}(kappa)cdotsup_{i<mathrm{cf}(kappa)}kappa^{mu_i} = mathrm{cf}(kappa)cdotkappa^lambda = kappa^lambda$$



      Is this approach a good idea? How would I go about finding the appropriate $mu_i$? I suppose it will have to make use of the assumption that $mathrm{cf}(kappa) leq lambda < kappa$, but I don't see how.



      I'd much appreciate any hints on how to proceed (either with my suggestion or another way).






      set-theory cardinals







      share|cite|improve this question











      $endgroup$




      I am trying to fill in the details for part (ii) of Theorem 5.15 in Jech's Set Theory:




      Theorem 5.15 If GCH holds and $kappa$ and $lambda$ are infinite cardinals, then



      (i) If $kappa leq lambda$, then $kappa^lambda = lambda^+$.



      (ii) If $mathrm{cf}(kappa) leq lambda < kappa$, then $kappa^lambda = kappa^+$.



      (iii) If $lambda < mathrm{cf}(kappa)$, then $kappa^lambda = kappa$.




      In the proof of (ii) he just states that it follows from these two lemmas:




      Lemma 5.7 If $|A| = kappa geq lambda$, then the set $[A]^lambda$ has cardinality $kappa^lambda.$




      (Here $[A]^lambda = {X subset A : |X| = lambda}.$)




      Lemma 5.8 If $lambda$ is an infinite cardinal and $kappa_i > 0$ for each $i < lambda$, then



      $$sum_{i<lambda}{kappa_i} = lambdacdotsup_{i<lambda}{kappa_i}.$$




      I am struggling to construct an explicit proof using these results and a few facts about cardinal arithmetic such as absorption for infinite cardinals and cardinalities of power sets.



      Since, assuming GCH, $kappa^+ = 2^kappa = |mathcal{P}(A)|$ for some $A$ with cardinality $kappa$, I thought I could come up with some sequence ${mu_i : i < mathrm{cf}(kappa)}$ such that



      $$kappa^+ = 2^kappa = |mathcal{P}(A)| = sum_{i < mathrm{cf}(kappa)} |[A]^{mu_i}| = sum_{i < mathrm{cf}(kappa)} kappa^{mu_i} = mathrm{cf}(kappa)cdotsup_{i<mathrm{cf}(kappa)}kappa^{mu_i} = mathrm{cf}(kappa)cdotkappa^lambda = kappa^lambda$$



      Is this approach a good idea? How would I go about finding the appropriate $mu_i$? I suppose it will have to make use of the assumption that $mathrm{cf}(kappa) leq lambda < kappa$, but I don't see how.



      I'd much appreciate any hints on how to proceed (either with my suggestion or another way).






      set-theory cardinals




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      edited Dec 4 '18 at 16:39









      Asaf Karagila♦

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      asked Dec 4 '18 at 16:35









      ryan221bryan221b

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          $begingroup$

          You fell into a trap that also caught me when I was just starting to read mathematical books. When you see a reference to (5.7) it does not mean Lemma 5.7, but rather the displayed formula whose tag is 5.7, which in this case is the formula:



          $$kappaleqkappa^lambdaleq 2^kappa.$$ Similarly, (5.8) refers to the inequality $$kappa<kappa^lambdaquadtext{ if }lambdageqoperatorname{cf}kappa.$$



          And indeed these are the inequalities needed to prove (i) and (ii).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh! That makes a lot more of sense. Thanks! I think it was more confusing as the line immediately above cites Lemma 5.6, then it cites (5.7) and (5.8).
            $endgroup$
            – ryan221b
            Dec 4 '18 at 16:55











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          $begingroup$

          You fell into a trap that also caught me when I was just starting to read mathematical books. When you see a reference to (5.7) it does not mean Lemma 5.7, but rather the displayed formula whose tag is 5.7, which in this case is the formula:



          $$kappaleqkappa^lambdaleq 2^kappa.$$ Similarly, (5.8) refers to the inequality $$kappa<kappa^lambdaquadtext{ if }lambdageqoperatorname{cf}kappa.$$



          And indeed these are the inequalities needed to prove (i) and (ii).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh! That makes a lot more of sense. Thanks! I think it was more confusing as the line immediately above cites Lemma 5.6, then it cites (5.7) and (5.8).
            $endgroup$
            – ryan221b
            Dec 4 '18 at 16:55
















          1












          $begingroup$

          You fell into a trap that also caught me when I was just starting to read mathematical books. When you see a reference to (5.7) it does not mean Lemma 5.7, but rather the displayed formula whose tag is 5.7, which in this case is the formula:



          $$kappaleqkappa^lambdaleq 2^kappa.$$ Similarly, (5.8) refers to the inequality $$kappa<kappa^lambdaquadtext{ if }lambdageqoperatorname{cf}kappa.$$



          And indeed these are the inequalities needed to prove (i) and (ii).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh! That makes a lot more of sense. Thanks! I think it was more confusing as the line immediately above cites Lemma 5.6, then it cites (5.7) and (5.8).
            $endgroup$
            – ryan221b
            Dec 4 '18 at 16:55














          1












          1








          1





          $begingroup$

          You fell into a trap that also caught me when I was just starting to read mathematical books. When you see a reference to (5.7) it does not mean Lemma 5.7, but rather the displayed formula whose tag is 5.7, which in this case is the formula:



          $$kappaleqkappa^lambdaleq 2^kappa.$$ Similarly, (5.8) refers to the inequality $$kappa<kappa^lambdaquadtext{ if }lambdageqoperatorname{cf}kappa.$$



          And indeed these are the inequalities needed to prove (i) and (ii).






          share|cite|improve this answer









          $endgroup$



          You fell into a trap that also caught me when I was just starting to read mathematical books. When you see a reference to (5.7) it does not mean Lemma 5.7, but rather the displayed formula whose tag is 5.7, which in this case is the formula:



          $$kappaleqkappa^lambdaleq 2^kappa.$$ Similarly, (5.8) refers to the inequality $$kappa<kappa^lambdaquadtext{ if }lambdageqoperatorname{cf}kappa.$$



          And indeed these are the inequalities needed to prove (i) and (ii).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 16:44









          Asaf Karagila♦Asaf Karagila

          302k32427757




          302k32427757












          • $begingroup$
            Oh! That makes a lot more of sense. Thanks! I think it was more confusing as the line immediately above cites Lemma 5.6, then it cites (5.7) and (5.8).
            $endgroup$
            – ryan221b
            Dec 4 '18 at 16:55


















          • $begingroup$
            Oh! That makes a lot more of sense. Thanks! I think it was more confusing as the line immediately above cites Lemma 5.6, then it cites (5.7) and (5.8).
            $endgroup$
            – ryan221b
            Dec 4 '18 at 16:55
















          $begingroup$
          Oh! That makes a lot more of sense. Thanks! I think it was more confusing as the line immediately above cites Lemma 5.6, then it cites (5.7) and (5.8).
          $endgroup$
          – ryan221b
          Dec 4 '18 at 16:55




          $begingroup$
          Oh! That makes a lot more of sense. Thanks! I think it was more confusing as the line immediately above cites Lemma 5.6, then it cites (5.7) and (5.8).
          $endgroup$
          – ryan221b
          Dec 4 '18 at 16:55


















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