Almost free modules, PCF theory bound on $2^{aleph_omega}$












0












$begingroup$


Why here (Almost Free Modules: Set-theoretic Methods by P.C. Eklof, A.H. Mekler), on the page 181, in the 3rd line there is
$$2^{aleph_omega}<aleph_{omega_2}$$? How the index $2$ in the r.h.s. was created?




The notation "pcf" stands for possible cofinality. Shelahintroduced a powerful tool into the study of cardinal arithmetic when he considered (for an infinite set $A$ of regular cardinals) the set $operatorname{pcf}(A)$, defined to be the set of all cardinals $lambda$ such that for some ultrafilter $U$ on $A$, $lambda$ is the cofinality of the ultraproduct of ordered sets $prod_{kappain A} kappa/U$. Among the striking results he proved (in ZFC) are: if $2^{aleph_n}<aleph_omega$ for all $ninomega$ then $2^{aleph_omega}<min{aleph_{(2^{aleph_0})^+},aleph_{omega_4}}$; in particular, if GCH holds below $aleph_omega$, then $2^{aleph_omega}<aleph_{omega_2}$. (See Shelah 1992, Shelah 1994 or Jech 1995 or Burke-Magidor 1990.)




page 180



page 181






set-theory cardinals







share|cite|improve this question











$endgroup$












  • $begingroup$
    could someone tell me what $omega_2$ is supposed to mean? I've never seen this notation before.
    $endgroup$
    – Pink Panther
    Dec 18 '18 at 21:41






  • 1




    $begingroup$
    @PinkPanther: It is the second uncountable cardinal, or the third infinite cardinal.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 22:07






  • 2




    $begingroup$
    I don't understand the question; the text you quoted gives the reason for the bound and provides references for the proof. It is a combination of two results of Shelah. One first appeared in the first edition of his "Proper forcing" book (it is not in the revised edition), and the other is in his "Cardinal arithmetic" book.
    $endgroup$
    – Andrés E. Caicedo
    Dec 19 '18 at 12:26










  • $begingroup$
    OK then. I just wondered if $2$ should be read as $4$, but you say that "no".
    $endgroup$
    – user122424
    Dec 19 '18 at 15:03








  • 2




    $begingroup$
    Yes, in fact you do not need the results from "Cardinal arithmetic", just the earlier result in the Proper forcing book.
    $endgroup$
    – Andrés E. Caicedo
    Dec 19 '18 at 16:35
















0












$begingroup$


Why here (Almost Free Modules: Set-theoretic Methods by P.C. Eklof, A.H. Mekler), on the page 181, in the 3rd line there is
$$2^{aleph_omega}<aleph_{omega_2}$$? How the index $2$ in the r.h.s. was created?




The notation "pcf" stands for possible cofinality. Shelahintroduced a powerful tool into the study of cardinal arithmetic when he considered (for an infinite set $A$ of regular cardinals) the set $operatorname{pcf}(A)$, defined to be the set of all cardinals $lambda$ such that for some ultrafilter $U$ on $A$, $lambda$ is the cofinality of the ultraproduct of ordered sets $prod_{kappain A} kappa/U$. Among the striking results he proved (in ZFC) are: if $2^{aleph_n}<aleph_omega$ for all $ninomega$ then $2^{aleph_omega}<min{aleph_{(2^{aleph_0})^+},aleph_{omega_4}}$; in particular, if GCH holds below $aleph_omega$, then $2^{aleph_omega}<aleph_{omega_2}$. (See Shelah 1992, Shelah 1994 or Jech 1995 or Burke-Magidor 1990.)




page 180



page 181






set-theory cardinals







share|cite|improve this question











$endgroup$












  • $begingroup$
    could someone tell me what $omega_2$ is supposed to mean? I've never seen this notation before.
    $endgroup$
    – Pink Panther
    Dec 18 '18 at 21:41






  • 1




    $begingroup$
    @PinkPanther: It is the second uncountable cardinal, or the third infinite cardinal.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 22:07






  • 2




    $begingroup$
    I don't understand the question; the text you quoted gives the reason for the bound and provides references for the proof. It is a combination of two results of Shelah. One first appeared in the first edition of his "Proper forcing" book (it is not in the revised edition), and the other is in his "Cardinal arithmetic" book.
    $endgroup$
    – Andrés E. Caicedo
    Dec 19 '18 at 12:26










  • $begingroup$
    OK then. I just wondered if $2$ should be read as $4$, but you say that "no".
    $endgroup$
    – user122424
    Dec 19 '18 at 15:03








  • 2




    $begingroup$
    Yes, in fact you do not need the results from "Cardinal arithmetic", just the earlier result in the Proper forcing book.
    $endgroup$
    – Andrés E. Caicedo
    Dec 19 '18 at 16:35














0












0








0





$begingroup$


Why here (Almost Free Modules: Set-theoretic Methods by P.C. Eklof, A.H. Mekler), on the page 181, in the 3rd line there is
$$2^{aleph_omega}<aleph_{omega_2}$$? How the index $2$ in the r.h.s. was created?




The notation "pcf" stands for possible cofinality. Shelahintroduced a powerful tool into the study of cardinal arithmetic when he considered (for an infinite set $A$ of regular cardinals) the set $operatorname{pcf}(A)$, defined to be the set of all cardinals $lambda$ such that for some ultrafilter $U$ on $A$, $lambda$ is the cofinality of the ultraproduct of ordered sets $prod_{kappain A} kappa/U$. Among the striking results he proved (in ZFC) are: if $2^{aleph_n}<aleph_omega$ for all $ninomega$ then $2^{aleph_omega}<min{aleph_{(2^{aleph_0})^+},aleph_{omega_4}}$; in particular, if GCH holds below $aleph_omega$, then $2^{aleph_omega}<aleph_{omega_2}$. (See Shelah 1992, Shelah 1994 or Jech 1995 or Burke-Magidor 1990.)




page 180



page 181






set-theory cardinals







share|cite|improve this question











$endgroup$




Why here (Almost Free Modules: Set-theoretic Methods by P.C. Eklof, A.H. Mekler), on the page 181, in the 3rd line there is
$$2^{aleph_omega}<aleph_{omega_2}$$? How the index $2$ in the r.h.s. was created?




The notation "pcf" stands for possible cofinality. Shelahintroduced a powerful tool into the study of cardinal arithmetic when he considered (for an infinite set $A$ of regular cardinals) the set $operatorname{pcf}(A)$, defined to be the set of all cardinals $lambda$ such that for some ultrafilter $U$ on $A$, $lambda$ is the cofinality of the ultraproduct of ordered sets $prod_{kappain A} kappa/U$. Among the striking results he proved (in ZFC) are: if $2^{aleph_n}<aleph_omega$ for all $ninomega$ then $2^{aleph_omega}<min{aleph_{(2^{aleph_0})^+},aleph_{omega_4}}$; in particular, if GCH holds below $aleph_omega$, then $2^{aleph_omega}<aleph_{omega_2}$. (See Shelah 1992, Shelah 1994 or Jech 1995 or Burke-Magidor 1990.)




page 180



page 181






set-theory cardinals




set-theory cardinals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 18:10









Lord_Farin

15.5k636108




15.5k636108










asked Dec 18 '18 at 21:13









user122424user122424

1,1182716




1,1182716












  • $begingroup$
    could someone tell me what $omega_2$ is supposed to mean? I've never seen this notation before.
    $endgroup$
    – Pink Panther
    Dec 18 '18 at 21:41






  • 1




    $begingroup$
    @PinkPanther: It is the second uncountable cardinal, or the third infinite cardinal.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 22:07






  • 2




    $begingroup$
    I don't understand the question; the text you quoted gives the reason for the bound and provides references for the proof. It is a combination of two results of Shelah. One first appeared in the first edition of his "Proper forcing" book (it is not in the revised edition), and the other is in his "Cardinal arithmetic" book.
    $endgroup$
    – Andrés E. Caicedo
    Dec 19 '18 at 12:26










  • $begingroup$
    OK then. I just wondered if $2$ should be read as $4$, but you say that "no".
    $endgroup$
    – user122424
    Dec 19 '18 at 15:03








  • 2




    $begingroup$
    Yes, in fact you do not need the results from "Cardinal arithmetic", just the earlier result in the Proper forcing book.
    $endgroup$
    – Andrés E. Caicedo
    Dec 19 '18 at 16:35


















  • $begingroup$
    could someone tell me what $omega_2$ is supposed to mean? I've never seen this notation before.
    $endgroup$
    – Pink Panther
    Dec 18 '18 at 21:41






  • 1




    $begingroup$
    @PinkPanther: It is the second uncountable cardinal, or the third infinite cardinal.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 22:07






  • 2




    $begingroup$
    I don't understand the question; the text you quoted gives the reason for the bound and provides references for the proof. It is a combination of two results of Shelah. One first appeared in the first edition of his "Proper forcing" book (it is not in the revised edition), and the other is in his "Cardinal arithmetic" book.
    $endgroup$
    – Andrés E. Caicedo
    Dec 19 '18 at 12:26










  • $begingroup$
    OK then. I just wondered if $2$ should be read as $4$, but you say that "no".
    $endgroup$
    – user122424
    Dec 19 '18 at 15:03








  • 2




    $begingroup$
    Yes, in fact you do not need the results from "Cardinal arithmetic", just the earlier result in the Proper forcing book.
    $endgroup$
    – Andrés E. Caicedo
    Dec 19 '18 at 16:35
















$begingroup$
could someone tell me what $omega_2$ is supposed to mean? I've never seen this notation before.
$endgroup$
– Pink Panther
Dec 18 '18 at 21:41




$begingroup$
could someone tell me what $omega_2$ is supposed to mean? I've never seen this notation before.
$endgroup$
– Pink Panther
Dec 18 '18 at 21:41




1




1




$begingroup$
@PinkPanther: It is the second uncountable cardinal, or the third infinite cardinal.
$endgroup$
– Asaf Karagila♦
Dec 18 '18 at 22:07




$begingroup$
@PinkPanther: It is the second uncountable cardinal, or the third infinite cardinal.
$endgroup$
– Asaf Karagila♦
Dec 18 '18 at 22:07




2




2




$begingroup$
I don't understand the question; the text you quoted gives the reason for the bound and provides references for the proof. It is a combination of two results of Shelah. One first appeared in the first edition of his "Proper forcing" book (it is not in the revised edition), and the other is in his "Cardinal arithmetic" book.
$endgroup$
– Andrés E. Caicedo
Dec 19 '18 at 12:26




$begingroup$
I don't understand the question; the text you quoted gives the reason for the bound and provides references for the proof. It is a combination of two results of Shelah. One first appeared in the first edition of his "Proper forcing" book (it is not in the revised edition), and the other is in his "Cardinal arithmetic" book.
$endgroup$
– Andrés E. Caicedo
Dec 19 '18 at 12:26












$begingroup$
OK then. I just wondered if $2$ should be read as $4$, but you say that "no".
$endgroup$
– user122424
Dec 19 '18 at 15:03






$begingroup$
OK then. I just wondered if $2$ should be read as $4$, but you say that "no".
$endgroup$
– user122424
Dec 19 '18 at 15:03






2




2




$begingroup$
Yes, in fact you do not need the results from "Cardinal arithmetic", just the earlier result in the Proper forcing book.
$endgroup$
– Andrés E. Caicedo
Dec 19 '18 at 16:35




$begingroup$
Yes, in fact you do not need the results from "Cardinal arithmetic", just the earlier result in the Proper forcing book.
$endgroup$
– Andrés E. Caicedo
Dec 19 '18 at 16:35










1 Answer
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$begingroup$

The $aleph_{omega_2}$ arises because under the assumption of GCH below $aleph_omega$, we have:



$$(2^{aleph_0})^+ = aleph_1^+ = omega_2$$



so that the minimum $min{aleph_{(2^{aleph_0})^+},aleph_{omega_4}}$ is actually not $aleph_{omega_4}$, but the other argument, which evaluates to $aleph_{omega_2}$.






share|cite|improve this answer









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    $begingroup$

    The $aleph_{omega_2}$ arises because under the assumption of GCH below $aleph_omega$, we have:



    $$(2^{aleph_0})^+ = aleph_1^+ = omega_2$$



    so that the minimum $min{aleph_{(2^{aleph_0})^+},aleph_{omega_4}}$ is actually not $aleph_{omega_4}$, but the other argument, which evaluates to $aleph_{omega_2}$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The $aleph_{omega_2}$ arises because under the assumption of GCH below $aleph_omega$, we have:



      $$(2^{aleph_0})^+ = aleph_1^+ = omega_2$$



      so that the minimum $min{aleph_{(2^{aleph_0})^+},aleph_{omega_4}}$ is actually not $aleph_{omega_4}$, but the other argument, which evaluates to $aleph_{omega_2}$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The $aleph_{omega_2}$ arises because under the assumption of GCH below $aleph_omega$, we have:



        $$(2^{aleph_0})^+ = aleph_1^+ = omega_2$$



        so that the minimum $min{aleph_{(2^{aleph_0})^+},aleph_{omega_4}}$ is actually not $aleph_{omega_4}$, but the other argument, which evaluates to $aleph_{omega_2}$.






        share|cite|improve this answer









        $endgroup$



        The $aleph_{omega_2}$ arises because under the assumption of GCH below $aleph_omega$, we have:



        $$(2^{aleph_0})^+ = aleph_1^+ = omega_2$$



        so that the minimum $min{aleph_{(2^{aleph_0})^+},aleph_{omega_4}}$ is actually not $aleph_{omega_4}$, but the other argument, which evaluates to $aleph_{omega_2}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 18:10









        Lord_FarinLord_Farin

        15.5k636108




        15.5k636108






























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