Almost free modules, PCF theory bound on $2^{aleph_omega}$
$begingroup$
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.)
set-theory cardinals
edited Dec 19 '18 at 18:10
Lord_Farin
15.5k636108
asked Dec 18 '18 at 21:13
user122424user122424
1,1182716
$endgroup$
|
show 1 more comment
$begingroup$
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.)
set-theory cardinals
edited Dec 19 '18 at 18:10
Lord_Farin
15.5k636108
asked Dec 18 '18 at 21:13
user122424user122424
1,1182716
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.)
set-theory cardinals
edited Dec 19 '18 at 18:10
Lord_Farin
15.5k636108
asked Dec 18 '18 at 21:13
user122424user122424
1,1182716
$endgroup$
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.)
set-theory cardinals
set-theory cardinals
edited Dec 19 '18 at 18:10
Lord_Farin
15.5k636108
asked Dec 18 '18 at 21:13
user122424user122424
1,1182716
edited Dec 19 '18 at 18:10
Lord_Farin
15.5k636108
asked Dec 18 '18 at 21:13
user122424user122424
1,1182716
edited Dec 19 '18 at 18:10
Lord_Farin
15.5k636108
edited Dec 19 '18 at 18:10
Lord_Farin
15.5k636108
edited Dec 19 '18 at 18:10
Lord_Farin
15.5k636108
15.5k636108
asked Dec 18 '18 at 21:13
user122424user122424
1,1182716
asked Dec 18 '18 at 21:13
user122424user122424
1,1182716
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
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 $aleph_{omega_2}$.
answered Dec 19 '18 at 18:10
Lord_FarinLord_Farin
15.5k636108
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add a comment |
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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 $aleph_{omega_2}$.
answered Dec 19 '18 at 18:10
Lord_FarinLord_Farin
15.5k636108
$endgroup$
add a comment |
$begingroup$
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 $aleph_{omega_2}$.
answered Dec 19 '18 at 18:10
Lord_FarinLord_Farin
15.5k636108
$endgroup$
add a comment |
$begingroup$
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 $aleph_{omega_2}$.
answered Dec 19 '18 at 18:10
Lord_FarinLord_Farin
15.5k636108
$endgroup$
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 $aleph_{omega_2}$.
answered Dec 19 '18 at 18:10
Lord_FarinLord_Farin
15.5k636108
answered Dec 19 '18 at 18:10
Lord_FarinLord_Farin
15.5k636108
answered Dec 19 '18 at 18:10
Lord_FarinLord_Farin
15.5k636108
answered Dec 19 '18 at 18:10
Lord_FarinLord_Farin
15.5k636108
15.5k636108
add a comment |
add a comment |
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$begingroup$
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