Mixing
Is every limit cardinal $alpha$ of the form $alpha=card(X)+card(P(X))+cardP(P(X))+…$, for some set $X$?
$begingroup$
A limit cardinal is a cardinal number $alpha$ such that if $beta<alpha,$ then there is a cardinal number $gamma$ with $beta<gamma<alpha.$
Now if $alpha$ is a limit cardinal then can we find a set $X$ which satisfies
$$alpha= card X + card P(X) + card P(P(X))+...?$$
set-theory
asked Jan 3 at 6:52
aliali
38418
$endgroup$
add a comment |
$begingroup$
A limit cardinal is a cardinal number $alpha$ such that if $beta<alpha,$ then there is a cardinal number $gamma$ with $beta<gamma<alpha.$
Now if $alpha$ is a limit cardinal then can we find a set $X$ which satisfies
$$alpha= card X + card P(X) + card P(P(X))+...?$$
set-theory
asked Jan 3 at 6:52
aliali
38418
$endgroup$
add a comment |
$begingroup$
A limit cardinal is a cardinal number $alpha$ such that if $beta<alpha,$ then there is a cardinal number $gamma$ with $beta<gamma<alpha.$
Now if $alpha$ is a limit cardinal then can we find a set $X$ which satisfies
$$alpha= card X + card P(X) + card P(P(X))+...?$$
set-theory
asked Jan 3 at 6:52
aliali
38418
$endgroup$
A limit cardinal is a cardinal number $alpha$ such that if $beta<alpha,$ then there is a cardinal number $gamma$ with $beta<gamma<alpha.$
Now if $alpha$ is a limit cardinal then can we find a set $X$ which satisfies
$$alpha= card X + card P(X) + card P(P(X))+...?$$
set-theory
set-theory
asked Jan 3 at 6:52
aliali
38418
asked Jan 3 at 6:52
aliali
38418
asked Jan 3 at 6:52
aliali
38418
asked Jan 3 at 6:52
aliali
38418
asked Jan 3 at 6:52
aliali
38418
38418
add a comment |
add a comment |
2 Answers
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$begingroup$
No. An uncountable cardinal with that property is a strong limit cardinal of countable cofinality. An uncountable cardinal $a$ such that $forall b<a, (2^b<a)$ is called a strong limit cardinal.
GCH is consistent with ZFC (Kurt Godel, 1930's), and GCH implies that all limit cardinals are strong limits.
It is consistent with ZFC that $2^{omega}>omega_{omega}$ (Paul Cohen, 1960's) so it is consistent with ZFC that the limit cardinal $omega_{omega}$ is not a strong limit.
In ZFC the recursively-defined beth function maps the class $On$ of ordinals into the class of cardinal ordinals: $beth(0)=omega,$ and $beth(b+1)=2^{beth(b)}$ , and if $0ne b=cup b$ then $beth(b)=cup_{cin b}beth(c).$ Consider $beth (omega_1)=cup_{cin omega_1}beth(c),$ which is a strong limit cardinal of cofinality $omega_1.$
answered Jan 3 at 7:48
DanielWainfleetDanielWainfleet
34.6k31648
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No. at least not without assuming GCH. Take $aleph_ω$ and look at the model of ZFC where $2^{aleph_0}=aleph_{ω+1}$.
Even assuming $GCH$, take $aleph_{ω_1}$, and assume that your proposition is true, you get that $mbox{cof}(aleph_{ω_1})=mbox{cof}(ω_1)=aleph_0$, which is false.
answered Jan 3 at 6:58
HoloHolo
5,60321030
$endgroup$
$begingroup$
It seems $α=ω+ω$ is a limit ordinal but not a limit cardinal.
$endgroup$
– ali
Jan 3 at 7:07
$begingroup$
@ali sorry, I misread, edit it
$endgroup$
– Holo
Jan 3 at 7:11
$begingroup$
@ali it should be good now
$endgroup$
– Holo
Jan 3 at 7:20
add a comment |
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2 Answers
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2 Answers
2
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active
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active
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$begingroup$
No. An uncountable cardinal with that property is a strong limit cardinal of countable cofinality. An uncountable cardinal $a$ such that $forall b<a, (2^b<a)$ is called a strong limit cardinal.
GCH is consistent with ZFC (Kurt Godel, 1930's), and GCH implies that all limit cardinals are strong limits.
It is consistent with ZFC that $2^{omega}>omega_{omega}$ (Paul Cohen, 1960's) so it is consistent with ZFC that the limit cardinal $omega_{omega}$ is not a strong limit.
In ZFC the recursively-defined beth function maps the class $On$ of ordinals into the class of cardinal ordinals: $beth(0)=omega,$ and $beth(b+1)=2^{beth(b)}$ , and if $0ne b=cup b$ then $beth(b)=cup_{cin b}beth(c).$ Consider $beth (omega_1)=cup_{cin omega_1}beth(c),$ which is a strong limit cardinal of cofinality $omega_1.$
answered Jan 3 at 7:48
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
add a comment |
$begingroup$
No. An uncountable cardinal with that property is a strong limit cardinal of countable cofinality. An uncountable cardinal $a$ such that $forall b<a, (2^b<a)$ is called a strong limit cardinal.
GCH is consistent with ZFC (Kurt Godel, 1930's), and GCH implies that all limit cardinals are strong limits.
It is consistent with ZFC that $2^{omega}>omega_{omega}$ (Paul Cohen, 1960's) so it is consistent with ZFC that the limit cardinal $omega_{omega}$ is not a strong limit.
In ZFC the recursively-defined beth function maps the class $On$ of ordinals into the class of cardinal ordinals: $beth(0)=omega,$ and $beth(b+1)=2^{beth(b)}$ , and if $0ne b=cup b$ then $beth(b)=cup_{cin b}beth(c).$ Consider $beth (omega_1)=cup_{cin omega_1}beth(c),$ which is a strong limit cardinal of cofinality $omega_1.$
answered Jan 3 at 7:48
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
add a comment |
$begingroup$
No. An uncountable cardinal with that property is a strong limit cardinal of countable cofinality. An uncountable cardinal $a$ such that $forall b<a, (2^b<a)$ is called a strong limit cardinal.
GCH is consistent with ZFC (Kurt Godel, 1930's), and GCH implies that all limit cardinals are strong limits.
It is consistent with ZFC that $2^{omega}>omega_{omega}$ (Paul Cohen, 1960's) so it is consistent with ZFC that the limit cardinal $omega_{omega}$ is not a strong limit.
In ZFC the recursively-defined beth function maps the class $On$ of ordinals into the class of cardinal ordinals: $beth(0)=omega,$ and $beth(b+1)=2^{beth(b)}$ , and if $0ne b=cup b$ then $beth(b)=cup_{cin b}beth(c).$ Consider $beth (omega_1)=cup_{cin omega_1}beth(c),$ which is a strong limit cardinal of cofinality $omega_1.$
answered Jan 3 at 7:48
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
No. An uncountable cardinal with that property is a strong limit cardinal of countable cofinality. An uncountable cardinal $a$ such that $forall b<a, (2^b<a)$ is called a strong limit cardinal.
GCH is consistent with ZFC (Kurt Godel, 1930's), and GCH implies that all limit cardinals are strong limits.
It is consistent with ZFC that $2^{omega}>omega_{omega}$ (Paul Cohen, 1960's) so it is consistent with ZFC that the limit cardinal $omega_{omega}$ is not a strong limit.
In ZFC the recursively-defined beth function maps the class $On$ of ordinals into the class of cardinal ordinals: $beth(0)=omega,$ and $beth(b+1)=2^{beth(b)}$ , and if $0ne b=cup b$ then $beth(b)=cup_{cin b}beth(c).$ Consider $beth (omega_1)=cup_{cin omega_1}beth(c),$ which is a strong limit cardinal of cofinality $omega_1.$
answered Jan 3 at 7:48
DanielWainfleetDanielWainfleet
34.6k31648
answered Jan 3 at 7:48
DanielWainfleetDanielWainfleet
34.6k31648
answered Jan 3 at 7:48
DanielWainfleetDanielWainfleet
34.6k31648
answered Jan 3 at 7:48
DanielWainfleetDanielWainfleet
34.6k31648
34.6k31648
add a comment |
add a comment |
$begingroup$
No. at least not without assuming GCH. Take $aleph_ω$ and look at the model of ZFC where $2^{aleph_0}=aleph_{ω+1}$.
Even assuming $GCH$, take $aleph_{ω_1}$, and assume that your proposition is true, you get that $mbox{cof}(aleph_{ω_1})=mbox{cof}(ω_1)=aleph_0$, which is false.
answered Jan 3 at 6:58
HoloHolo
5,60321030
$endgroup$
$begingroup$
It seems $α=ω+ω$ is a limit ordinal but not a limit cardinal.
$endgroup$
– ali
Jan 3 at 7:07
$begingroup$
@ali sorry, I misread, edit it
$endgroup$
– Holo
Jan 3 at 7:11
$begingroup$
@ali it should be good now
$endgroup$
– Holo
Jan 3 at 7:20
add a comment |
$begingroup$
No. at least not without assuming GCH. Take $aleph_ω$ and look at the model of ZFC where $2^{aleph_0}=aleph_{ω+1}$.
Even assuming $GCH$, take $aleph_{ω_1}$, and assume that your proposition is true, you get that $mbox{cof}(aleph_{ω_1})=mbox{cof}(ω_1)=aleph_0$, which is false.
answered Jan 3 at 6:58
HoloHolo
5,60321030
$endgroup$
$begingroup$
It seems $α=ω+ω$ is a limit ordinal but not a limit cardinal.
$endgroup$
– ali
Jan 3 at 7:07
$begingroup$
@ali sorry, I misread, edit it
$endgroup$
– Holo
Jan 3 at 7:11
$begingroup$
@ali it should be good now
$endgroup$
– Holo
Jan 3 at 7:20
add a comment |
$begingroup$
No. at least not without assuming GCH. Take $aleph_ω$ and look at the model of ZFC where $2^{aleph_0}=aleph_{ω+1}$.
Even assuming $GCH$, take $aleph_{ω_1}$, and assume that your proposition is true, you get that $mbox{cof}(aleph_{ω_1})=mbox{cof}(ω_1)=aleph_0$, which is false.
answered Jan 3 at 6:58
HoloHolo
5,60321030
$endgroup$
No. at least not without assuming GCH. Take $aleph_ω$ and look at the model of ZFC where $2^{aleph_0}=aleph_{ω+1}$.
Even assuming $GCH$, take $aleph_{ω_1}$, and assume that your proposition is true, you get that $mbox{cof}(aleph_{ω_1})=mbox{cof}(ω_1)=aleph_0$, which is false.
answered Jan 3 at 6:58
HoloHolo
5,60321030
edited Jan 3 at 7:16
answered Jan 3 at 6:58
HoloHolo
5,60321030
answered Jan 3 at 6:58
HoloHolo
5,60321030
answered Jan 3 at 6:58
HoloHolo
5,60321030
5,60321030
$begingroup$
It seems $α=ω+ω$ is a limit ordinal but not a limit cardinal.
$endgroup$
– ali
Jan 3 at 7:07
$begingroup$
@ali sorry, I misread, edit it
$endgroup$
– Holo
Jan 3 at 7:11
$begingroup$
@ali it should be good now
$endgroup$
– Holo
Jan 3 at 7:20
add a comment |
$begingroup$
It seems $α=ω+ω$ is a limit ordinal but not a limit cardinal.
$endgroup$
– ali
Jan 3 at 7:07
$begingroup$
@ali sorry, I misread, edit it
$endgroup$
– Holo
Jan 3 at 7:11
$begingroup$
@ali it should be good now
$endgroup$
– Holo
Jan 3 at 7:20
$begingroup$
It seems $α=ω+ω$ is a limit ordinal but not a limit cardinal.
$endgroup$
– ali
Jan 3 at 7:07
$begingroup$
It seems $α=ω+ω$ is a limit ordinal but not a limit cardinal.
$endgroup$
– ali
Jan 3 at 7:07
$begingroup$
@ali sorry, I misread, edit it
$endgroup$
– Holo
Jan 3 at 7:11
$begingroup$
@ali sorry, I misread, edit it
$endgroup$
– Holo
Jan 3 at 7:11
$begingroup$
@ali it should be good now
$endgroup$
– Holo
Jan 3 at 7:20
$begingroup$
@ali it should be good now
$endgroup$
– Holo
Jan 3 at 7:20
add a comment |
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