How can we determine $2^kappa$ for singular $kappa$ assuming that $2^lambda=lambda^+$ whenever...
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
edited Dec 26 '18 at 22:43
Andrés E. Caicedo
64.7k8158246
asked Dec 26 '18 at 18:36
Jan T.
61
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Jan T. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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...
set-theory cardinals
edited Dec 26 '18 at 22:43
Andrés E. Caicedo
64.7k8158246
asked Dec 26 '18 at 18:36
Jan T.
61
New contributor
Jan T. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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
edited Dec 26 '18 at 22:43
Andrés E. Caicedo
64.7k8158246
asked Dec 26 '18 at 18:36
Jan T.
61
New contributor
Jan T. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
edited Dec 26 '18 at 22:43
Andrés E. Caicedo
64.7k8158246
asked Dec 26 '18 at 18:36
Jan T.
61
New contributor
Jan T. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 26 '18 at 22:43
Andrés E. Caicedo
64.7k8158246
asked Dec 26 '18 at 18:36
Jan T.
61
New contributor
Jan T. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 26 '18 at 22:43
Andrés E. Caicedo
64.7k8158246
edited Dec 26 '18 at 22:43
Andrés E. Caicedo
64.7k8158246
edited Dec 26 '18 at 22:43
Andrés E. Caicedo
64.7k8158246
64.7k8158246
asked Dec 26 '18 at 18:36
Jan T.
61
New contributor
Jan T. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 26 '18 at 18:36
Jan T.
61
asked Dec 26 '18 at 18:36
Jan T.
61
61
New contributor
Jan T. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jan T. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
add a comment |
add a comment |
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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.)
answered Dec 27 '18 at 2:31
Andreas Blass
49.3k351106
add a comment |
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1 Answer
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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.)
answered Dec 27 '18 at 2:31
Andreas Blass
49.3k351106
add a comment |
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.)
answered Dec 27 '18 at 2:31
Andreas Blass
49.3k351106
add a comment |
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.)
answered Dec 27 '18 at 2:31
Andreas Blass
49.3k351106
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.)
answered Dec 27 '18 at 2:31
Andreas Blass
49.3k351106
answered Dec 27 '18 at 2:31
Andreas Blass
49.3k351106
answered Dec 27 '18 at 2:31
Andreas Blass
49.3k351106
answered Dec 27 '18 at 2:31
Andreas Blass
49.3k351106
49.3k351106
add a comment |
add a comment |
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