How can we determine $2^kappa$ for singular $kappa$ assuming that $2^lambda=lambda^+$ whenever...
$begingroup$
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
65.4k8158249
asked Dec 26 '18 at 18:36
Jan T.Jan T.
82
$endgroup$
add a comment |
$begingroup$
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
65.4k8158249
asked Dec 26 '18 at 18:36
Jan T.Jan T.
82
$endgroup$
add a comment |
$begingroup$
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
65.4k8158249
asked Dec 26 '18 at 18:36
Jan T.Jan T.
82
$endgroup$
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
65.4k8158249
asked Dec 26 '18 at 18:36
Jan T.Jan T.
82
edited Dec 26 '18 at 22:43
Andrés E. Caicedo
65.4k8158249
asked Dec 26 '18 at 18:36
Jan T.Jan T.
82
edited Dec 26 '18 at 22:43
Andrés E. Caicedo
65.4k8158249
edited Dec 26 '18 at 22:43
Andrés E. Caicedo
65.4k8158249
edited Dec 26 '18 at 22:43
Andrés E. Caicedo
65.4k8158249
65.4k8158249
asked Dec 26 '18 at 18:36
Jan T.Jan T.
82
asked Dec 26 '18 at 18:36
Jan T.Jan T.
82
asked Dec 26 '18 at 18:36
Jan T.Jan T.
82
82
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 BlassAndreas Blass
49.8k451108
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053197%2fhow-can-we-determine-2-kappa-for-singular-kappa-assuming-that-2-lambda%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 BlassAndreas Blass
49.8k451108
$endgroup$
add a comment |
$begingroup$
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 BlassAndreas Blass
49.8k451108
$endgroup$
add a comment |
$begingroup$
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 BlassAndreas Blass
49.8k451108
$endgroup$
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 BlassAndreas Blass
49.8k451108
answered Dec 27 '18 at 2:31
Andreas BlassAndreas Blass
49.8k451108
answered Dec 27 '18 at 2:31
Andreas BlassAndreas Blass
49.8k451108
answered Dec 27 '18 at 2:31
Andreas BlassAndreas Blass
49.8k451108
49.8k451108
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053197%2fhow-can-we-determine-2-kappa-for-singular-kappa-assuming-that-2-lambda%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
