Mixing
Re. Jech Set Theory, Theorem 7.8 ($2^{aleph_0} = aleph_1 implies exists: text{Ramsey ultrafilter}$)
$begingroup$
Theorem 7.8 in Jech's Set Theory states that if $2^{aleph_0} = aleph_1$, there exists a Ramsey ultrafilter. The proof is constructive: We enumerate all partitions of $omega$ (denoted $mathcal{A}_alpha$, where $alpha = 1, 2, ldots < omega_1$) and define $X_{alpha + 1} subseteq X_alpha$ as either a subset of some $A in mathcal{A}_alpha$ or such that $|X_{alpha+1} cap A| leq 1 forall A in mathcal{A}_alpha$. If $alpha$ is a limit ordinal, then $X_alpha$ is such that $X_alpha - X_beta$ is finite for all $beta < alpha$. The desired Ramsey ultrafilter is then given by ${X: X_alpha subseteq X, text{for some $alpha$}}$.
My question is regarding the assumption that $2^{aleph_0} = aleph_1$; I don't see for example why the suggested construction could not also be applied if $2^{aleph_0} = aleph_n$, where $n in mathbb{N}$. Am I right in assuming that the proposed assumption is just a (weak?) sufficient condition?
set-theory filters
edited Jan 29 at 0:03
Karl Kronenfeld
4,40511525
asked Jan 28 at 23:25
user480881user480881
1118
$endgroup$
add a comment |
$begingroup$
Theorem 7.8 in Jech's Set Theory states that if $2^{aleph_0} = aleph_1$, there exists a Ramsey ultrafilter. The proof is constructive: We enumerate all partitions of $omega$ (denoted $mathcal{A}_alpha$, where $alpha = 1, 2, ldots < omega_1$) and define $X_{alpha + 1} subseteq X_alpha$ as either a subset of some $A in mathcal{A}_alpha$ or such that $|X_{alpha+1} cap A| leq 1 forall A in mathcal{A}_alpha$. If $alpha$ is a limit ordinal, then $X_alpha$ is such that $X_alpha - X_beta$ is finite for all $beta < alpha$. The desired Ramsey ultrafilter is then given by ${X: X_alpha subseteq X, text{for some $alpha$}}$.
My question is regarding the assumption that $2^{aleph_0} = aleph_1$; I don't see for example why the suggested construction could not also be applied if $2^{aleph_0} = aleph_n$, where $n in mathbb{N}$. Am I right in assuming that the proposed assumption is just a (weak?) sufficient condition?
set-theory filters
edited Jan 29 at 0:03
Karl Kronenfeld
4,40511525
asked Jan 28 at 23:25
user480881user480881
1118
$endgroup$
2
$begingroup$
In the proof, Jech explicitly says that the existence of $X_{alpha}$ depends on the fact that $alpha$ is countable.
$endgroup$
– Karl Kronenfeld
Jan 29 at 0:05
$begingroup$
There had been several questions about this already. Did you look around the site?
$endgroup$
– Asaf Karagila♦
Jan 29 at 8:17
$begingroup$
Dear Karl and Asaf, Thank you for your comments. I have clarified my question in a comment to Eric's answer.
$endgroup$
– user480881
Feb 4 at 11:42
add a comment |
$begingroup$
Theorem 7.8 in Jech's Set Theory states that if $2^{aleph_0} = aleph_1$, there exists a Ramsey ultrafilter. The proof is constructive: We enumerate all partitions of $omega$ (denoted $mathcal{A}_alpha$, where $alpha = 1, 2, ldots < omega_1$) and define $X_{alpha + 1} subseteq X_alpha$ as either a subset of some $A in mathcal{A}_alpha$ or such that $|X_{alpha+1} cap A| leq 1 forall A in mathcal{A}_alpha$. If $alpha$ is a limit ordinal, then $X_alpha$ is such that $X_alpha - X_beta$ is finite for all $beta < alpha$. The desired Ramsey ultrafilter is then given by ${X: X_alpha subseteq X, text{for some $alpha$}}$.
My question is regarding the assumption that $2^{aleph_0} = aleph_1$; I don't see for example why the suggested construction could not also be applied if $2^{aleph_0} = aleph_n$, where $n in mathbb{N}$. Am I right in assuming that the proposed assumption is just a (weak?) sufficient condition?
set-theory filters
edited Jan 29 at 0:03
Karl Kronenfeld
4,40511525
asked Jan 28 at 23:25
user480881user480881
1118
$endgroup$
Theorem 7.8 in Jech's Set Theory states that if $2^{aleph_0} = aleph_1$, there exists a Ramsey ultrafilter. The proof is constructive: We enumerate all partitions of $omega$ (denoted $mathcal{A}_alpha$, where $alpha = 1, 2, ldots < omega_1$) and define $X_{alpha + 1} subseteq X_alpha$ as either a subset of some $A in mathcal{A}_alpha$ or such that $|X_{alpha+1} cap A| leq 1 forall A in mathcal{A}_alpha$. If $alpha$ is a limit ordinal, then $X_alpha$ is such that $X_alpha - X_beta$ is finite for all $beta < alpha$. The desired Ramsey ultrafilter is then given by ${X: X_alpha subseteq X, text{for some $alpha$}}$.
My question is regarding the assumption that $2^{aleph_0} = aleph_1$; I don't see for example why the suggested construction could not also be applied if $2^{aleph_0} = aleph_n$, where $n in mathbb{N}$. Am I right in assuming that the proposed assumption is just a (weak?) sufficient condition?
set-theory filters
set-theory filters
edited Jan 29 at 0:03
Karl Kronenfeld
4,40511525
asked Jan 28 at 23:25
user480881user480881
1118
edited Jan 29 at 0:03
Karl Kronenfeld
4,40511525
asked Jan 28 at 23:25
user480881user480881
1118
edited Jan 29 at 0:03
Karl Kronenfeld
4,40511525
edited Jan 29 at 0:03
Karl Kronenfeld
4,40511525
edited Jan 29 at 0:03
Karl Kronenfeld
4,40511525
4,40511525
asked Jan 28 at 23:25
user480881user480881
1118
asked Jan 28 at 23:25
user480881user480881
1118
asked Jan 28 at 23:25
user480881user480881
1118
1118
2
$begingroup$
In the proof, Jech explicitly says that the existence of $X_{alpha}$ depends on the fact that $alpha$ is countable.
$endgroup$
– Karl Kronenfeld
Jan 29 at 0:05
$begingroup$
There had been several questions about this already. Did you look around the site?
$endgroup$
– Asaf Karagila♦
Jan 29 at 8:17
$begingroup$
Dear Karl and Asaf, Thank you for your comments. I have clarified my question in a comment to Eric's answer.
$endgroup$
– user480881
Feb 4 at 11:42
add a comment |
2
$begingroup$
In the proof, Jech explicitly says that the existence of $X_{alpha}$ depends on the fact that $alpha$ is countable.
$endgroup$
– Karl Kronenfeld
Jan 29 at 0:05
$begingroup$
There had been several questions about this already. Did you look around the site?
$endgroup$
– Asaf Karagila♦
Jan 29 at 8:17
$begingroup$
Dear Karl and Asaf, Thank you for your comments. I have clarified my question in a comment to Eric's answer.
$endgroup$
– user480881
Feb 4 at 11:42
2
2
$begingroup$
In the proof, Jech explicitly says that the existence of $X_{alpha}$ depends on the fact that $alpha$ is countable.
$endgroup$
– Karl Kronenfeld
Jan 29 at 0:05
$begingroup$
In the proof, Jech explicitly says that the existence of $X_{alpha}$ depends on the fact that $alpha$ is countable.
$endgroup$
– Karl Kronenfeld
Jan 29 at 0:05
$begingroup$
There had been several questions about this already. Did you look around the site?
$endgroup$
– Asaf Karagila♦
Jan 29 at 8:17
$begingroup$
There had been several questions about this already. Did you look around the site?
$endgroup$
– Asaf Karagila♦
Jan 29 at 8:17
$begingroup$
Dear Karl and Asaf, Thank you for your comments. I have clarified my question in a comment to Eric's answer.
$endgroup$
– user480881
Feb 4 at 11:42
$begingroup$
Dear Karl and Asaf, Thank you for your comments. I have clarified my question in a comment to Eric's answer.
$endgroup$
– user480881
Feb 4 at 11:42
add a comment |
1 Answer
1
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oldest
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$begingroup$
The construction very crucially uses the assumption that $2^{aleph_0}=aleph_1$ in limit steps. To choose $X_alpha$ such that $X_alpha-X_beta$ is finite for all $beta<alpha$, you must use the fact that there are only countably many such $beta$, so that you can build $X_alpha$ by a diagonal construction so that it is infinite and yet eventually contained in each $X_beta$.
(The argument can be generalized to use weaker assumptions than $2^{aleph_0}=aleph_1$; for instance, Martin's axiom suffices. But the result is not provable in ZFC alone, and I don't know why you think it would be relevant to assume something like $2^{aleph_0}=aleph_n$ for $ninmathbb{N}$.)
answered Jan 29 at 0:07
Eric WofseyEric Wofsey
191k14216349
$endgroup$
$begingroup$
Dear Eric, Thank you for your answer and please excuse the late reply. To make my question more precise: I don't claim that $2^{aleph_0} = aleph_n$ is any way necessary. I am merely pondering the following thought: Suppose for example that $2^{aleph_0} = aleph_2$. What if I construct $X_{omega_1}$ by applying the diagonal construction (of $X_omega$) to ${X_omega, X_{omega + 1}, ldots}$? As far as I can see, the finite-intersection property still holds, and so does the property that $|{X_{omega_1} - X_beta}| < infty, forall beta < omega_1$. Am I missing something?
$endgroup$
– user480881
Feb 4 at 11:38
$begingroup$
What do you mean by "applying the diagonal construction"? How would you apply it to $aleph_1$ sets instead of countably many? The whole point of the "diagonal" is that it uses the fact that ${X_beta:beta<alpha}$, like $omega$, is countable.
$endgroup$
– Eric Wofsey
Feb 4 at 16:21
$begingroup$
Dear Eric, Having returned to Jech's Set Theory after being occupied with different matters for a while, I now see that my misunderstanding stems from the terribly embarrassing fact of me for some reason believing that $omega_1 = 2omega$ (and hence assuming that the construction of $X_omega$ can be continued...). Sorry for wasting your time.
$endgroup$
– user480881
Feb 6 at 11:52
add a comment |
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1 Answer
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$begingroup$
The construction very crucially uses the assumption that $2^{aleph_0}=aleph_1$ in limit steps. To choose $X_alpha$ such that $X_alpha-X_beta$ is finite for all $beta<alpha$, you must use the fact that there are only countably many such $beta$, so that you can build $X_alpha$ by a diagonal construction so that it is infinite and yet eventually contained in each $X_beta$.
(The argument can be generalized to use weaker assumptions than $2^{aleph_0}=aleph_1$; for instance, Martin's axiom suffices. But the result is not provable in ZFC alone, and I don't know why you think it would be relevant to assume something like $2^{aleph_0}=aleph_n$ for $ninmathbb{N}$.)
answered Jan 29 at 0:07
Eric WofseyEric Wofsey
191k14216349
$endgroup$
$begingroup$
Dear Eric, Thank you for your answer and please excuse the late reply. To make my question more precise: I don't claim that $2^{aleph_0} = aleph_n$ is any way necessary. I am merely pondering the following thought: Suppose for example that $2^{aleph_0} = aleph_2$. What if I construct $X_{omega_1}$ by applying the diagonal construction (of $X_omega$) to ${X_omega, X_{omega + 1}, ldots}$? As far as I can see, the finite-intersection property still holds, and so does the property that $|{X_{omega_1} - X_beta}| < infty, forall beta < omega_1$. Am I missing something?
$endgroup$
– user480881
Feb 4 at 11:38
$begingroup$
What do you mean by "applying the diagonal construction"? How would you apply it to $aleph_1$ sets instead of countably many? The whole point of the "diagonal" is that it uses the fact that ${X_beta:beta<alpha}$, like $omega$, is countable.
$endgroup$
– Eric Wofsey
Feb 4 at 16:21
$begingroup$
Dear Eric, Having returned to Jech's Set Theory after being occupied with different matters for a while, I now see that my misunderstanding stems from the terribly embarrassing fact of me for some reason believing that $omega_1 = 2omega$ (and hence assuming that the construction of $X_omega$ can be continued...). Sorry for wasting your time.
$endgroup$
– user480881
Feb 6 at 11:52
add a comment |
$begingroup$
The construction very crucially uses the assumption that $2^{aleph_0}=aleph_1$ in limit steps. To choose $X_alpha$ such that $X_alpha-X_beta$ is finite for all $beta<alpha$, you must use the fact that there are only countably many such $beta$, so that you can build $X_alpha$ by a diagonal construction so that it is infinite and yet eventually contained in each $X_beta$.
(The argument can be generalized to use weaker assumptions than $2^{aleph_0}=aleph_1$; for instance, Martin's axiom suffices. But the result is not provable in ZFC alone, and I don't know why you think it would be relevant to assume something like $2^{aleph_0}=aleph_n$ for $ninmathbb{N}$.)
answered Jan 29 at 0:07
Eric WofseyEric Wofsey
191k14216349
$endgroup$
$begingroup$
Dear Eric, Thank you for your answer and please excuse the late reply. To make my question more precise: I don't claim that $2^{aleph_0} = aleph_n$ is any way necessary. I am merely pondering the following thought: Suppose for example that $2^{aleph_0} = aleph_2$. What if I construct $X_{omega_1}$ by applying the diagonal construction (of $X_omega$) to ${X_omega, X_{omega + 1}, ldots}$? As far as I can see, the finite-intersection property still holds, and so does the property that $|{X_{omega_1} - X_beta}| < infty, forall beta < omega_1$. Am I missing something?
$endgroup$
– user480881
Feb 4 at 11:38
$begingroup$
What do you mean by "applying the diagonal construction"? How would you apply it to $aleph_1$ sets instead of countably many? The whole point of the "diagonal" is that it uses the fact that ${X_beta:beta<alpha}$, like $omega$, is countable.
$endgroup$
– Eric Wofsey
Feb 4 at 16:21
$begingroup$
Dear Eric, Having returned to Jech's Set Theory after being occupied with different matters for a while, I now see that my misunderstanding stems from the terribly embarrassing fact of me for some reason believing that $omega_1 = 2omega$ (and hence assuming that the construction of $X_omega$ can be continued...). Sorry for wasting your time.
$endgroup$
– user480881
Feb 6 at 11:52
add a comment |
$begingroup$
The construction very crucially uses the assumption that $2^{aleph_0}=aleph_1$ in limit steps. To choose $X_alpha$ such that $X_alpha-X_beta$ is finite for all $beta<alpha$, you must use the fact that there are only countably many such $beta$, so that you can build $X_alpha$ by a diagonal construction so that it is infinite and yet eventually contained in each $X_beta$.
(The argument can be generalized to use weaker assumptions than $2^{aleph_0}=aleph_1$; for instance, Martin's axiom suffices. But the result is not provable in ZFC alone, and I don't know why you think it would be relevant to assume something like $2^{aleph_0}=aleph_n$ for $ninmathbb{N}$.)
answered Jan 29 at 0:07
Eric WofseyEric Wofsey
191k14216349
$endgroup$
The construction very crucially uses the assumption that $2^{aleph_0}=aleph_1$ in limit steps. To choose $X_alpha$ such that $X_alpha-X_beta$ is finite for all $beta<alpha$, you must use the fact that there are only countably many such $beta$, so that you can build $X_alpha$ by a diagonal construction so that it is infinite and yet eventually contained in each $X_beta$.
(The argument can be generalized to use weaker assumptions than $2^{aleph_0}=aleph_1$; for instance, Martin's axiom suffices. But the result is not provable in ZFC alone, and I don't know why you think it would be relevant to assume something like $2^{aleph_0}=aleph_n$ for $ninmathbb{N}$.)
answered Jan 29 at 0:07
Eric WofseyEric Wofsey
191k14216349
answered Jan 29 at 0:07
Eric WofseyEric Wofsey
191k14216349
answered Jan 29 at 0:07
Eric WofseyEric Wofsey
191k14216349
answered Jan 29 at 0:07
Eric WofseyEric Wofsey
191k14216349
191k14216349
$begingroup$
Dear Eric, Thank you for your answer and please excuse the late reply. To make my question more precise: I don't claim that $2^{aleph_0} = aleph_n$ is any way necessary. I am merely pondering the following thought: Suppose for example that $2^{aleph_0} = aleph_2$. What if I construct $X_{omega_1}$ by applying the diagonal construction (of $X_omega$) to ${X_omega, X_{omega + 1}, ldots}$? As far as I can see, the finite-intersection property still holds, and so does the property that $|{X_{omega_1} - X_beta}| < infty, forall beta < omega_1$. Am I missing something?
$endgroup$
– user480881
Feb 4 at 11:38
$begingroup$
What do you mean by "applying the diagonal construction"? How would you apply it to $aleph_1$ sets instead of countably many? The whole point of the "diagonal" is that it uses the fact that ${X_beta:beta<alpha}$, like $omega$, is countable.
$endgroup$
– Eric Wofsey
Feb 4 at 16:21
$begingroup$
Dear Eric, Having returned to Jech's Set Theory after being occupied with different matters for a while, I now see that my misunderstanding stems from the terribly embarrassing fact of me for some reason believing that $omega_1 = 2omega$ (and hence assuming that the construction of $X_omega$ can be continued...). Sorry for wasting your time.
$endgroup$
– user480881
Feb 6 at 11:52
add a comment |
$begingroup$
Dear Eric, Thank you for your answer and please excuse the late reply. To make my question more precise: I don't claim that $2^{aleph_0} = aleph_n$ is any way necessary. I am merely pondering the following thought: Suppose for example that $2^{aleph_0} = aleph_2$. What if I construct $X_{omega_1}$ by applying the diagonal construction (of $X_omega$) to ${X_omega, X_{omega + 1}, ldots}$? As far as I can see, the finite-intersection property still holds, and so does the property that $|{X_{omega_1} - X_beta}| < infty, forall beta < omega_1$. Am I missing something?
$endgroup$
– user480881
Feb 4 at 11:38
$begingroup$
What do you mean by "applying the diagonal construction"? How would you apply it to $aleph_1$ sets instead of countably many? The whole point of the "diagonal" is that it uses the fact that ${X_beta:beta<alpha}$, like $omega$, is countable.
$endgroup$
– Eric Wofsey
Feb 4 at 16:21
$begingroup$
Dear Eric, Having returned to Jech's Set Theory after being occupied with different matters for a while, I now see that my misunderstanding stems from the terribly embarrassing fact of me for some reason believing that $omega_1 = 2omega$ (and hence assuming that the construction of $X_omega$ can be continued...). Sorry for wasting your time.
$endgroup$
– user480881
Feb 6 at 11:52
$begingroup$
Dear Eric, Thank you for your answer and please excuse the late reply. To make my question more precise: I don't claim that $2^{aleph_0} = aleph_n$ is any way necessary. I am merely pondering the following thought: Suppose for example that $2^{aleph_0} = aleph_2$. What if I construct $X_{omega_1}$ by applying the diagonal construction (of $X_omega$) to ${X_omega, X_{omega + 1}, ldots}$? As far as I can see, the finite-intersection property still holds, and so does the property that $|{X_{omega_1} - X_beta}| < infty, forall beta < omega_1$. Am I missing something?
$endgroup$
– user480881
Feb 4 at 11:38
$begingroup$
Dear Eric, Thank you for your answer and please excuse the late reply. To make my question more precise: I don't claim that $2^{aleph_0} = aleph_n$ is any way necessary. I am merely pondering the following thought: Suppose for example that $2^{aleph_0} = aleph_2$. What if I construct $X_{omega_1}$ by applying the diagonal construction (of $X_omega$) to ${X_omega, X_{omega + 1}, ldots}$? As far as I can see, the finite-intersection property still holds, and so does the property that $|{X_{omega_1} - X_beta}| < infty, forall beta < omega_1$. Am I missing something?
$endgroup$
– user480881
Feb 4 at 11:38
$begingroup$
What do you mean by "applying the diagonal construction"? How would you apply it to $aleph_1$ sets instead of countably many? The whole point of the "diagonal" is that it uses the fact that ${X_beta:beta<alpha}$, like $omega$, is countable.
$endgroup$
– Eric Wofsey
Feb 4 at 16:21
$begingroup$
What do you mean by "applying the diagonal construction"? How would you apply it to $aleph_1$ sets instead of countably many? The whole point of the "diagonal" is that it uses the fact that ${X_beta:beta<alpha}$, like $omega$, is countable.
$endgroup$
– Eric Wofsey
Feb 4 at 16:21
$begingroup$
Dear Eric, Having returned to Jech's Set Theory after being occupied with different matters for a while, I now see that my misunderstanding stems from the terribly embarrassing fact of me for some reason believing that $omega_1 = 2omega$ (and hence assuming that the construction of $X_omega$ can be continued...). Sorry for wasting your time.
$endgroup$
– user480881
Feb 6 at 11:52
$begingroup$
Dear Eric, Having returned to Jech's Set Theory after being occupied with different matters for a while, I now see that my misunderstanding stems from the terribly embarrassing fact of me for some reason believing that $omega_1 = 2omega$ (and hence assuming that the construction of $X_omega$ can be continued...). Sorry for wasting your time.
$endgroup$
– user480881
Feb 6 at 11:52
add a comment |
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$begingroup$
In the proof, Jech explicitly says that the existence of $X_{alpha}$ depends on the fact that $alpha$ is countable.
$endgroup$
– Karl Kronenfeld
Jan 29 at 0:05
$begingroup$
There had been several questions about this already. Did you look around the site?
$endgroup$
– Asaf Karagila♦
Jan 29 at 8:17
$begingroup$
Dear Karl and Asaf, Thank you for your comments. I have clarified my question in a comment to Eric's answer.
$endgroup$
– user480881
Feb 4 at 11:42