Proof of MM implies non-stationary ideal on $aleph_1$ is $aleph_2$ saturated.











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I am trying to understand the proof of thm 37.16 of Jech on page 687.
I don't understand the first 4 lines, why does that suffice to proof the theorem? I don't see how they are related. It sais the following:



Assume MM and let ${A_i : i in W } $ be a maximal almost disjoint collection of stationary subsets of $omega_1$. We shall find a set $Zsubset W$ of size $leq aleph_1$ such that $sum_{iin Z} A_i$ contains a closed unbounded set. This implies the theorem.



Here the sum is the diagonal union. In the comments it was suggested that almost disjoint in this case means two different $A_i$'s have a non-stationary intersection.



Thanks!










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  • 1




    Since not everyone memorized Jech's 700+ pages book by heart, it is a good idea to at least properly state the theorem and the parts which are unclear. It is not plagiarism if this is for educational purposes, which it is.
    – Asaf Karagila
    Nov 15 at 12:09










  • Fair enough! There is Jech online ofcourse, but you are right.
    – T. Jacobs
    Nov 15 at 13:33






  • 1




    When asking random strangers online for help, one shouldn't have them bother to go to various lengths in order to understand the question you're really asking.
    – Asaf Karagila
    Nov 15 at 13:43






  • 1




    I think that almost disjoint here means that the intersection is non-stationary, especially if the sum is the diagonal union.
    – Asaf Karagila
    Nov 15 at 13:44










  • @AsafKaragila, Good to know my guess for the almost disjointness was false. will edit it.
    – T. Jacobs
    Nov 15 at 13:45















up vote
1
down vote

favorite












I am trying to understand the proof of thm 37.16 of Jech on page 687.
I don't understand the first 4 lines, why does that suffice to proof the theorem? I don't see how they are related. It sais the following:



Assume MM and let ${A_i : i in W } $ be a maximal almost disjoint collection of stationary subsets of $omega_1$. We shall find a set $Zsubset W$ of size $leq aleph_1$ such that $sum_{iin Z} A_i$ contains a closed unbounded set. This implies the theorem.



Here the sum is the diagonal union. In the comments it was suggested that almost disjoint in this case means two different $A_i$'s have a non-stationary intersection.



Thanks!










share|cite|improve this question




















  • 1




    Since not everyone memorized Jech's 700+ pages book by heart, it is a good idea to at least properly state the theorem and the parts which are unclear. It is not plagiarism if this is for educational purposes, which it is.
    – Asaf Karagila
    Nov 15 at 12:09










  • Fair enough! There is Jech online ofcourse, but you are right.
    – T. Jacobs
    Nov 15 at 13:33






  • 1




    When asking random strangers online for help, one shouldn't have them bother to go to various lengths in order to understand the question you're really asking.
    – Asaf Karagila
    Nov 15 at 13:43






  • 1




    I think that almost disjoint here means that the intersection is non-stationary, especially if the sum is the diagonal union.
    – Asaf Karagila
    Nov 15 at 13:44










  • @AsafKaragila, Good to know my guess for the almost disjointness was false. will edit it.
    – T. Jacobs
    Nov 15 at 13:45













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I am trying to understand the proof of thm 37.16 of Jech on page 687.
I don't understand the first 4 lines, why does that suffice to proof the theorem? I don't see how they are related. It sais the following:



Assume MM and let ${A_i : i in W } $ be a maximal almost disjoint collection of stationary subsets of $omega_1$. We shall find a set $Zsubset W$ of size $leq aleph_1$ such that $sum_{iin Z} A_i$ contains a closed unbounded set. This implies the theorem.



Here the sum is the diagonal union. In the comments it was suggested that almost disjoint in this case means two different $A_i$'s have a non-stationary intersection.



Thanks!










share|cite|improve this question















I am trying to understand the proof of thm 37.16 of Jech on page 687.
I don't understand the first 4 lines, why does that suffice to proof the theorem? I don't see how they are related. It sais the following:



Assume MM and let ${A_i : i in W } $ be a maximal almost disjoint collection of stationary subsets of $omega_1$. We shall find a set $Zsubset W$ of size $leq aleph_1$ such that $sum_{iin Z} A_i$ contains a closed unbounded set. This implies the theorem.



Here the sum is the diagonal union. In the comments it was suggested that almost disjoint in this case means two different $A_i$'s have a non-stationary intersection.



Thanks!







set-theory forcing






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 15 at 13:46

























asked Nov 15 at 11:59









T. Jacobs

62




62








  • 1




    Since not everyone memorized Jech's 700+ pages book by heart, it is a good idea to at least properly state the theorem and the parts which are unclear. It is not plagiarism if this is for educational purposes, which it is.
    – Asaf Karagila
    Nov 15 at 12:09










  • Fair enough! There is Jech online ofcourse, but you are right.
    – T. Jacobs
    Nov 15 at 13:33






  • 1




    When asking random strangers online for help, one shouldn't have them bother to go to various lengths in order to understand the question you're really asking.
    – Asaf Karagila
    Nov 15 at 13:43






  • 1




    I think that almost disjoint here means that the intersection is non-stationary, especially if the sum is the diagonal union.
    – Asaf Karagila
    Nov 15 at 13:44










  • @AsafKaragila, Good to know my guess for the almost disjointness was false. will edit it.
    – T. Jacobs
    Nov 15 at 13:45














  • 1




    Since not everyone memorized Jech's 700+ pages book by heart, it is a good idea to at least properly state the theorem and the parts which are unclear. It is not plagiarism if this is for educational purposes, which it is.
    – Asaf Karagila
    Nov 15 at 12:09










  • Fair enough! There is Jech online ofcourse, but you are right.
    – T. Jacobs
    Nov 15 at 13:33






  • 1




    When asking random strangers online for help, one shouldn't have them bother to go to various lengths in order to understand the question you're really asking.
    – Asaf Karagila
    Nov 15 at 13:43






  • 1




    I think that almost disjoint here means that the intersection is non-stationary, especially if the sum is the diagonal union.
    – Asaf Karagila
    Nov 15 at 13:44










  • @AsafKaragila, Good to know my guess for the almost disjointness was false. will edit it.
    – T. Jacobs
    Nov 15 at 13:45








1




1




Since not everyone memorized Jech's 700+ pages book by heart, it is a good idea to at least properly state the theorem and the parts which are unclear. It is not plagiarism if this is for educational purposes, which it is.
– Asaf Karagila
Nov 15 at 12:09




Since not everyone memorized Jech's 700+ pages book by heart, it is a good idea to at least properly state the theorem and the parts which are unclear. It is not plagiarism if this is for educational purposes, which it is.
– Asaf Karagila
Nov 15 at 12:09












Fair enough! There is Jech online ofcourse, but you are right.
– T. Jacobs
Nov 15 at 13:33




Fair enough! There is Jech online ofcourse, but you are right.
– T. Jacobs
Nov 15 at 13:33




1




1




When asking random strangers online for help, one shouldn't have them bother to go to various lengths in order to understand the question you're really asking.
– Asaf Karagila
Nov 15 at 13:43




When asking random strangers online for help, one shouldn't have them bother to go to various lengths in order to understand the question you're really asking.
– Asaf Karagila
Nov 15 at 13:43




1




1




I think that almost disjoint here means that the intersection is non-stationary, especially if the sum is the diagonal union.
– Asaf Karagila
Nov 15 at 13:44




I think that almost disjoint here means that the intersection is non-stationary, especially if the sum is the diagonal union.
– Asaf Karagila
Nov 15 at 13:44












@AsafKaragila, Good to know my guess for the almost disjointness was false. will edit it.
– T. Jacobs
Nov 15 at 13:45




@AsafKaragila, Good to know my guess for the almost disjointness was false. will edit it.
– T. Jacobs
Nov 15 at 13:45










1 Answer
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up vote
1
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Let's consider the Boolean algebra $Bbb B=mathcal P(omega_1)/rm NS$ (one can also just consider this as a notion of forcing). Saturation simply means that it has the $aleph_2$-chain condition.



So take a maximal family of mutually stationary sets, which is exactly a maximal antichain in $Bbb B$. If there is a family of size $aleph_1$ whose union contains a club, that means that this family would already constitute a partition of $1_{Bbb B}$ (in forcing terms: this family would induce a maxiaml antichain already), because $1_{Bbb B}$ is exactly the club filter on $omega_1$.



In other words, it means that every maximal antichain in $Bbb B$ is of size $<aleph_2$, which is exactly what we want to prove.






share|cite|improve this answer























  • So.. Unfortunatly I am not yet familiar with Boolean algebra's. I don't understand what is going on. Though, I was planning to have a closer look at it anyway, which means I will probably understand it some point in the future. A more elementary proof would be great as well.
    – T. Jacobs
    Nov 15 at 13:50












  • Well. You were reading in Jech, so I figured you at least know the basics, because Jech is doing almost everything he can in terms of Boolean algebras. Specifically, forcing (and MM is a forcing axiom...).
    – Asaf Karagila
    Nov 15 at 13:51












  • Do you know anything about forcing?
    – Asaf Karagila
    Nov 15 at 13:52










  • Yes, I did quite a bit of forcing. Including product forcing, iterated forcing... Though never learned that via Jech/the boolen algebra way.
    – T. Jacobs
    Nov 15 at 13:53










  • Well. Think about this just in terms of forcing. Where a maximal antichain is an antichain whose only supremum is $1$, the maximum condition.
    – Asaf Karagila
    Nov 15 at 13:54











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Let's consider the Boolean algebra $Bbb B=mathcal P(omega_1)/rm NS$ (one can also just consider this as a notion of forcing). Saturation simply means that it has the $aleph_2$-chain condition.



So take a maximal family of mutually stationary sets, which is exactly a maximal antichain in $Bbb B$. If there is a family of size $aleph_1$ whose union contains a club, that means that this family would already constitute a partition of $1_{Bbb B}$ (in forcing terms: this family would induce a maxiaml antichain already), because $1_{Bbb B}$ is exactly the club filter on $omega_1$.



In other words, it means that every maximal antichain in $Bbb B$ is of size $<aleph_2$, which is exactly what we want to prove.






share|cite|improve this answer























  • So.. Unfortunatly I am not yet familiar with Boolean algebra's. I don't understand what is going on. Though, I was planning to have a closer look at it anyway, which means I will probably understand it some point in the future. A more elementary proof would be great as well.
    – T. Jacobs
    Nov 15 at 13:50












  • Well. You were reading in Jech, so I figured you at least know the basics, because Jech is doing almost everything he can in terms of Boolean algebras. Specifically, forcing (and MM is a forcing axiom...).
    – Asaf Karagila
    Nov 15 at 13:51












  • Do you know anything about forcing?
    – Asaf Karagila
    Nov 15 at 13:52










  • Yes, I did quite a bit of forcing. Including product forcing, iterated forcing... Though never learned that via Jech/the boolen algebra way.
    – T. Jacobs
    Nov 15 at 13:53










  • Well. Think about this just in terms of forcing. Where a maximal antichain is an antichain whose only supremum is $1$, the maximum condition.
    – Asaf Karagila
    Nov 15 at 13:54















up vote
1
down vote













Let's consider the Boolean algebra $Bbb B=mathcal P(omega_1)/rm NS$ (one can also just consider this as a notion of forcing). Saturation simply means that it has the $aleph_2$-chain condition.



So take a maximal family of mutually stationary sets, which is exactly a maximal antichain in $Bbb B$. If there is a family of size $aleph_1$ whose union contains a club, that means that this family would already constitute a partition of $1_{Bbb B}$ (in forcing terms: this family would induce a maxiaml antichain already), because $1_{Bbb B}$ is exactly the club filter on $omega_1$.



In other words, it means that every maximal antichain in $Bbb B$ is of size $<aleph_2$, which is exactly what we want to prove.






share|cite|improve this answer























  • So.. Unfortunatly I am not yet familiar with Boolean algebra's. I don't understand what is going on. Though, I was planning to have a closer look at it anyway, which means I will probably understand it some point in the future. A more elementary proof would be great as well.
    – T. Jacobs
    Nov 15 at 13:50












  • Well. You were reading in Jech, so I figured you at least know the basics, because Jech is doing almost everything he can in terms of Boolean algebras. Specifically, forcing (and MM is a forcing axiom...).
    – Asaf Karagila
    Nov 15 at 13:51












  • Do you know anything about forcing?
    – Asaf Karagila
    Nov 15 at 13:52










  • Yes, I did quite a bit of forcing. Including product forcing, iterated forcing... Though never learned that via Jech/the boolen algebra way.
    – T. Jacobs
    Nov 15 at 13:53










  • Well. Think about this just in terms of forcing. Where a maximal antichain is an antichain whose only supremum is $1$, the maximum condition.
    – Asaf Karagila
    Nov 15 at 13:54













up vote
1
down vote










up vote
1
down vote









Let's consider the Boolean algebra $Bbb B=mathcal P(omega_1)/rm NS$ (one can also just consider this as a notion of forcing). Saturation simply means that it has the $aleph_2$-chain condition.



So take a maximal family of mutually stationary sets, which is exactly a maximal antichain in $Bbb B$. If there is a family of size $aleph_1$ whose union contains a club, that means that this family would already constitute a partition of $1_{Bbb B}$ (in forcing terms: this family would induce a maxiaml antichain already), because $1_{Bbb B}$ is exactly the club filter on $omega_1$.



In other words, it means that every maximal antichain in $Bbb B$ is of size $<aleph_2$, which is exactly what we want to prove.






share|cite|improve this answer














Let's consider the Boolean algebra $Bbb B=mathcal P(omega_1)/rm NS$ (one can also just consider this as a notion of forcing). Saturation simply means that it has the $aleph_2$-chain condition.



So take a maximal family of mutually stationary sets, which is exactly a maximal antichain in $Bbb B$. If there is a family of size $aleph_1$ whose union contains a club, that means that this family would already constitute a partition of $1_{Bbb B}$ (in forcing terms: this family would induce a maxiaml antichain already), because $1_{Bbb B}$ is exactly the club filter on $omega_1$.



In other words, it means that every maximal antichain in $Bbb B$ is of size $<aleph_2$, which is exactly what we want to prove.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 15 at 13:56

























answered Nov 15 at 13:47









Asaf Karagila

300k32421751




300k32421751












  • So.. Unfortunatly I am not yet familiar with Boolean algebra's. I don't understand what is going on. Though, I was planning to have a closer look at it anyway, which means I will probably understand it some point in the future. A more elementary proof would be great as well.
    – T. Jacobs
    Nov 15 at 13:50












  • Well. You were reading in Jech, so I figured you at least know the basics, because Jech is doing almost everything he can in terms of Boolean algebras. Specifically, forcing (and MM is a forcing axiom...).
    – Asaf Karagila
    Nov 15 at 13:51












  • Do you know anything about forcing?
    – Asaf Karagila
    Nov 15 at 13:52










  • Yes, I did quite a bit of forcing. Including product forcing, iterated forcing... Though never learned that via Jech/the boolen algebra way.
    – T. Jacobs
    Nov 15 at 13:53










  • Well. Think about this just in terms of forcing. Where a maximal antichain is an antichain whose only supremum is $1$, the maximum condition.
    – Asaf Karagila
    Nov 15 at 13:54


















  • So.. Unfortunatly I am not yet familiar with Boolean algebra's. I don't understand what is going on. Though, I was planning to have a closer look at it anyway, which means I will probably understand it some point in the future. A more elementary proof would be great as well.
    – T. Jacobs
    Nov 15 at 13:50












  • Well. You were reading in Jech, so I figured you at least know the basics, because Jech is doing almost everything he can in terms of Boolean algebras. Specifically, forcing (and MM is a forcing axiom...).
    – Asaf Karagila
    Nov 15 at 13:51












  • Do you know anything about forcing?
    – Asaf Karagila
    Nov 15 at 13:52










  • Yes, I did quite a bit of forcing. Including product forcing, iterated forcing... Though never learned that via Jech/the boolen algebra way.
    – T. Jacobs
    Nov 15 at 13:53










  • Well. Think about this just in terms of forcing. Where a maximal antichain is an antichain whose only supremum is $1$, the maximum condition.
    – Asaf Karagila
    Nov 15 at 13:54
















So.. Unfortunatly I am not yet familiar with Boolean algebra's. I don't understand what is going on. Though, I was planning to have a closer look at it anyway, which means I will probably understand it some point in the future. A more elementary proof would be great as well.
– T. Jacobs
Nov 15 at 13:50






So.. Unfortunatly I am not yet familiar with Boolean algebra's. I don't understand what is going on. Though, I was planning to have a closer look at it anyway, which means I will probably understand it some point in the future. A more elementary proof would be great as well.
– T. Jacobs
Nov 15 at 13:50














Well. You were reading in Jech, so I figured you at least know the basics, because Jech is doing almost everything he can in terms of Boolean algebras. Specifically, forcing (and MM is a forcing axiom...).
– Asaf Karagila
Nov 15 at 13:51






Well. You were reading in Jech, so I figured you at least know the basics, because Jech is doing almost everything he can in terms of Boolean algebras. Specifically, forcing (and MM is a forcing axiom...).
– Asaf Karagila
Nov 15 at 13:51














Do you know anything about forcing?
– Asaf Karagila
Nov 15 at 13:52




Do you know anything about forcing?
– Asaf Karagila
Nov 15 at 13:52












Yes, I did quite a bit of forcing. Including product forcing, iterated forcing... Though never learned that via Jech/the boolen algebra way.
– T. Jacobs
Nov 15 at 13:53




Yes, I did quite a bit of forcing. Including product forcing, iterated forcing... Though never learned that via Jech/the boolen algebra way.
– T. Jacobs
Nov 15 at 13:53












Well. Think about this just in terms of forcing. Where a maximal antichain is an antichain whose only supremum is $1$, the maximum condition.
– Asaf Karagila
Nov 15 at 13:54




Well. Think about this just in terms of forcing. Where a maximal antichain is an antichain whose only supremum is $1$, the maximum condition.
– Asaf Karagila
Nov 15 at 13:54


















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