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!
set-theory forcing
asked Nov 15 at 11:59
T. Jacobs
62
add a comment |
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!
set-theory forcing
asked Nov 15 at 11:59
T. Jacobs
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
add a comment |
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!
set-theory forcing
asked Nov 15 at 11:59
T. Jacobs
62
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
set-theory forcing
asked Nov 15 at 11:59
T. Jacobs
62
asked Nov 15 at 11:59
T. Jacobs
62
edited Nov 15 at 13:46
asked Nov 15 at 11:59
T. Jacobs
62
asked Nov 15 at 11:59
T. Jacobs
62
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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.
answered Nov 15 at 13:47
Asaf Karagila♦
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
|
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Nov 15 at 13:47
Asaf Karagila♦
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
|
show 5 more comments
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.
answered Nov 15 at 13:47
Asaf Karagila♦
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
|
show 5 more comments
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.
answered Nov 15 at 13:47
Asaf Karagila♦
300k32421751
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.
answered Nov 15 at 13:47
Asaf Karagila♦
300k32421751
edited Nov 15 at 13:56
answered Nov 15 at 13:47
Asaf Karagila♦
300k32421751
answered Nov 15 at 13:47
Asaf Karagila♦
300k32421751
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
|
show 5 more comments
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
|
show 5 more comments
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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