When to use Iterated Forcing The Next CEO of Stack OverflowIs the axiom of universes 'harmless'?Do we usually assume the consistency of ZFC when we use forcing?Dependent choice does not imply “the reals are well-ordered”; citation?Existence of Certain Names in Iterated ForcingBest Less-Famous Texts for ForcingA question about $aleph_1$-dense sets and the basis problem for uncountable linear orderingsExercise of Iterated Forcing from Kunen's bookProvability of 'generalized continuum hypothesis implies axiom of choice' over ZFRudimentary results in iterated forcing.Standard fact about iterated forcing
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When to use Iterated Forcing
The Next CEO of Stack OverflowIs the axiom of universes 'harmless'?Do we usually assume the consistency of ZFC when we use forcing?Dependent choice does not imply “the reals are well-ordered”; citation?Existence of Certain Names in Iterated ForcingBest Less-Famous Texts for ForcingA question about $aleph_1$-dense sets and the basis problem for uncountable linear orderingsExercise of Iterated Forcing from Kunen's bookProvability of 'generalized continuum hypothesis implies axiom of choice' over ZFRudimentary results in iterated forcing.Standard fact about iterated forcing
$begingroup$
Iterated forcing is useful for separating cardinal characteristics, and to prove the consistency of Martin's axiom. My question is, what is it about these problems determine that a normal forcing proof can't work and that we require multiple forcings?
What about a consistency result determines that it need be proved through iterated forcing rather than ordinary forcing?
set-theory forcing
$endgroup$
add a comment |
$begingroup$
Iterated forcing is useful for separating cardinal characteristics, and to prove the consistency of Martin's axiom. My question is, what is it about these problems determine that a normal forcing proof can't work and that we require multiple forcings?
What about a consistency result determines that it need be proved through iterated forcing rather than ordinary forcing?
set-theory forcing
$endgroup$
$begingroup$
Always. All forcing is iterated forcing.
$endgroup$
– Asaf Karagila♦
2 days ago
add a comment |
$begingroup$
Iterated forcing is useful for separating cardinal characteristics, and to prove the consistency of Martin's axiom. My question is, what is it about these problems determine that a normal forcing proof can't work and that we require multiple forcings?
What about a consistency result determines that it need be proved through iterated forcing rather than ordinary forcing?
set-theory forcing
$endgroup$
Iterated forcing is useful for separating cardinal characteristics, and to prove the consistency of Martin's axiom. My question is, what is it about these problems determine that a normal forcing proof can't work and that we require multiple forcings?
What about a consistency result determines that it need be proved through iterated forcing rather than ordinary forcing?
set-theory forcing
set-theory forcing
asked 2 days ago
MarmosetMarmoset
406
406
$begingroup$
Always. All forcing is iterated forcing.
$endgroup$
– Asaf Karagila♦
2 days ago
add a comment |
$begingroup$
Always. All forcing is iterated forcing.
$endgroup$
– Asaf Karagila♦
2 days ago
$begingroup$
Always. All forcing is iterated forcing.
$endgroup$
– Asaf Karagila♦
2 days ago
$begingroup$
Always. All forcing is iterated forcing.
$endgroup$
– Asaf Karagila♦
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There's no criteria that requires to be satisfied in order to use iterated forcing. Not only that, but since iterating forcing is the same as taking a single forcing extension (using the iteration poset), the question sort of falls flat on itself.
Even worse, with the exception of a certain class of "minimal" generic extension, most (in some sense) forcing notions are in fact an iteration, since they can be decomposed to iteration of one or more subforcings. For example, adding a Cohen real can be thought of as adding two Cohen reals one after the other. And collapsing $omega_1$ can be thought as first adding a Cohen real, then adding a branch to the Suslin tree added by that Cohen real, and then collapsing $omega_1$.
So why do we even use iterated forcing?
Because it's convenient. Because it is easier to break down a large problem into smaller problems, and then deal with them, one at a time. When forcing Martin's Axiom, for example, it is easier to deal with the forcing notions one step at a time, rather than trying to somehow capture all of the existing ones, and the ones who would come, simultaneously.
Even worse. The iterative approach to Martin's Axiom is pure magic. Every limit step adds Cohen reals. Every Cohen real adds a Suslin tree. Martin's Axiom implies that there are no Suslin trees.
How does it happen? Well. Because of the very nature of the iteration, at each step we anticipate "a problem", and we solve it.
Other times we might want to construct an object via forcing, but our starting model would require to have certain objects which are not guaranteed to exist. Or perhaps the construction would require a certain degree of genericity over the model, so first adding something new to work with is a good thing. In those approaches we start with $V$, we extend it once with a preparation (which itself may or may not be an iteration, e.g. Martin's Axiom or indestructibility of large cardinals), and then perform one or two extensions to obtain a final model.
Yes, we can describe the whole thing as a single forcing poset. But why? It will offer no better result, and will only increase the difficulty when trying to describe your objects or reason as to why they have this or that property.
For this reason exactly it is sometimes convenient to think about a Cohen real as a subset of $omega$, sometimes as a binary sequence in $2^omega$, and sometimes as a general sequence in $omega^omega$. But sometimes it's easier to think about a single Cohen real as infinitely many different Cohen reals instead, exactly for that reason.
$endgroup$
$begingroup$
This is a great answer, I had never thought about the fact that adding a Cohen real is essentially iterated forcing in itself. Thanks!
$endgroup$
– Marmoset
2 days ago
add a comment |
$begingroup$
Maybe I can add a natural example to Asaf's great answer. In the case of establishing Borel's conjecture, iterated forcing naturally comes up in the construction. Borel's conjecture states that all strong measure zero sets of reals are countable (it is not important what that means). As it turns out, this statement is independent of ZFC. It is not too hard to construct a counterexample to it using CH. To show that BC is consistent, you need forcing. But how do you force BC? Richard Laver found a poset (now called Laver forcing) that is easy to describe and has the following property:
If $V[g]$ is a generic extension of $V$ via Laver forcing and $Ain V$ is a set of reals which is strong measure zero in $V[g]$, then $A$ is countable (both in $V$ and $V[g]$). (Note that the evaluation of the statement "$A$ is strong measure zero" may change from $V$ to $V[g]$). Now you have made some progress, there are no "old" counterexamples to BC anymore. But there still might be new counterexamples! It is now clear that one should try to iterate this process to make the new counterexamples vanish in the limit. This turns out to work, however it is not enough to iterate this just $omega$ or $omega_1$-many times, it is important to do it $omega_2$-many times (otherwise CH might still hold in the extension). Moreover one has to choose the support correctly, in this case it is countable support for technical reasons.
$endgroup$
1
$begingroup$
I think this is a bit problematic as written. I assume you are using the term "ordinary forcing" in a purely informal fashion, but it is really to informal to be useful. Why would adding $aleph_mathfrak c^+$ Cohen reals not be considered ordinary?
$endgroup$
– Andrés E. Caicedo
2 days ago
$begingroup$
In fact, it reads as if you are claiming that there is a well understood and reasonably wide and varied class of "ordinary forcing notions", and that no forcing in this class can force $lnotmathsfCH$ and (perhaps) that no forcing in this class is forcing isomorphic to an iteration.
$endgroup$
– Andrés E. Caicedo
2 days ago
$begingroup$
I agree that this is problematic. "Ordinary forcing" is the term OP used to phrase their question and I simply tried to make sense of this to appeal to OPs intuition.
$endgroup$
– Andreas Lietz
2 days ago
1
$begingroup$
I have edited out this section to avoid further confusion.
$endgroup$
– Andreas Lietz
2 days ago
add a comment |
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2 Answers
2
active
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votes
2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
There's no criteria that requires to be satisfied in order to use iterated forcing. Not only that, but since iterating forcing is the same as taking a single forcing extension (using the iteration poset), the question sort of falls flat on itself.
Even worse, with the exception of a certain class of "minimal" generic extension, most (in some sense) forcing notions are in fact an iteration, since they can be decomposed to iteration of one or more subforcings. For example, adding a Cohen real can be thought of as adding two Cohen reals one after the other. And collapsing $omega_1$ can be thought as first adding a Cohen real, then adding a branch to the Suslin tree added by that Cohen real, and then collapsing $omega_1$.
So why do we even use iterated forcing?
Because it's convenient. Because it is easier to break down a large problem into smaller problems, and then deal with them, one at a time. When forcing Martin's Axiom, for example, it is easier to deal with the forcing notions one step at a time, rather than trying to somehow capture all of the existing ones, and the ones who would come, simultaneously.
Even worse. The iterative approach to Martin's Axiom is pure magic. Every limit step adds Cohen reals. Every Cohen real adds a Suslin tree. Martin's Axiom implies that there are no Suslin trees.
How does it happen? Well. Because of the very nature of the iteration, at each step we anticipate "a problem", and we solve it.
Other times we might want to construct an object via forcing, but our starting model would require to have certain objects which are not guaranteed to exist. Or perhaps the construction would require a certain degree of genericity over the model, so first adding something new to work with is a good thing. In those approaches we start with $V$, we extend it once with a preparation (which itself may or may not be an iteration, e.g. Martin's Axiom or indestructibility of large cardinals), and then perform one or two extensions to obtain a final model.
Yes, we can describe the whole thing as a single forcing poset. But why? It will offer no better result, and will only increase the difficulty when trying to describe your objects or reason as to why they have this or that property.
For this reason exactly it is sometimes convenient to think about a Cohen real as a subset of $omega$, sometimes as a binary sequence in $2^omega$, and sometimes as a general sequence in $omega^omega$. But sometimes it's easier to think about a single Cohen real as infinitely many different Cohen reals instead, exactly for that reason.
$endgroup$
$begingroup$
This is a great answer, I had never thought about the fact that adding a Cohen real is essentially iterated forcing in itself. Thanks!
$endgroup$
– Marmoset
2 days ago
add a comment |
$begingroup$
There's no criteria that requires to be satisfied in order to use iterated forcing. Not only that, but since iterating forcing is the same as taking a single forcing extension (using the iteration poset), the question sort of falls flat on itself.
Even worse, with the exception of a certain class of "minimal" generic extension, most (in some sense) forcing notions are in fact an iteration, since they can be decomposed to iteration of one or more subforcings. For example, adding a Cohen real can be thought of as adding two Cohen reals one after the other. And collapsing $omega_1$ can be thought as first adding a Cohen real, then adding a branch to the Suslin tree added by that Cohen real, and then collapsing $omega_1$.
So why do we even use iterated forcing?
Because it's convenient. Because it is easier to break down a large problem into smaller problems, and then deal with them, one at a time. When forcing Martin's Axiom, for example, it is easier to deal with the forcing notions one step at a time, rather than trying to somehow capture all of the existing ones, and the ones who would come, simultaneously.
Even worse. The iterative approach to Martin's Axiom is pure magic. Every limit step adds Cohen reals. Every Cohen real adds a Suslin tree. Martin's Axiom implies that there are no Suslin trees.
How does it happen? Well. Because of the very nature of the iteration, at each step we anticipate "a problem", and we solve it.
Other times we might want to construct an object via forcing, but our starting model would require to have certain objects which are not guaranteed to exist. Or perhaps the construction would require a certain degree of genericity over the model, so first adding something new to work with is a good thing. In those approaches we start with $V$, we extend it once with a preparation (which itself may or may not be an iteration, e.g. Martin's Axiom or indestructibility of large cardinals), and then perform one or two extensions to obtain a final model.
Yes, we can describe the whole thing as a single forcing poset. But why? It will offer no better result, and will only increase the difficulty when trying to describe your objects or reason as to why they have this or that property.
For this reason exactly it is sometimes convenient to think about a Cohen real as a subset of $omega$, sometimes as a binary sequence in $2^omega$, and sometimes as a general sequence in $omega^omega$. But sometimes it's easier to think about a single Cohen real as infinitely many different Cohen reals instead, exactly for that reason.
$endgroup$
$begingroup$
This is a great answer, I had never thought about the fact that adding a Cohen real is essentially iterated forcing in itself. Thanks!
$endgroup$
– Marmoset
2 days ago
add a comment |
$begingroup$
There's no criteria that requires to be satisfied in order to use iterated forcing. Not only that, but since iterating forcing is the same as taking a single forcing extension (using the iteration poset), the question sort of falls flat on itself.
Even worse, with the exception of a certain class of "minimal" generic extension, most (in some sense) forcing notions are in fact an iteration, since they can be decomposed to iteration of one or more subforcings. For example, adding a Cohen real can be thought of as adding two Cohen reals one after the other. And collapsing $omega_1$ can be thought as first adding a Cohen real, then adding a branch to the Suslin tree added by that Cohen real, and then collapsing $omega_1$.
So why do we even use iterated forcing?
Because it's convenient. Because it is easier to break down a large problem into smaller problems, and then deal with them, one at a time. When forcing Martin's Axiom, for example, it is easier to deal with the forcing notions one step at a time, rather than trying to somehow capture all of the existing ones, and the ones who would come, simultaneously.
Even worse. The iterative approach to Martin's Axiom is pure magic. Every limit step adds Cohen reals. Every Cohen real adds a Suslin tree. Martin's Axiom implies that there are no Suslin trees.
How does it happen? Well. Because of the very nature of the iteration, at each step we anticipate "a problem", and we solve it.
Other times we might want to construct an object via forcing, but our starting model would require to have certain objects which are not guaranteed to exist. Or perhaps the construction would require a certain degree of genericity over the model, so first adding something new to work with is a good thing. In those approaches we start with $V$, we extend it once with a preparation (which itself may or may not be an iteration, e.g. Martin's Axiom or indestructibility of large cardinals), and then perform one or two extensions to obtain a final model.
Yes, we can describe the whole thing as a single forcing poset. But why? It will offer no better result, and will only increase the difficulty when trying to describe your objects or reason as to why they have this or that property.
For this reason exactly it is sometimes convenient to think about a Cohen real as a subset of $omega$, sometimes as a binary sequence in $2^omega$, and sometimes as a general sequence in $omega^omega$. But sometimes it's easier to think about a single Cohen real as infinitely many different Cohen reals instead, exactly for that reason.
$endgroup$
There's no criteria that requires to be satisfied in order to use iterated forcing. Not only that, but since iterating forcing is the same as taking a single forcing extension (using the iteration poset), the question sort of falls flat on itself.
Even worse, with the exception of a certain class of "minimal" generic extension, most (in some sense) forcing notions are in fact an iteration, since they can be decomposed to iteration of one or more subforcings. For example, adding a Cohen real can be thought of as adding two Cohen reals one after the other. And collapsing $omega_1$ can be thought as first adding a Cohen real, then adding a branch to the Suslin tree added by that Cohen real, and then collapsing $omega_1$.
So why do we even use iterated forcing?
Because it's convenient. Because it is easier to break down a large problem into smaller problems, and then deal with them, one at a time. When forcing Martin's Axiom, for example, it is easier to deal with the forcing notions one step at a time, rather than trying to somehow capture all of the existing ones, and the ones who would come, simultaneously.
Even worse. The iterative approach to Martin's Axiom is pure magic. Every limit step adds Cohen reals. Every Cohen real adds a Suslin tree. Martin's Axiom implies that there are no Suslin trees.
How does it happen? Well. Because of the very nature of the iteration, at each step we anticipate "a problem", and we solve it.
Other times we might want to construct an object via forcing, but our starting model would require to have certain objects which are not guaranteed to exist. Or perhaps the construction would require a certain degree of genericity over the model, so first adding something new to work with is a good thing. In those approaches we start with $V$, we extend it once with a preparation (which itself may or may not be an iteration, e.g. Martin's Axiom or indestructibility of large cardinals), and then perform one or two extensions to obtain a final model.
Yes, we can describe the whole thing as a single forcing poset. But why? It will offer no better result, and will only increase the difficulty when trying to describe your objects or reason as to why they have this or that property.
For this reason exactly it is sometimes convenient to think about a Cohen real as a subset of $omega$, sometimes as a binary sequence in $2^omega$, and sometimes as a general sequence in $omega^omega$. But sometimes it's easier to think about a single Cohen real as infinitely many different Cohen reals instead, exactly for that reason.
answered 2 days ago
Asaf Karagila♦Asaf Karagila
307k33439771
307k33439771
$begingroup$
This is a great answer, I had never thought about the fact that adding a Cohen real is essentially iterated forcing in itself. Thanks!
$endgroup$
– Marmoset
2 days ago
add a comment |
$begingroup$
This is a great answer, I had never thought about the fact that adding a Cohen real is essentially iterated forcing in itself. Thanks!
$endgroup$
– Marmoset
2 days ago
$begingroup$
This is a great answer, I had never thought about the fact that adding a Cohen real is essentially iterated forcing in itself. Thanks!
$endgroup$
– Marmoset
2 days ago
$begingroup$
This is a great answer, I had never thought about the fact that adding a Cohen real is essentially iterated forcing in itself. Thanks!
$endgroup$
– Marmoset
2 days ago
add a comment |
$begingroup$
Maybe I can add a natural example to Asaf's great answer. In the case of establishing Borel's conjecture, iterated forcing naturally comes up in the construction. Borel's conjecture states that all strong measure zero sets of reals are countable (it is not important what that means). As it turns out, this statement is independent of ZFC. It is not too hard to construct a counterexample to it using CH. To show that BC is consistent, you need forcing. But how do you force BC? Richard Laver found a poset (now called Laver forcing) that is easy to describe and has the following property:
If $V[g]$ is a generic extension of $V$ via Laver forcing and $Ain V$ is a set of reals which is strong measure zero in $V[g]$, then $A$ is countable (both in $V$ and $V[g]$). (Note that the evaluation of the statement "$A$ is strong measure zero" may change from $V$ to $V[g]$). Now you have made some progress, there are no "old" counterexamples to BC anymore. But there still might be new counterexamples! It is now clear that one should try to iterate this process to make the new counterexamples vanish in the limit. This turns out to work, however it is not enough to iterate this just $omega$ or $omega_1$-many times, it is important to do it $omega_2$-many times (otherwise CH might still hold in the extension). Moreover one has to choose the support correctly, in this case it is countable support for technical reasons.
$endgroup$
1
$begingroup$
I think this is a bit problematic as written. I assume you are using the term "ordinary forcing" in a purely informal fashion, but it is really to informal to be useful. Why would adding $aleph_mathfrak c^+$ Cohen reals not be considered ordinary?
$endgroup$
– Andrés E. Caicedo
2 days ago
$begingroup$
In fact, it reads as if you are claiming that there is a well understood and reasonably wide and varied class of "ordinary forcing notions", and that no forcing in this class can force $lnotmathsfCH$ and (perhaps) that no forcing in this class is forcing isomorphic to an iteration.
$endgroup$
– Andrés E. Caicedo
2 days ago
$begingroup$
I agree that this is problematic. "Ordinary forcing" is the term OP used to phrase their question and I simply tried to make sense of this to appeal to OPs intuition.
$endgroup$
– Andreas Lietz
2 days ago
1
$begingroup$
I have edited out this section to avoid further confusion.
$endgroup$
– Andreas Lietz
2 days ago
add a comment |
$begingroup$
Maybe I can add a natural example to Asaf's great answer. In the case of establishing Borel's conjecture, iterated forcing naturally comes up in the construction. Borel's conjecture states that all strong measure zero sets of reals are countable (it is not important what that means). As it turns out, this statement is independent of ZFC. It is not too hard to construct a counterexample to it using CH. To show that BC is consistent, you need forcing. But how do you force BC? Richard Laver found a poset (now called Laver forcing) that is easy to describe and has the following property:
If $V[g]$ is a generic extension of $V$ via Laver forcing and $Ain V$ is a set of reals which is strong measure zero in $V[g]$, then $A$ is countable (both in $V$ and $V[g]$). (Note that the evaluation of the statement "$A$ is strong measure zero" may change from $V$ to $V[g]$). Now you have made some progress, there are no "old" counterexamples to BC anymore. But there still might be new counterexamples! It is now clear that one should try to iterate this process to make the new counterexamples vanish in the limit. This turns out to work, however it is not enough to iterate this just $omega$ or $omega_1$-many times, it is important to do it $omega_2$-many times (otherwise CH might still hold in the extension). Moreover one has to choose the support correctly, in this case it is countable support for technical reasons.
$endgroup$
1
$begingroup$
I think this is a bit problematic as written. I assume you are using the term "ordinary forcing" in a purely informal fashion, but it is really to informal to be useful. Why would adding $aleph_mathfrak c^+$ Cohen reals not be considered ordinary?
$endgroup$
– Andrés E. Caicedo
2 days ago
$begingroup$
In fact, it reads as if you are claiming that there is a well understood and reasonably wide and varied class of "ordinary forcing notions", and that no forcing in this class can force $lnotmathsfCH$ and (perhaps) that no forcing in this class is forcing isomorphic to an iteration.
$endgroup$
– Andrés E. Caicedo
2 days ago
$begingroup$
I agree that this is problematic. "Ordinary forcing" is the term OP used to phrase their question and I simply tried to make sense of this to appeal to OPs intuition.
$endgroup$
– Andreas Lietz
2 days ago
1
$begingroup$
I have edited out this section to avoid further confusion.
$endgroup$
– Andreas Lietz
2 days ago
add a comment |
$begingroup$
Maybe I can add a natural example to Asaf's great answer. In the case of establishing Borel's conjecture, iterated forcing naturally comes up in the construction. Borel's conjecture states that all strong measure zero sets of reals are countable (it is not important what that means). As it turns out, this statement is independent of ZFC. It is not too hard to construct a counterexample to it using CH. To show that BC is consistent, you need forcing. But how do you force BC? Richard Laver found a poset (now called Laver forcing) that is easy to describe and has the following property:
If $V[g]$ is a generic extension of $V$ via Laver forcing and $Ain V$ is a set of reals which is strong measure zero in $V[g]$, then $A$ is countable (both in $V$ and $V[g]$). (Note that the evaluation of the statement "$A$ is strong measure zero" may change from $V$ to $V[g]$). Now you have made some progress, there are no "old" counterexamples to BC anymore. But there still might be new counterexamples! It is now clear that one should try to iterate this process to make the new counterexamples vanish in the limit. This turns out to work, however it is not enough to iterate this just $omega$ or $omega_1$-many times, it is important to do it $omega_2$-many times (otherwise CH might still hold in the extension). Moreover one has to choose the support correctly, in this case it is countable support for technical reasons.
$endgroup$
Maybe I can add a natural example to Asaf's great answer. In the case of establishing Borel's conjecture, iterated forcing naturally comes up in the construction. Borel's conjecture states that all strong measure zero sets of reals are countable (it is not important what that means). As it turns out, this statement is independent of ZFC. It is not too hard to construct a counterexample to it using CH. To show that BC is consistent, you need forcing. But how do you force BC? Richard Laver found a poset (now called Laver forcing) that is easy to describe and has the following property:
If $V[g]$ is a generic extension of $V$ via Laver forcing and $Ain V$ is a set of reals which is strong measure zero in $V[g]$, then $A$ is countable (both in $V$ and $V[g]$). (Note that the evaluation of the statement "$A$ is strong measure zero" may change from $V$ to $V[g]$). Now you have made some progress, there are no "old" counterexamples to BC anymore. But there still might be new counterexamples! It is now clear that one should try to iterate this process to make the new counterexamples vanish in the limit. This turns out to work, however it is not enough to iterate this just $omega$ or $omega_1$-many times, it is important to do it $omega_2$-many times (otherwise CH might still hold in the extension). Moreover one has to choose the support correctly, in this case it is countable support for technical reasons.
edited 2 days ago
answered 2 days ago
Andreas LietzAndreas Lietz
39817
39817
1
$begingroup$
I think this is a bit problematic as written. I assume you are using the term "ordinary forcing" in a purely informal fashion, but it is really to informal to be useful. Why would adding $aleph_mathfrak c^+$ Cohen reals not be considered ordinary?
$endgroup$
– Andrés E. Caicedo
2 days ago
$begingroup$
In fact, it reads as if you are claiming that there is a well understood and reasonably wide and varied class of "ordinary forcing notions", and that no forcing in this class can force $lnotmathsfCH$ and (perhaps) that no forcing in this class is forcing isomorphic to an iteration.
$endgroup$
– Andrés E. Caicedo
2 days ago
$begingroup$
I agree that this is problematic. "Ordinary forcing" is the term OP used to phrase their question and I simply tried to make sense of this to appeal to OPs intuition.
$endgroup$
– Andreas Lietz
2 days ago
1
$begingroup$
I have edited out this section to avoid further confusion.
$endgroup$
– Andreas Lietz
2 days ago
add a comment |
1
$begingroup$
I think this is a bit problematic as written. I assume you are using the term "ordinary forcing" in a purely informal fashion, but it is really to informal to be useful. Why would adding $aleph_mathfrak c^+$ Cohen reals not be considered ordinary?
$endgroup$
– Andrés E. Caicedo
2 days ago
$begingroup$
In fact, it reads as if you are claiming that there is a well understood and reasonably wide and varied class of "ordinary forcing notions", and that no forcing in this class can force $lnotmathsfCH$ and (perhaps) that no forcing in this class is forcing isomorphic to an iteration.
$endgroup$
– Andrés E. Caicedo
2 days ago
$begingroup$
I agree that this is problematic. "Ordinary forcing" is the term OP used to phrase their question and I simply tried to make sense of this to appeal to OPs intuition.
$endgroup$
– Andreas Lietz
2 days ago
1
$begingroup$
I have edited out this section to avoid further confusion.
$endgroup$
– Andreas Lietz
2 days ago
1
1
$begingroup$
I think this is a bit problematic as written. I assume you are using the term "ordinary forcing" in a purely informal fashion, but it is really to informal to be useful. Why would adding $aleph_mathfrak c^+$ Cohen reals not be considered ordinary?
$endgroup$
– Andrés E. Caicedo
2 days ago
$begingroup$
I think this is a bit problematic as written. I assume you are using the term "ordinary forcing" in a purely informal fashion, but it is really to informal to be useful. Why would adding $aleph_mathfrak c^+$ Cohen reals not be considered ordinary?
$endgroup$
– Andrés E. Caicedo
2 days ago
$begingroup$
In fact, it reads as if you are claiming that there is a well understood and reasonably wide and varied class of "ordinary forcing notions", and that no forcing in this class can force $lnotmathsfCH$ and (perhaps) that no forcing in this class is forcing isomorphic to an iteration.
$endgroup$
– Andrés E. Caicedo
2 days ago
$begingroup$
In fact, it reads as if you are claiming that there is a well understood and reasonably wide and varied class of "ordinary forcing notions", and that no forcing in this class can force $lnotmathsfCH$ and (perhaps) that no forcing in this class is forcing isomorphic to an iteration.
$endgroup$
– Andrés E. Caicedo
2 days ago
$begingroup$
I agree that this is problematic. "Ordinary forcing" is the term OP used to phrase their question and I simply tried to make sense of this to appeal to OPs intuition.
$endgroup$
– Andreas Lietz
2 days ago
$begingroup$
I agree that this is problematic. "Ordinary forcing" is the term OP used to phrase their question and I simply tried to make sense of this to appeal to OPs intuition.
$endgroup$
– Andreas Lietz
2 days ago
1
1
$begingroup$
I have edited out this section to avoid further confusion.
$endgroup$
– Andreas Lietz
2 days ago
$begingroup$
I have edited out this section to avoid further confusion.
$endgroup$
– Andreas Lietz
2 days ago
add a comment |
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$begingroup$
Always. All forcing is iterated forcing.
$endgroup$
– Asaf Karagila♦
2 days ago