intuition for PFA
$begingroup$
Kunen (Set theory 2011) says on the page 307:
The Proper Forcing Axiom (PFA) is the assertion that $MA_{mathbb P}(aleph_1)$ holds for all proper $mathbb P$.
My question is, what kind of axiom this is, i.e. on what level it is
supposed to hold. Can we added to the axioms of ZFC, like CH, say?
What is the consistency strength of PFA? Is it implied by some LC or rather $V=L$?
set-theory axioms forcing large-cardinals
edited Dec 23 '18 at 18:25
Andrés E. Caicedo
65.7k8160250
asked Dec 23 '18 at 17:13
user122424user122424
1,1582717
$endgroup$
add a comment |
$begingroup$
Kunen (Set theory 2011) says on the page 307:
The Proper Forcing Axiom (PFA) is the assertion that $MA_{mathbb P}(aleph_1)$ holds for all proper $mathbb P$.
My question is, what kind of axiom this is, i.e. on what level it is
supposed to hold. Can we added to the axioms of ZFC, like CH, say?
What is the consistency strength of PFA? Is it implied by some LC or rather $V=L$?
set-theory axioms forcing large-cardinals
edited Dec 23 '18 at 18:25
Andrés E. Caicedo
65.7k8160250
asked Dec 23 '18 at 17:13
user122424user122424
1,1582717
$endgroup$
add a comment |
$begingroup$
Kunen (Set theory 2011) says on the page 307:
The Proper Forcing Axiom (PFA) is the assertion that $MA_{mathbb P}(aleph_1)$ holds for all proper $mathbb P$.
My question is, what kind of axiom this is, i.e. on what level it is
supposed to hold. Can we added to the axioms of ZFC, like CH, say?
What is the consistency strength of PFA? Is it implied by some LC or rather $V=L$?
set-theory axioms forcing large-cardinals
edited Dec 23 '18 at 18:25
Andrés E. Caicedo
65.7k8160250
asked Dec 23 '18 at 17:13
user122424user122424
1,1582717
$endgroup$
Kunen (Set theory 2011) says on the page 307:
The Proper Forcing Axiom (PFA) is the assertion that $MA_{mathbb P}(aleph_1)$ holds for all proper $mathbb P$.
My question is, what kind of axiom this is, i.e. on what level it is
supposed to hold. Can we added to the axioms of ZFC, like CH, say?
What is the consistency strength of PFA? Is it implied by some LC or rather $V=L$?
set-theory axioms forcing large-cardinals
set-theory axioms forcing large-cardinals
edited Dec 23 '18 at 18:25
Andrés E. Caicedo
65.7k8160250
asked Dec 23 '18 at 17:13
user122424user122424
1,1582717
edited Dec 23 '18 at 18:25
Andrés E. Caicedo
65.7k8160250
asked Dec 23 '18 at 17:13
user122424user122424
1,1582717
edited Dec 23 '18 at 18:25
Andrés E. Caicedo
65.7k8160250
edited Dec 23 '18 at 18:25
Andrés E. Caicedo
65.7k8160250
edited Dec 23 '18 at 18:25
Andrés E. Caicedo
65.7k8160250
65.7k8160250
asked Dec 23 '18 at 17:13
user122424user122424
1,1582717
asked Dec 23 '18 at 17:13
user122424user122424
1,1582717
asked Dec 23 '18 at 17:13
user122424user122424
1,1582717
1,1582717
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First of all, note that since PFA implies that CH fails (indeed, it implies $2^{aleph_0}=aleph_2$, but this is harder to show), PFA contradicts V=L.
Moving to more interesting facts, the consistency strength of (ZFC+)PFA is quite large. The best known upper bound is a supercompact cardinal, which is completely incompatible with $L$ - not only does $L$ not have supercompacts, no forcing extension of $L$ has supercompacts either. PFA is indeed conjectured to have consistency strength exactly a supercompact, although this is a major open problem; see this MO question for a discussion of the subject.
Finally, in response to your question "Is PFA implied by some large cardinal?" the answer is no (as far as we understand - "large cardinal" isn't a precise term). The issue is that large cardinals don't decide the continuum hypothesis (again, this is heuristic), whereas PFA does (namely, it implies its negation), so large cardinals can't outright imply PFA. Large cardinals can imply the existence of inner models or forcing extensions in which a given combinatorial property holds. At least the latter is true in the case of PFA - given a supercompact, we can force PFA (similarly to how we force MA+$neg$CH, incidentally) - and if I recall correctly so is the former, but I can't find a citation at the moment.
answered Dec 23 '18 at 17:28
Noah SchweberNoah Schweber
126k10151290
$endgroup$
3
$begingroup$
In general, forcing extensions don't tend to create large cardinals. So if $L$ did not have a kind of large cardinal it is unexpected that a forcing extension would.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 18:20
$begingroup$
Yes, I think there are significant ways in which to justify the "complete incompatibility" you mention. That forcing does not add large cardinals by accident is definitely not one of them. :-)
$endgroup$
– Andrés E. Caicedo
Dec 23 '18 at 19:44
add a comment |
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1 Answer
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1 Answer
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$begingroup$
First of all, note that since PFA implies that CH fails (indeed, it implies $2^{aleph_0}=aleph_2$, but this is harder to show), PFA contradicts V=L.
Moving to more interesting facts, the consistency strength of (ZFC+)PFA is quite large. The best known upper bound is a supercompact cardinal, which is completely incompatible with $L$ - not only does $L$ not have supercompacts, no forcing extension of $L$ has supercompacts either. PFA is indeed conjectured to have consistency strength exactly a supercompact, although this is a major open problem; see this MO question for a discussion of the subject.
Finally, in response to your question "Is PFA implied by some large cardinal?" the answer is no (as far as we understand - "large cardinal" isn't a precise term). The issue is that large cardinals don't decide the continuum hypothesis (again, this is heuristic), whereas PFA does (namely, it implies its negation), so large cardinals can't outright imply PFA. Large cardinals can imply the existence of inner models or forcing extensions in which a given combinatorial property holds. At least the latter is true in the case of PFA - given a supercompact, we can force PFA (similarly to how we force MA+$neg$CH, incidentally) - and if I recall correctly so is the former, but I can't find a citation at the moment.
answered Dec 23 '18 at 17:28
Noah SchweberNoah Schweber
126k10151290
$endgroup$
3
$begingroup$
In general, forcing extensions don't tend to create large cardinals. So if $L$ did not have a kind of large cardinal it is unexpected that a forcing extension would.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 18:20
$begingroup$
Yes, I think there are significant ways in which to justify the "complete incompatibility" you mention. That forcing does not add large cardinals by accident is definitely not one of them. :-)
$endgroup$
– Andrés E. Caicedo
Dec 23 '18 at 19:44
add a comment |
$begingroup$
First of all, note that since PFA implies that CH fails (indeed, it implies $2^{aleph_0}=aleph_2$, but this is harder to show), PFA contradicts V=L.
Moving to more interesting facts, the consistency strength of (ZFC+)PFA is quite large. The best known upper bound is a supercompact cardinal, which is completely incompatible with $L$ - not only does $L$ not have supercompacts, no forcing extension of $L$ has supercompacts either. PFA is indeed conjectured to have consistency strength exactly a supercompact, although this is a major open problem; see this MO question for a discussion of the subject.
Finally, in response to your question "Is PFA implied by some large cardinal?" the answer is no (as far as we understand - "large cardinal" isn't a precise term). The issue is that large cardinals don't decide the continuum hypothesis (again, this is heuristic), whereas PFA does (namely, it implies its negation), so large cardinals can't outright imply PFA. Large cardinals can imply the existence of inner models or forcing extensions in which a given combinatorial property holds. At least the latter is true in the case of PFA - given a supercompact, we can force PFA (similarly to how we force MA+$neg$CH, incidentally) - and if I recall correctly so is the former, but I can't find a citation at the moment.
answered Dec 23 '18 at 17:28
Noah SchweberNoah Schweber
126k10151290
$endgroup$
3
$begingroup$
In general, forcing extensions don't tend to create large cardinals. So if $L$ did not have a kind of large cardinal it is unexpected that a forcing extension would.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 18:20
$begingroup$
Yes, I think there are significant ways in which to justify the "complete incompatibility" you mention. That forcing does not add large cardinals by accident is definitely not one of them. :-)
$endgroup$
– Andrés E. Caicedo
Dec 23 '18 at 19:44
add a comment |
$begingroup$
First of all, note that since PFA implies that CH fails (indeed, it implies $2^{aleph_0}=aleph_2$, but this is harder to show), PFA contradicts V=L.
Moving to more interesting facts, the consistency strength of (ZFC+)PFA is quite large. The best known upper bound is a supercompact cardinal, which is completely incompatible with $L$ - not only does $L$ not have supercompacts, no forcing extension of $L$ has supercompacts either. PFA is indeed conjectured to have consistency strength exactly a supercompact, although this is a major open problem; see this MO question for a discussion of the subject.
Finally, in response to your question "Is PFA implied by some large cardinal?" the answer is no (as far as we understand - "large cardinal" isn't a precise term). The issue is that large cardinals don't decide the continuum hypothesis (again, this is heuristic), whereas PFA does (namely, it implies its negation), so large cardinals can't outright imply PFA. Large cardinals can imply the existence of inner models or forcing extensions in which a given combinatorial property holds. At least the latter is true in the case of PFA - given a supercompact, we can force PFA (similarly to how we force MA+$neg$CH, incidentally) - and if I recall correctly so is the former, but I can't find a citation at the moment.
answered Dec 23 '18 at 17:28
Noah SchweberNoah Schweber
126k10151290
$endgroup$
First of all, note that since PFA implies that CH fails (indeed, it implies $2^{aleph_0}=aleph_2$, but this is harder to show), PFA contradicts V=L.
Moving to more interesting facts, the consistency strength of (ZFC+)PFA is quite large. The best known upper bound is a supercompact cardinal, which is completely incompatible with $L$ - not only does $L$ not have supercompacts, no forcing extension of $L$ has supercompacts either. PFA is indeed conjectured to have consistency strength exactly a supercompact, although this is a major open problem; see this MO question for a discussion of the subject.
Finally, in response to your question "Is PFA implied by some large cardinal?" the answer is no (as far as we understand - "large cardinal" isn't a precise term). The issue is that large cardinals don't decide the continuum hypothesis (again, this is heuristic), whereas PFA does (namely, it implies its negation), so large cardinals can't outright imply PFA. Large cardinals can imply the existence of inner models or forcing extensions in which a given combinatorial property holds. At least the latter is true in the case of PFA - given a supercompact, we can force PFA (similarly to how we force MA+$neg$CH, incidentally) - and if I recall correctly so is the former, but I can't find a citation at the moment.
answered Dec 23 '18 at 17:28
Noah SchweberNoah Schweber
126k10151290
answered Dec 23 '18 at 17:28
Noah SchweberNoah Schweber
126k10151290
answered Dec 23 '18 at 17:28
Noah SchweberNoah Schweber
126k10151290
answered Dec 23 '18 at 17:28
Noah SchweberNoah Schweber
126k10151290
126k10151290
3
$begingroup$
In general, forcing extensions don't tend to create large cardinals. So if $L$ did not have a kind of large cardinal it is unexpected that a forcing extension would.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 18:20
$begingroup$
Yes, I think there are significant ways in which to justify the "complete incompatibility" you mention. That forcing does not add large cardinals by accident is definitely not one of them. :-)
$endgroup$
– Andrés E. Caicedo
Dec 23 '18 at 19:44
add a comment |
3
$begingroup$
In general, forcing extensions don't tend to create large cardinals. So if $L$ did not have a kind of large cardinal it is unexpected that a forcing extension would.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 18:20
$begingroup$
Yes, I think there are significant ways in which to justify the "complete incompatibility" you mention. That forcing does not add large cardinals by accident is definitely not one of them. :-)
$endgroup$
– Andrés E. Caicedo
Dec 23 '18 at 19:44
3
3
$begingroup$
In general, forcing extensions don't tend to create large cardinals. So if $L$ did not have a kind of large cardinal it is unexpected that a forcing extension would.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 18:20
$begingroup$
In general, forcing extensions don't tend to create large cardinals. So if $L$ did not have a kind of large cardinal it is unexpected that a forcing extension would.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 18:20
$begingroup$
Yes, I think there are significant ways in which to justify the "complete incompatibility" you mention. That forcing does not add large cardinals by accident is definitely not one of them. :-)
$endgroup$
– Andrés E. Caicedo
Dec 23 '18 at 19:44
$begingroup$
Yes, I think there are significant ways in which to justify the "complete incompatibility" you mention. That forcing does not add large cardinals by accident is definitely not one of them. :-)
$endgroup$
– Andrés E. Caicedo
Dec 23 '18 at 19:44
add a comment |
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