We know $PFA$ implies $2^{\aleph_0}=\aleph_2$.

Q1. What does $PFA$ say about other values of continuum function? Does proper forcing axiom carry any further information about values of continuum function in other cardinals greater than $\aleph_{0}$? In the other words, is there any other known non-trivial result in the form "$ZFC+PFA\vdash 2^{\aleph_{\alpha}}=\aleph_{\beta}$" for some ordinals $\alpha , \beta$, or all other situations are consistent with $PFA$ like Easton's theorem in presence of some large cardinal?

In the direction of my previous question on Godel's program for deciding $CH$ and $GCH$ using adding large cardinal axioms to $ZFC$ and based on the impact of $PFA$ on $CH$ which is compatible with Godel's conjecture on refuting $CH$ and $GCH$ using large cardinal axioms. One can consider existence of another large cardinal tree in the shadow of the current standard tree of large cardinal assumptions including $PFA$ in a rank near supercompacts. (The equiconsistency of PFA and supercompacts is not proved yet but we assume it for straightforwardness of the below diagram.)

Maybe Godel's conjecture and program are true if we replace the standard tree of large cardinals with another tree with equiconsistent steps. Please note the following imaginary large cardinal ladder which includes two parallel trees of "large cardinal" assumptions:

Q2. Can we complete the middle part of the following ladder by some axioms like $A$, $B$, $C$, ..., $X$ which are equiconsistent with usual large cardinal axioms and also refute $CH$, $GCH$ and $V=L$ in direction of Godel's program?

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Update. Using above approach to large cardinal axioms (inspired by success of $PFA$ in deciding $CH$ and its possible equiconsistency with existence of supercompact cardinals) we can restate Godel's program in deciding independent statements of set theory as follows:

For every independent statement $\sigma$ from $ZFC$ there is a (standard) large cardinal axiom $I$ and a statement $J$ (a non-standard large cardinal axiom) such that:

(a) Consistency of $ZFC+I$ implies consistency of $ZFC+J$.

(b) $ZFC+J$ decides $\sigma$ (i.e. $ZFC+J\vdash \sigma$ or $ZFC+J\vdash \neg\sigma$)

In the other words one can decide any independent statement of mathematics using some standard/non-standard large cardinal axiom.

Q3. Is the above thesis true?

I would like to thank Asaf Karagila for his comment in my previous question that was the main guide for this question.

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    $\begingroup$ It known that $PFA$ (and even $MA$) also implies $2^{\aleph_1} = \aleph_2$. For higher regular cardinals you can force whatever you want. For singular - it seems like an interesting question... $\endgroup$ – Yair Hayut Jan 23 '14 at 6:19
  • $\begingroup$ @YairHayut: The singular case of Easton's theorem is not easy in presence of large cardinals too. Maybe one can find useful related results in Moti Gitik's papers. $\endgroup$ – user45878 Jan 23 '14 at 6:46
  • $\begingroup$ Quick observation: it is not known that PFA is equiconsistent with a supercompact. $\endgroup$ – Noah Schweber Jan 23 '14 at 8:10
  • $\begingroup$ @NoahS: Dear Noah please note the sentence in the parentheses. $\endgroup$ – user45878 Jan 23 '14 at 10:50
  • $\begingroup$ math.stackexchange.com/questions/21097/… $\endgroup$ – Asaf Karagila Jan 23 '14 at 10:53

First note that PFA implies SCH. On the other hand, PFA is indestructible under $\aleph_2-$directed closed forcings, so we can force any pattern of Easton theorem for regular cardinals $\geq \aleph_2.$ By SCH the power function on singular cardinals is completely determined by its behavior on regular cardinals

  • $\begingroup$ Thank you very much Mohammad. Your answer is very useful. $\endgroup$ – user45878 Jan 23 '14 at 7:14
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    $\begingroup$ Dear Jitra, I think most set theorists beleive in $2^{\aleph_0}=\aleph_2.$ I strongly suggest the following paper "Rectangular axioms, perfect set properties and decomposition". The paper is based on some attempts of Godel. $\endgroup$ – Mohammad Golshani Jan 23 '14 at 8:45
  • $\begingroup$ Godel in his paper "What is Cantor's Continuum Problem?" suggests this one: $2^{\aleph_0}=\text{The least weakly inaccessible cardinal}$. We know the size of continuum could be even larger such as a real valued measurable cardinal. $\endgroup$ – user45878 Jan 23 '14 at 10:57
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    $\begingroup$ @Mohammad, I think your remark about "most set theorists" may be over-stated, since I think a more common view is that the situation is basically unsettled. $\endgroup$ – Joel David Hamkins Jan 23 '14 at 14:19

If we start with an indestructibly supercompact cardinal $\kappa$, then we can do Easton forcing to get any reasonable continuum function on the regular cardinals $\geq \kappa$ and retain the supercompactness of $\kappa$. Then we can force PFA via the original Baumgartner way with a $\kappa$-c.c. forcing of size $\kappa$. Using these facts, it is then easy to show that no cardinals above $\kappa$ are collapsed, and the same cardinal arithmetic will hold after forcing PFA.

  • $\begingroup$ Very nice answer Monroe. Can you introduce some reference or paper for further explanations? $\endgroup$ – user45878 Jan 23 '14 at 6:26

Question 2 admits a trivial answer, because any assertion implies $0\neq 1$, provably so in ZFC, and so for $A$ you can just use "there is an inaccessible cardinal". Similarly, the existence of $0^\sharp$ already implies $V\neq L$, and so you can use $B=$ "$0^\sharp$ exists". And for C, you can use "there is a measurable cardinal and $\neg$CH"; this is equiconsistent with a measurable cardinal, and it implies $\neg$CH. For $X$, you can just use the Reinhardt cardinal again, since any statement implies $0\neq 1$ in ZFC.

Similarly, question 3 also has such a kind of unsatisfactory simple answer. Suppose that $\sigma$ is any statement at all, and $I$ is any large cardinal axiom. Since any particular model of ZFC+I will either satisfy $\sigma$ or satisfy $\neg\sigma$, we can deduce from Con(ZFC+I) that one of Con(ZFC$+I+\sigma$) or Con(ZFC$+I+\neg\sigma$) holds. Let $J$ be either $I+\sigma$ or $I+\neg\sigma$ accordingly, so we get property (a) as a material implication, and property (b) holds because $J$ settles $\sigma$ explicitly.

  • $\begingroup$ But I suppose for question 3, one really wants the consistency implication to be provable, and not merely true. $\endgroup$ – Joel David Hamkins Jan 23 '14 at 14:45
  • $\begingroup$ Thank you very much. By the questions 2, 3 I mean something like a new characterization for large cardinal axioms with a possibly better direct implication power. A new set of large cardinal axioms which are natural in definition and powerful in decision about basic essential mathematical questions like $CH$. $\endgroup$ – user45878 Jan 23 '14 at 17:54

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