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edited Nov 20 at 14:51

As Andres and François noted

First, let me remark that the particular way that you've formalized posed the question is open to has several problematic issues of formalization. For exampleOne issue, ZFC noted by François, Andres and Andreas, is consistent that it doesn't make sense to speak about proving an assertion with an ordinal parameter (one would instead want to speak of definitions of particular ordinals). Another issue is that for all we know, we may be living in a universe with ZFC + $\neg\text{Con}(\text{ZFC})$, and any model of that theory thinks that both in this universe everything is provable, so even if we are able to resolve the first issue nevertheless the sets $S$ and $S'$ are will be empty, since this model thinks everything is provable. I would say that your question is considering the provability of

So let me propose a propertymore semantic, when it is really alternative version of the truth question, which to my of thinking gets at the property issue in a model that which I believe you would seem to want to considerare interested.

So let's drop the provability angle,

Question. If $\alpha$ is an ordinal and consider instead the continuum $2^{\aleph_0}$ can be $\aleph_\alpha$ in a purely semantic version forcing extension of the question, namelyuniverse, do the possibe values of then can the continuum be $\aleph_\alpha$ in a forcing extension depend on whether we of the universe in which also want the OPP to hold in that extensionholds?

The answer is noyes, and so in this sense the same cardinals can be realized OPP imposes no additional constraints on the value of the continuum. In this question and in any casethe theorem below, I am speaking about possibly proper class forcing, and this is required, since if the OPP fails unboundedly often, it will require proper class forcing to force OPP again.

Theorem. If the universe $V$ is a model of ZFC+GCHsatisifes ZFC, then for any cardinal ordinal $\delta$ \alpha$, the following are equivalent:

There is a forcing extension $V[G]$ in which the continuum is $\delta$. 2^\omega=\aleph_\alpha$.

There is a forcing extension $V[G]$ in which the continuum is $\delta$ 2^\omega=\aleph_\alpha$ and the OPP holds.

The cardinal

Either $\delta$ \alpha$ is a successor ordinal or $\alpha$ has uncountable cofinality.

Proof. Clearly 2 implies 1, and 1 implies 3by Konig's theorem. So assume Suppose 3 holds, and we I argue for 2. Simply choose Fix any ordinal $\alpha$ as in $3$. First, we may simply force the GCH by the canonical forcing of the GCH. This forcing (which may be a proper class), is countably closed and hence preserves the property of having uncountable cofinality. So $3$ still holds about $\alpha$ in the extension with GCH. We may now simply apply Easton's theorem, using an Easton function $E$ with the right value at that takes $\aleph_0$, such as \aleph_0$ to the function with current $E(\aleph_0)=\delta=\aleph_\alpha$ \aleph_\alpha$, and more generally which takes $E(\aleph_\beta)=\aleph_{\alpha+\beta+1}$ for \aleph_\beta$ to $\beta\gt 0$. \aleph_{\alpha+\beta+1}$. (But any strictly increasing Easton function will do, and there are many variations.) Note that $\alpha+\beta=\beta$ once $\alpha\cdot\omega\leq\beta$, and so this pattern is eventually the GCH pattern.) The point is that there are numerous strictly increasing Easton functions having a desired value for $\aleph_0$, provided that that value satisfies the cofinality requirement of Konig's theorempattern. It follows by By Easton's theorem that , there is a further forcing extension $V[G]$ in which $E$ is 2^{\aleph_0}=\aleph_\alpha$ and the continuum function $\kappa\mapsto 2^\kappa$ for regular cardinals $\kappa$. Since is given by $E$ E$, which is strictly increasing, we obtain so the OPP in $V[G]$, as desired, while ensuring that the continuum is $\delta$. So $2$ holds. QED

In particular, over for any model of ZFC+GCHordinal $\alpha$ that you care to define, we then you can provably force that the continuum is to become $\aleph_\alpha$ for a particular ordinal $\alpha$ if and only if we you can do so while preserving also ensuring the OPP.

Perhaps this formulation is closer to what you want,

Notice that in terms the proof of the ordinal index theorem, the value of $\aleph_\alpha$ may have changed, during the cardinalforcing of the GCH, and here since this will collapse cardinals if the GCH did not already hold. So there is another version of the question, which is about cardinals, rather than about ordinals. If we don't need any start with the GCH, then a similar conclusion can be made.

Theorem. If $V$ is any a model of ZFCZFC+GCH, then for any ordinal cardinal $\alpha$, \delta$ the following are equivalent:

There is a forcing extension $V[G]$ in which the continuum is $2^\omega=\aleph_\alpha$.\delta$.

There is a forcing extension $V[G]$ in which the continuum is $2^\omega=\aleph_\alpha$ \delta$ and the OPP holds.

Either $\alpha$ is a successor ordinal or

The cardinal $\alpha$ \delta$ has uncountable cofinality.

Proof. Clearly 2 implies 1, and 1 implies 3

The proof is essentially the same as above. Suppose The nontrivial part is 3 holds, and I argue for implies 2. If $\alpha$ is a successor ordinal, then we which can simply force the GCH and then apply be achieved via Easton's theorem as in by using a strictly increasing Easton function $E$ with the previous theorem to get 2property that $E(\aleph_0)=\aleph_\alpha$. There are many choices of such $E$, such as $E(\aleph_\beta)=\aleph_{\alpha+\beta+1}$, as above. If Any such $\alpha$ E$ will ensure the right value for $2^{\aleph_0}$ and, because it is a limit ordinal with uncountable cofinalitystrictly increasing, then will also achieve the OPP. In this case, since we may first force started with the GCHabove $\aleph_0$. This , one requires only set-sized forcingis countably closed and hence preserves .

By the fact that $\alpha$ has uncountable cofinalityway, there seems to be alternative terminology to refer to what you call the OPP. NowFor example, we may apply Easton's theorem in my paper, "Is the dream solution of the continuum hypothesis attainable?", I refer to make $2^\kappa$ increasingthe power set size axiom, denoted, PSA, and much bigger than (this is the new) $\aleph_\alpha$, thereby getting same as what you call OPP. Finally, we may add $\aleph_\alpha$ many Cohen reals to This axiom also achieve $2^\omega=\aleph_\alpha$ while retaining appears in the OPPMO question on reasonable-sounding statements that are independent of ZFC. QED

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edited Nov 20 at 11:44

Joel David Hamkins
75.1k●7●187●343

As Andres and François noted, the particular way that you've formalized the question is open to problematic issues. For example, ZFC is consistent with $\neg\text{Con}(\text{ZFC})$, and any model of that theory thinks that both $S$ and $S'$ are empty, since this model thinks everything is provable. I would say that your question is considering the provability of a property, when it is really the truth of the property in a model that you would seem to want to consider.

So let's drop the provability angle, and consider instead a purely semantic version of the question, namely, do the possibe values of the continuum in a forcing extension depend on whether we also want the OPP to hold in that extension?

The answer is no, the same cardinals can be realized in any case.

Theorem. If $V$ is a model of ZFC+GCH, then for any cardinal $\delta$ the following are equivalent:

There is a forcing extension $V[G]$ in which the continuum is $\delta$.
There is a forcing extension $V[G]$ in which the continuum is $\delta$ and the OPP holds.
The cardinal $\delta$ has uncountable cofinality.

Proof. Clearly 2 implies 1, and 1 implies 3 by Konig's theorem. So assume 3, and we argue for 2. Simply choose any Easton function $E$ with the right value at $\aleph_0$, such as the function with $E(\aleph_0)=\delta=\aleph_\alpha$ and $E(\aleph_\beta)=\aleph_{\alpha+\beta+1}$ for $\beta\gt 0$. (Note that $\alpha+\beta=\beta$ once $\alpha\cdot\omega\leq\beta$, and so this pattern is eventually the GCH pattern.) The point is that there are numerous strictly increasing Easton functions having a desired value for $\aleph_0$, provided that that value satisfies the cofinality requirement of Konig's theorem. It follows by Easton's theorem that there is a forcing extension $V[G]$ in which $E$ is the continuum function $\kappa\mapsto 2^\kappa$ for regular cardinals $\kappa$. Since $E$ is strictly increasing, we obtain the OPP in $V[G]$, as desired, while ensuring that the continuum is $\delta$. So $2$ holds. QED

In particular, over any model of ZFC+GCH, we can force that the continuum is $\aleph_\alpha$ for a particular ordinal $\alpha$ if and only if we can do so while preserving the OPP.

Perhaps this formulation is closer to what you want, in terms of the ordinal index of the cardinal, and here we don't need any GCH.

Theorem. If $V$ is any model of ZFC, then for any ordinal $\alpha$, the following are equivalent:

There is a forcing extension in which $2^\omega=\aleph_\alpha$.
There is a forcing extension in which $2^\omega=\aleph_\alpha$ and the OPP holds.
Either $\alpha$ is a successor ordinal or $\alpha$ has uncountable cofinality.

Proof. Clearly 2 implies 1, and 1 implies 3. Suppose 1 3 holds, and I argue for 2. It follows that either If $\alpha$ is a successor ordinalor that it has uncountable cofinality. In the first case, then we could can simply force the GCH and then apply Easton's theorem as in the previous theorem to get 2. If $\alpha$ is a limit ordinal , then by 1 it must have with uncountable cofinality. In this case, then we may first force the GCH above $\aleph_0$. This forcing is countably closed and hence preserves the fact that $\alpha$ has uncountable cofinality. Now, we may apply Easton's theorem to make $2^\kappa$ increasing, and much bigger than (the new) $\aleph_\alpha$, thereby getting OPP. Finally, we may add $\aleph_\alpha$ many Cohen reals to also achieve $2^\omega=\aleph_\alpha$ while retaining the OPP. QED

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edited Nov 20 at 11:20

Joel David Hamkins
75.1k●7●187●343

Proof. Clearly 2 implies 1, and 1 implies 3 by Konig's theorem. So assume 3, and we argue for 2. Let $E$ be an Simply choose any Easton function $E$ with the property that right value at $E(\aleph_0)=\delta=\aleph_\alpha$, \aleph_0$, such as the function with $E(\aleph_0)=\delta=\aleph_\alpha$ and more generally $E(\aleph_\beta)=\aleph_{\alpha+\beta}$, which might also be denoted E(\aleph_\beta)=\aleph_{\alpha+\beta+1}$ for $\delta^{+^\beta}$. This \beta\gt 0$. The point is that there are numerous strictly increasing Easton functions having a desired value for $\aleph_0$, provided that that value satisfies the requirements cofinality requirement of Konig's theorem. It follows by Easton's theorem, and so that there is a forcing extension $V[G]$ in which $E$ is the continuum function $\kappa\mapsto 2^\kappa$ for regular cardinals $\kappa$. Since $E$ is strictly increasing, we obtain the OPP in $V[G]$, as desired, while ensuring that the continuum is $\delta$. So $2$ holds. QED

If we only

Perhaps this formulation is closer to what you want($1\Leftrightarrow 2$), then in terms of the ordinal index of the cardinal, and here we don't need to require any GCHhere.The point

Theorem. If $V$ is any model of ZFC, then for any ordinal $\alpha$, the following are equivalent:

There is a forcing extension in which $2^\omega=\aleph_\alpha$.

There is a forcing extension in which $2^\omega=\aleph_\alpha$ and the OPP holds.

Proof. Clearly 2 implies 1. Suppose 1 holds. It follows that either $\alpha$ is a successor ordinal or that if it has uncountable cofinality. In the first case, we can could simply force the GCH and then apply Easton's theorem as in the previous theorem to get 2. If $2^\omega=\delta$, \alpha$ is a limit ordinal, then we can combine by 1 it must have uncountable cofinality. In this with case, we may first force the GCH above $\aleph_0$. This forcing is countably closed and hence preserves the fact that $\alpha$ has uncountable cofinality. Now, we may apply Easton's theorem to ensure make $2^\kappa$ increasing, and much bigger than (the new) $\aleph_\alpha$, thereby getting OPPon . Finally, we may add $\aleph_\alpha$ many Cohen reals to also achieve $2^\omega=\aleph_\alpha$ while retaining the other cardinals using Easton's methodOPP. QED

Two minor typos, I believe.

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edited Nov 20 at 5:42

Asaf Karagila
7,198●1●14●44

The answer is no, the same cardinals can be realized in any case.

Theorem. If $V$ is a model of ZFC+GCH, then for any cardinal $\delta$ the following are equivalent:

There is a forcing extension $V[G]$ in which the continuum is $\delta$.
There is a forcing extension $V[G]$ in which the continuum is $\delta$ and the OPP holds.
The cardinal $\delta$ has uncountable cofinality.

Proof. Clearly 2 implies 1, and 1 implies 3 by Konig's theorem. So assume 3, and we argue for 2. Let $E$ be an Easton function with the property that $E(\aleph_0)=delta=\aleph_\alpha$, E(\aleph_0)=\delta=\aleph_\alpha$, and more generally $E(\aleph_\beta)=\aleph_{\alpha+\beta}$, which might also be denoted $\delta^{+^\beta}$. This satisfies the requirements of Easton's theorem, and so there is a forcing extension $V[G]$ in which $E$ is the continuum function $\kappa\mapsto 2^\kappa$ for regular cardinals $\kappa$. Since $E$ is strictly increasing, we obtain the OPP in $V[G]$, as desired, while ensuring that the continuum is $\delta$. So $2$ holds. QED

In particular, over any model of ZFC+GHCZFC+GCH, we can force that the continuum is $\aleph_\alpha$ for a particular ordinal $\alpha$ if and only if we can do so while preserving the OPP.

If we only want ($1\Leftrightarrow 2$), then we don't need to require GCH here. The point is that if we can force $2^\omega=\delta$, then we can combine this with forcing to ensure OPP on the other cardinals using Easton's method.

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answered Nov 20 at 4:19

Joel David Hamkins
75.1k●7●187●343

The answer is no, the same cardinals can be realized in any case.

Theorem. If $V$ is a model of ZFC+GCH, then for any cardinal $\delta$ the following are equivalent:

There is a forcing extension $V[G]$ in which the continuum is $\delta$.
There is a forcing extension $V[G]$ in which the continuum is $\delta$ and the OPP holds.
The cardinal $\delta$ has uncountable cofinality.

Proof. Clearly 2 implies 1, and 1 implies 3 by Konig's theorem. So assume 3, and we argue for 2. Let $E$ be an Easton function with the property that $E(\aleph_0)=delta=\aleph_\alpha$, and more generally $E(\aleph_\beta)=\aleph_{\alpha+\beta}$, which might also be denoted $\delta^{+^\beta}$. This satisfies the requirements of Easton's theorem, and so there is a forcing extension $V[G]$ in which $E$ is the continuum function $\kappa\mapsto 2^\kappa$ for regular cardinals $\kappa$. Since $E$ is strictly increasing, we obtain the OPP in $V[G]$, as desired, while ensuring that the continuum is $\delta$. So $2$ holds. QED

In particular, over any model of ZFC+GHC, we can force that the continuum is $\aleph_\alpha$ for a particular ordinal $\alpha$ if and only if we can do so while preserving the OPP.