# Is it consistent with ZFC that for all ordinals $\alpha, \beta < \omega$ it holds that $2^{\aleph_\alpha} = 2^{\aleph_\beta}$?

Let $\gamma=\omega$ (the first transfinite ordinal). Is it consistent with ZFC that for all ordinals $\alpha, \beta < \gamma$ it holds that $2^{\aleph_\alpha} = 2^{\aleph_\beta}$? If yes, can the bound $\gamma$ be increased here and how much?

Update: In what sense the bound $\gamma$ can be made arbitrarily high? If $\beta$ is the initial ordinal of $\beth_1$, then it cannot be that $2^{\aleph_0}=2^{\aleph_\beta}$, right?

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In your update $\gamma$ is a free variable. I assume it comes from the answers, but please point out exactly what it signifies. – Asaf Karagila Dec 6 '11 at 19:10
“Changing the bound” only makes sense if you are comparing two particular models of set theory with the same ordinals (and, better yet, cardinals), otherwise the bounds have no identity. What Chad means is that given a model of set theory containing an ordinal $\gamma$, the model has a transitive extension with the same ordinals and cardinals which satisfies $2^{\aleph_\alpha}=2^{\aleph_\beta}$ for all $\alpha,\beta< \gamma$. Obviously, the new model will have a different cardinal power function, hence also the value of $\beth_1=2^{\alepha_0}$ will be different from the ground model. – Emil Jeřábek Dec 6 '11 at 19:34
@Asaf Karagila: Yes, $\gamma$ comes from the answers. Sorry for not explaining that, and thanks for pointing that out. I reformulated my question to make it more clear. – Vladimir Reshetnikov Dec 6 '11 at 20:53

## 3 Answers

Yes. Start with a model of GCH and add $\aleph_{\omega+1}$ Cohen reals. Then $2^{\aleph_n}=\aleph_{\omega+1}$ for all $n<\omega$. You can get the bound $\gamma$ arbitrarily high within the ordinal hierarchy by adding $\kappa$ Cohen reals instead, where $\kappa$ is a regular cardinal greater than $\aleph_\gamma$. (I think that's all correct.)

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Yes. A fairly general answer to such questions is provided by Easton’s theorem: if the ground model satisfies GCH and $F$ is a class function from a subclass of regular cardinals to cardinals such that $F$ is nondecreasing and $\kappa< \operatorname{cf}(F(\kappa))$ for each $\kappa\in\operatorname{dom}(F)$, then one can construct a forcing extension with the same cardinals and cofinalities which satisfies $2^\kappa=F(\kappa)$ for each $\kappa\in\operatorname{dom}(F)$.

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It is worth noting, for those unfamiliar with the theorem itself, that the resulting extension may also change the behaviour of $2^\kappa$ for $\kappa\notin F$ as well. – Asaf Karagila Dec 6 '11 at 13:08
It may also be worth noting that, if you use Easton's construction to get some desired values for $2^\kappa$ for all regular $\kappa$, then the values of $2^\kappa$ for singular $\kappa$ are automatically the smallest ones permitted by your choices for regular $\kappa$. In particular, if you make $2^\kappa$ constant for all regular $\kappa$ up to some chosen $\gamma$, then (in Easton models) $2^\kappa$ will also take the same constant value at singular cardinals below $\gamma$. – Andreas Blass Dec 6 '11 at 15:14
In any model of set theory, if $2^\kappa$ is constant for all regular $\kappa\le\gamma$, then it takes the same constant value at singulars below $\gamma$ since it is nondecreasing. – Emil Jeřábek Dec 6 '11 at 15:44

To your edit, note that for every ordinal $\alpha$ it holds that $\alpha\le\aleph_\alpha$. This is because there are $\alpha$ many cardinals below $\aleph_\alpha$.

Since the function $\kappa\mapsto 2^\kappa$ is strictly increasing, we have if so that $2^{\aleph_0}=\beta<2^{\beta}\le 2^{\aleph_\beta}$.

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