Is it consistent with ZFC that for all ordinals $\alpha, \beta < \omega$ it holds that $2^{\aleph_\alpha} = 2^{\aleph_\beta}$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:42:42Z http://mathoverflow.net/feeds/question/82769 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82769/is-it-consistent-with-zfc-that-for-all-ordinals-alpha-beta-omega-it-holds Is it consistent with ZFC that for all ordinals $\alpha, \beta < \omega$ it holds that $2^{\aleph_\alpha} = 2^{\aleph_\beta}$? Vladimir Reshetnikov 2011-12-06T05:03:22Z 2011-12-06T20:51:58Z <p>Let $\gamma=\omega$ (the first transfinite ordinal). Is it consistent with ZFC that for all ordinals $\alpha, \beta &lt; \gamma$ it holds that $2^{\aleph_\alpha} = 2^{\aleph_\beta}$? If yes, can the bound $\gamma$ be increased here and how much?</p> <hr> <p>Update: In what sense the bound $\gamma$ can be made <em>arbitrarily</em> high? If $\beta$ is the initial ordinal of $\beth_1$, then it cannot be that $2^{\aleph_0}=2^{\aleph_\beta}$, right?</p> http://mathoverflow.net/questions/82769/is-it-consistent-with-zfc-that-for-all-ordinals-alpha-beta-omega-it-holds/82771#82771 Answer by Chad Groft for Is it consistent with ZFC that for all ordinals $\alpha, \beta < \omega$ it holds that $2^{\aleph_\alpha} = 2^{\aleph_\beta}$? Chad Groft 2011-12-06T05:24:17Z 2011-12-06T05:24:17Z <p>Yes. Start with a model of GCH and add $\aleph_{\omega+1}$ Cohen reals. Then $2^{\aleph_n}=\aleph_{\omega+1}$ for all $n&lt;\omega$. 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.)</p> http://mathoverflow.net/questions/82769/is-it-consistent-with-zfc-that-for-all-ordinals-alpha-beta-omega-it-holds/82782#82782 Answer by Emil Jeřábek for Is it consistent with ZFC that for all ordinals $\alpha, \beta < \omega$ it holds that $2^{\aleph_\alpha} = 2^{\aleph_\beta}$? Emil Jeřábek 2011-12-06T11:05:37Z 2011-12-06T11:41:53Z <p>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 $\kappa&lt; \operatorname{cf}(F(\kappa))$ 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 $\kappa\in\operatorname{dom}(F)$.</p> http://mathoverflow.net/questions/82769/is-it-consistent-with-zfc-that-for-all-ordinals-alpha-beta-omega-it-holds/82815#82815 Answer by Asaf Karagila for Is it consistent with ZFC that for all ordinals $\alpha, \beta < \omega$ it holds that $2^{\aleph_\alpha} = 2^{\aleph_\beta}$? Asaf Karagila 2011-12-06T19:16:51Z 2011-12-06T19:16:51Z <p>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 $\aleph_\alpha$.</p> <p>Since the function $\kappa\mapsto 2^\kappa$ is strictly increasing, we have if so that $2^{\aleph_0}=\beta&lt;2^{\beta}\le 2^{\aleph_\beta}$.</p>