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It is consistent in $\Z$ that there is an uncountable cardinal $\kappa$ such that for no cardinal $\lambda$ we have $2^\lambda = \kappa$: Take any model in which $2^{\aleph_0} = \aleph_2$, and $\kappa = \aleph_1$.

Is there a model in $\Z$ such that for every cardinal $\kappa > \aleph_0$ there is a cardinal $\lambda$ such that $2^\lambda = \kappa$?

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By Konig's lemma $cf(2^\kappa) > \kappa,$ so for example $\aleph_\omega$ can never be of the form $2^\kappa$ for any $\kappa.$

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    $\begingroup$ You don't even need Kőnig's theorem to answer this question. If $\aleph_\omega$ is in the range of the exponential function, then $\aleph_\omega=2^{\aleph_n}$ for some $n\lt\omega$, so at most $n$ of the infinite cardinals below $\aleph_\omega$ can be in the range of the exponential function. $\endgroup$
    – bof
    Commented Feb 12 at 9:40
  • $\begingroup$ @bof Interesting observation! $\endgroup$
    – Mohammad Golshani
    Commented Feb 12 at 11:10
  • $\begingroup$ Note also that one doesn't even need the axiom of choice, since if $\aleph_\omega$ is the size of $P(X)$, then $X$ is well-orderable, hence $\aleph_n$ for some $n$. $\endgroup$
    – Joel David Hamkins
    Commented Feb 12 at 18:07
  • $\begingroup$ Note that the result you used is generally called König's theorem (as in @bof's comment). König's lemma is the one about paths through finitely branching trees. $\endgroup$
    – Andreas Blass
    Commented Feb 12 at 18:20
  • $\begingroup$ @AndreasBlass Forgive the pedantry but I believe the Hungarian mathematician Kőnig spelled his German surname with a Hungarian ő (as in Erdős). That's the phonetic spelling in Hungarian, which always distinguishes between long and short vowels. $\endgroup$
    – bof
    Commented Feb 12 at 22:49

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