## Is there always an uncountable $\kappa$ such that $\kappa^{\aleph_0}=2^\kappa$?

The cardinal equation $\kappa^{\aleph_0}=2^\kappa$ is satisfied by $\kappa=\aleph_0$.

It is also satisfied by any $\kappa$ for which $MA(\kappa)$ holds.

Under $GCH$, the equation is satisfied by $\kappa$ if and only if $cof(\kappa)=\aleph_0$.

So my question is:

Is it consistent with $ZFC$ that the only solution to $\kappa^{\aleph_0}=2^\kappa$ is $\kappa=\aleph_0$?

I´m sorry if this is too basic, but I just don´t see it.

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Ramiro, in general if $\kappa$ is strong limit, $\kappa^{cf(\kappa)}=2^\kappa$. – Andres Caicedo Feb 22 2012 at 2:49
Andres, thanks! I knew it was rather basic. – Ramiro de la Vega Feb 22 2012 at 11:53

No, it is provable in ZFC that there are many $\kappa$ for which $2^\kappa=\kappa^\omega$. For example, let $\kappa=\beth_\omega$, and the same argument will work with any strong limit cardinal of cofinality $\omega$. Observe that every subset of $\kappa=\beth_\omega$ is determined by the sequence of its intersections with all the $\beth_n$'s. Thus, $2^\kappa\leq \Pi_n P(\beth_n)\leq \Pi_n\beth_{n+1}\leq\kappa^\omega$. But on the other hand, $\kappa^\omega\leq\kappa^\kappa\leq (2^\kappa)^\kappa=2^\kappa$. So this is a case where $2^\kappa=\kappa^\omega$.