# If $\aleph_\lambda>2^{\aleph_0}$ is a limit cardinal, then $\aleph_\lambda^{\aleph_0}=\aleph_\lambda$?

I know that for successor cardinals, the result holds; i.e. if $\aleph_\alpha$ is a successor cardinal, then because this implies that it is a regular cardinal, then we can use this to show that if $\aleph_\alpha\geq 2^{\aleph_0}$, then $\aleph_\alpha^{\aleph_0}=\aleph_\alpha$. The proof of this can be found in For every $n < \omega$, $\aleph_n^{\aleph_0} = \max(\aleph_n,\aleph_0^{\aleph_0})$, and depends entirely on the fact that $\aleph_\alpha$ is regular.

Clearly, if $\aleph_\lambda$ is a limit cardinal, then the proof in the above cannot be used. It doesn't seem like assuming the the set of all limit ordinals for which the result is false is non-empty and then picking the least one is fruitful in finding a contradiction, since the fact that it is a limit ordinal stops us from easily looking elements smaller than it, though looking at ordinals smaller than the cofinality seems to be similar in concept to that of the successor cardinal case. However, to me it seems that in order to use this in much the same manner as in the above proof, the cofinality would need to be strictly larger than $\aleph_0$, which seems like a strong assumption to make.

Is this identity true, and if so, what approach should/could be used to prove it?

This is not true. König's theorem tells us that for all cardinals $\kappa$, $\kappa < \kappa^{\text{cf}(\kappa)}$. So for any $\kappa$ with $\text{cf}(\kappa) = \aleph_0$, we have $\kappa^{\aleph_0} > \kappa$. And there are plenty of cardinals greater than $2^{\aleph_0}$ with cofinality $\aleph_0$. For an explicit example, if CH holds, take $\aleph_{\omega}$.
• @bof, or simply put $\beth_\omega$. – Asaf Karagila Oct 5 '13 at 7:09