# Landing between $\beth_\lambda$ and $\beth_{\lambda+1}.$

Main Question. Is it consistent with ZFC that there exists a limit ordinal $\lambda$ and a cardinal number $\nu$ satisfying $$\beth_\lambda < \beth_\lambda^\nu < \beth_{\lambda+1}?$$

I am also interested in the following variant.

Secondary Question. Is it consistent with ZFC that there exists a limit ordinal $\lambda$ such that for some cardinal $\kappa < \beth_\lambda$, we can find a cardinal number $\nu$ satisfying the following? $$\beth_\lambda < \kappa^\nu < \beth_{\lambda+1}$$

Clearly neither statement is a theorem of ZFC, since GCH implies that the answers to both questions are "no."

The second question admits an easy no.

If $\nu<\beth_\lambda$, then $\kappa^\nu\leq2^\kappa\cdot2^\nu$ both of which are smaller than $\beth_\lambda$. If $\beth_\lambda\leq\nu$ then we have $$\beth_{\lambda+1}=2^{\beth_\lambda}\leq2^\nu\leq\kappa^\nu.$$

In either case we can't "land in the gap".

The first question also admits a negative answer, although in slightly greater efforts.

Note that $\beth_\lambda$ is a strong limit cardinal, therefore $2^{<\beth_\lambda}=\beth_\lambda$. From this it follows that $\beth_\lambda^{\operatorname{cf}(\lambda)}=\beth_{\lambda+1}$. And you should be able to finish the proof on your own.

• Is $κ^ν≤2^κ⋅2^ν$ a general principle? – goblin GONE May 5 '14 at 12:08
• @user18921: It's an easy consequence of cardinal arithmetic. Note that $$\kappa^\nu\leq(\kappa+\nu)^{\kappa+\nu}=2^{\kappa+\nu}=2^\kappa\cdot2^\nu.$$ – Asaf Karagila May 5 '14 at 12:16
• Very nice. $\;\!$ – goblin GONE May 5 '14 at 12:16