In the wonderful book Introduction to cardinal arithmetic (Holz, Steffens, Weitz) on the page 78 I have come across the following assertion:
Let $\kappa$ be a singular cardinal. $F$ is a function from $\operatorname{cf}(\kappa)$ into $\operatorname{cf}(\kappa)$. $a(i)$ is a sequence unbounded in $\kappa$ ($i < \operatorname{cf}(\kappa)$) such that the set $$\{ i < \operatorname{cf}(\kappa) : 2^{a(i)} \leq a(i)^{+F(i)}\}$$ is unbounded in $\operatorname{cf}(\kappa)$. ($a(i)^+$ is the successor cardinal of $a(i)$). Then $\kappa$ is a strong limit cardinal.
I can prove it only assuming that $a(i)$ is not decreasing plus $\kappa$ is $\operatorname{cf}(\kappa)$-strong. Can anyone prove the assertion without the assumption? Thanks in advance.
It is not true in general without the non-decreasing assumption.
For a counterexample, pick any singular $\kappa$ and let $a(i)=0$ and $F(i)=1$ for all even $i$. This will ensure that $2^{a(i)}\leq a(i)^{+F(i)}$ for these $i$, since $2^0=1=0^+$, and so the collection of $i$ will be unbounded. Now, specify $a(i)$ for odd $i$ to make it unbounded in $\kappa$.
Meanwhile, if one assumes that $a(i)$ is non-decreasing, then indeed $\kappa$ must be a strong limit. To see this, consider any $\delta<\kappa$. Pick some $i$ with $\delta\leq a(i)$ and $2^{a(i)}\leq a(i)^{+F(i)}$. Since $\kappa$ has cofinality $\text{cf}(\kappa)$ and $a(i)<\kappa$, it follows that $a(i)^{+F(i)}<\kappa$. And so $2^\delta<\kappa$, as desired.