Show that $\operatorname{cf}(2^{<\kappa})\geq\operatorname{cf}(\kappa)$

Question

I saw in my previous question that we may not have equality between $\operatorname{cf}(2^{<\kappa})$ and $\operatorname{cf}(\kappa)$, but I conjecture we still have $\operatorname{cf}(\kappa)\leq\operatorname{cf}(2^{<\kappa})$. Does anyone know if this is true (in ZFC) and if so, how I could prove it? Recall that a formula for $2^{<\kappa}$ is $$2^{<\kappa}=\sup\{2^\theta:\theta<\kappa\text{ and }\theta\text{ is a cardinal}\}.$$

Answer 1

If $2^{<\kappa} = 2^{\lambda}$ for some $\lambda<\kappa$, then for all $\lambda'$ with $\lambda\leq\lambda'<\kappa$, we have $2^\lambda\leq 2^{\lambda'}\leq 2^{<\kappa}$, so $2^{<\kappa} = 2^{\lambda'}$. Now $\DeclareMathOperator{\cf}{cf}\cf(2^{<\kappa}) = \cf(2^{\lambda'})>\lambda'$, so $\cf(2^{<\kappa})\geq\kappa\geq\cf(\kappa)$.

In particular, this deals with the case where $\kappa = \lambda^+$ is a successor, since then $2^{<\kappa}=2^\lambda$.

Now we may assume $\kappa$ is a limit cardinal, so $\kappa=\aleph_\alpha$ with $\alpha$ a limit ordinal, and $\cf(\kappa)=\cf(\alpha)$. We may also assume $2^{<\kappa} > 2^{\lambda}$ for all $\lambda<\kappa$. So the function $\beta\mapsto 2^{\aleph_\beta}$ is a monotone increasing cofinal function $\alpha\to 2^{<\kappa}$. It follows that $\cf(2^{<\kappa})= \cf(\alpha) = \cf(\kappa)$.