Cofinality of $2^{<\kappa}$ with $\kappa$ infinite cardinal

Question

Given $\kappa$ an infinite cardinal, it holds that $$2^{<\kappa} = \sup_{\lambda \in \kappa \cap \text{Card}} 2^\lambda$$ but then should't we have that $\text{cof}(2^{<\kappa}) \le \kappa$ ? I mean, isn't the function $$\begin{align}f:\kappa &\longrightarrow 2^{<\kappa} \\ \alpha &\longmapsto 2^{|\alpha|}\end{align}$$ cofinal in $2^{<\kappa}$ ?I know that this isn't true but I can't quite figure out what is the problem with what I've written above. Can you point it out? Thanks

Answer 1

I assume you want to require that $\kappa$ is a limit cardinal; otherwise, if $\kappa=\rho^+$ then $2^{<\kappa}=2^{\rho}$ and you have no information about its cofinality beyond knowing that it is at least $\kappa$ (because $2^\rho$ has cofinality larger than $\rho$).

Under the assumption that $\kappa$ is limit, it could be that the map $\lambda\mapsto 2^\lambda$ ($\lambda<\kappa$) is eventually constant, In that case, this is the value of $2^{<\kappa}$ and you cannot control its cofinality in terms of $\kappa$. However, note that the cofinality of $2^{<\kappa}$ is at least $\kappa$ in this case, since for any $\lambda$ the cofinality of $2^\lambda$ is larger than $\lambda$, so the cofinality of $2^{<\kappa}$ is larger than $\lambda$ for all $\lambda<\kappa$. Note also that this case may happen. For example, we could have (that is, it is relatively consistent with $\mathsf{ZFC}$ that) $2^{\aleph_0}=2^{\aleph_\omega}=\aleph_{\omega+1}$. In this case, $2^{<\aleph_\omega}=2^{\aleph_0}$ has cofinality $\aleph_{\omega+1}$.

On the other hand, if $\lambda\mapsto 2^\lambda$ ($\lambda<\kappa$) is not eventually constant, then the cofinality of $2^{<\kappa}$ is indeed the cofinality of $\kappa$.

(So, in all cases, the cofinality of $2^{<\kappa}$ is at least the cofinality of $\kappa$.)