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

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

• How do you know that this isn't true? Feb 10 '20 at 18:20
• Well, I'm studying a theorem that assumes cof$(2^{<\kappa}) > \kappa$ and $\kappa$ singular and it shows that $2^\kappa = 2^{<\kappa}$... Feb 10 '20 at 18:22
• It could be that the map $\lambda\mapsto 2^\lambda$ ($\lambda<\kappa$) is eventually constant. This is the case you have not considered. Feb 10 '20 at 18:23
• ohh I see, it could reach $2^{<\kappa}$ at a certain point... right? Feb 10 '20 at 18:25
• Right, that's the issue. Feb 10 '20 at 18:25

## 1 Answer

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 $$\kappa$$.)