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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

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  • $\begingroup$ How do you know that this isn't true? $\endgroup$ Commented Feb 10, 2020 at 18:20
  • $\begingroup$ Well, I'm studying a theorem that assumes cof$(2^{<\kappa}) > \kappa$ and $\kappa$ singular and it shows that $2^\kappa = 2^{<\kappa}$... $\endgroup$
    – Lorenzo
    Commented Feb 10, 2020 at 18:22
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    $\begingroup$ It could be that the map $\lambda\mapsto 2^\lambda$ ($\lambda<\kappa$) is eventually constant. This is the case you have not considered. $\endgroup$ Commented Feb 10, 2020 at 18:23
  • $\begingroup$ ohh I see, it could reach $2^{<\kappa}$ at a certain point... right? $\endgroup$
    – Lorenzo
    Commented Feb 10, 2020 at 18:25
  • $\begingroup$ Right, that's the issue. $\endgroup$ Commented Feb 10, 2020 at 18:25

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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 $\kappa$.)

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