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I am reading Indecomposable ultrafilters and $0^\sharp$ by Jack Silver (pp.357-363 in Proceedings of the Tarski Symposium (Proceedings of Symposia in Pure Mathematics XXV), 1974).

He gives the definition of a indecomposable ultrafiler over $\kappa$ as:

$\mathcal{U}$ is a uniform ultrafilter over $\kappa$, and, whenever $\kappa=\bigcup_{\alpha<\lambda}I_\alpha$ where $\lambda<\kappa$, there is some countable subset $C$ of $\lambda$ such that $\bigcup_{\alpha \in C} I_\alpha \in \mathcal{U}$.

I.e. "any decomposition of $\kappa$ lies, countably, in $\mathcal{U}$", so to speak.

Then he gives an equivalent definition:

$\mathcal{U}$ is a uniform ultrafilter over $\kappa$, and $D$ is $\lambda$ descendingly complete for any regular $\lambda$ such that $\omega < \lambda < \kappa$ (i.e., if $\langle S_\alpha\mid\alpha<\lambda\rangle$ is a descending sequence of members of $\mathcal{U}$, then $\bigcap_{\alpha<\lambda} S_\alpha\in\mathcal{U}$).

The thing is: I am not able to see this equivalence. Any advice?

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  • $\begingroup$ Is there a part that you can prove? $\endgroup$ – Asaf Karagila Feb 5 '17 at 4:57
  • $\begingroup$ Nope, there isn't :( $\endgroup$ – edgar alonso Feb 5 '17 at 20:51
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I think it's solved:

($\rightarrow$)Suppose indecomposability. Let $\langle I_\alpha:\alpha<\lambda\rangle$ be descending, $\omega<\lambda<\kappa$ regular and $I_\alpha \in D$ for $\alpha<\lambda$. If $\bigcap_{\alpha<\lambda}I_\alpha\notin\ D$ then $\kappa\setminus\bigcap_{\alpha<\lambda}I_\alpha=\bigcup_{\alpha<\lambda}\kappa\setminus I_\alpha\in D$. So, by the previous equivalence we can extract a $C\subset\lambda$ countable with: \begin{equation} \bigcup_{\alpha\in C}\kappa\setminus I_\alpha\in D. \end{equation} But $\langle \kappa\setminus I_\alpha:\alpha<\lambda\rangle$ is an increasing sequence, so $\bigcup_{\alpha\in C}\kappa\setminus I_\alpha= \kappa\setminus I_\gamma$, where $\gamma=\max C$, and thanks to regularity it is $<\lambda$. So $I_\gamma \notin D$ a contradiction.

($\leftarrow$)Now suppose $\lambda$-descending completeness for $\omega<\lambda<\kappa$ regular. Let $\lambda<\kappa$ and $\langle I_\alpha:\alpha<\lambda\rangle$ be such that $\bigcup_{\alpha<\lambda}I_\alpha=\kappa$. Without lost of generality we can take $\langle I_\alpha:\alpha<\lambda\rangle$ increasing --properly\footnote{If not we redefine $E_\alpha=\bigcup_{\beta\leq\alpha}I_\beta$, which is increasing.}. Now we suppose that $D$ is not indecomposable; therefore $I_\alpha\notin D$, $\alpha<\lambda$. Being $D$ an ultrafilter we know: $\kappa\setminus I_\alpha\in D$ and $\langle\kappa\setminus I_\alpha:\alpha<\lambda\rangle$ is descending --properly-- therefore $\bigcap_{\alpha<\lambda}\kappa\setminus I_\alpha\in D$ and $\neq\kappa$.

First suppose $cf(\lambda)=\omega$. Let $\langle \alpha_i:i<\omega\rangle$ be cofinal in $\lambda$. So that $\bigcap_{i<\omega}\kappa\setminus I_{\alpha_i}=\bigcap_{\alpha<\lambda}\kappa\setminus I_\alpha\in D$. But, because $\langle\kappa\setminus I_\alpha\rangle_{\alpha<\lambda}$ is descending properly, we know \begin{equation} \kappa\setminus(\bigcap_{i<\omega}\kappa\setminus I_{\alpha_i})\neq\kappa; \end{equation} but \begin{equation} \kappa\setminus(\bigcap_{i<\omega}\kappa\setminus I_{\alpha_i})=\bigcup_{i<\omega}I_{\alpha_i}=\bigcup_{\alpha<\lambda}I_{\alpha}=\kappa. \end{equation} a contradiction.

Now we assume $cf(\lambda)>\omega$. We remember that \begin{equation} E=E_\omega^\lambda=\{\eta<\lambda:cf(\eta)=\omega\} \end{equation} is stationary in $\lambda$. As before, for each $\eta\in E_\omega^\lambda$ we obtain $C_\eta\subset\eta$ countable and cofinal, for which $\bigcup_{\alpha\in C_\eta}I_\alpha\notin D$, so \begin{equation} \bigcap_{\alpha\in C_\eta}\kappa\setminus I_\alpha = \bigcap_{\alpha<\eta}\kappa\setminus I_\alpha \in D. \end{equation} Therefore $\langle D_\eta :\eta\in E\rangle$, $D_\eta=\bigcap_{\alpha\in C_\eta}\kappa\setminus I_\alpha$, is a descending sequence in $D$, and $\bigcap_{\eta\in E}D_\eta=\bigcap_{\alpha<\lambda}\kappa\setminus I_{\alpha}\in D$, but $\neq\kappa$. Exactly as before we obtain a contradiction.

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