I'm reading some contents on set-theory for my own interest, and I stumbled upon some questions I cannot solve yet.

Let $\mathcal{F}$ be a $\kappa$-complete non-principal ultrafilter on $\kappa$. Then for every $\alpha < \kappa$, we define the tail set to be $C_\alpha = \{\beta < \kappa, \ \alpha < \beta\}$. Show that for every $\alpha$, $C_\alpha$ belongs to $\mathcal{F}$.

My intuitions : suppose there exist $\alpha$ such that $C_\alpha \notin \mathcal{F}$. Then because this is an ultrafilter on $\kappa$, we have $\left(\kappa-C_\alpha\right) \in \mathcal{F}$. I could go two directions here :

  1. Using the fact that $\mathcal{F}$ is $\kappa$-complete, I need to to build some less-than-$\kappa$ intersection, contradicting the fact that $\mathcal{F}$ is non-principal. But I don't think that's the way to go.
  2. Or using the fact that $\mathcal{F}$ is non-principal, I should try to build a partition of $\kappa$, $\{X_\lambda, \lambda< \kappa\}$, from this $C_\alpha$, where all $X_\lambda \notin \mathcal{F}$, contradicting the fact that $\mathcal{F}$ is $\kappa$-complete.

Any suggestion? Thanks

Edit Let us first consider the case $\kappa = \omega$. $C_\alpha = \{ \beta < \omega : \alpha < \beta\}$. Suppose there exist $\alpha$ such that $C_\alpha \notin \mathcal{F}$. Note that $C_{\alpha+1} \subset C_\alpha$. Therefore $C_\alpha \notin \mathcal{U}$ implies $C_{\alpha+1} \notin \mathcal{F}$. Because $\mathcal{F}$ is an ultrafilter, then $\omega-C_{\alpha+1}, \omega-C_{\alpha} \in \mathcal{F}$. By $\omega$-completeness of $\mathcal{F}$ (or direct definition of filters) :
$$ \{\alpha+1\} = \left(\omega-C_{\alpha+1}\right)\cap\left(\omega-C_{\alpha}\right) \in \mathcal{F}$$ $\mathcal{F}$ contains a singleton, contradicting the fact that $\mathcal{F}$ is non-principal.

Not sure to see how this extend to any $\kappa$. I can see that if $C_\alpha \notin \mathcal{F}$, then for any $\lambda > \alpha$, $C_\lambda \notin \mathcal{F}$. I'm not comfortable yet working with infinite cardinal, this is my first intrusion in this world!

  • 2
    $\begingroup$ Do this for $\kappa=\omega$ and you will see what to do in general. The case $\kappa=\omega$ is simple; and any ultrafilter is $\omega$-complete, so the completeness requirement does not add anything in this case. $\endgroup$ – Andrés E. Caicedo Mar 6 at 12:53
  • $\begingroup$ I edited the question with your suggestion. not sure how to follow. $\endgroup$ – Thomas Lesgourgues Mar 6 at 13:57

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