# Tail set on $\mathcal{F}$ non-principal a $\kappa$-complete ultrafilter on $\kappa$

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!

• 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. – Andrés E. Caicedo Mar 6 at 12:53
• I edited the question with your suggestion. not sure how to follow. – Thomas Lesgourgues Mar 6 at 13:57