# 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. Mar 6, 2019 at 12:53
• I edited the question with your suggestion. not sure how to follow. Mar 6, 2019 at 13:57

This is a possible solution:

It is assumed, towards a contradiction, that there is some $$\alpha<\kappa$$ such that $$C_\alpha\not\in\mathcal{F}$$. Since $$\mathcal{F}$$ is an ultrafilter, this means that $$\kappa-C_\alpha\in\mathcal{F}$$ or, what is the same, $$\alpha\in\mathcal{F}$$. $$\mathcal{F}$$ is also non-principal, then, $$\bigcap\mathcal{F}\not\in\mathcal{F}$$, so, it is properly contained in every element of $$\mathcal{F}$$ and, in particular, $$\bigcap\mathcal{F}\subsetneq\alpha$$. This means that there is some $$\beta\in\alpha$$ such that $$\beta\not\in\bigcap\mathcal{F}$$. Therefore, there must be at least one $$X\in\mathcal{F}$$ such that $$\beta\not\in X$$ (in that way, $$\beta$$ disappears from $$\bigcap\mathcal{F}$$). It is clear that $$\alpha\cap X\in\mathcal{F}$$ but $$\beta\not\in \alpha\cap X$$. Again, since $$\mathcal{F}$$ is non-principal, $$\bigcap\mathcal{F}\subsetneq \alpha\cap X$$. In this way, it is possible to define the following sets:

• If $$\alpha'=0$$, $$X_0=\alpha\cap X$$.

• If $$\alpha'=\beta+1$$, $$X_{\alpha'}=X_\beta\cap Z$$, where $$Z\in\mathcal{F}$$ and such that, for some $$\gamma\in X_\beta-\bigcap\mathcal{F}$$, $$\gamma\not\in Z$$ (such $$Z$$ exists for the same reason as $$X$$ exists).

• If $$\alpha'$$ is a limit ordinal, define $$W_{\alpha'}=\bigcap_{\delta<\alpha'}X_\delta$$. Since $$\mathcal{F}$$ is $$\kappa$$-complete and $$|\alpha'|<\kappa$$ (this is because, as will be shown later, values of $$\alpha'$$ greater than $$\alpha$$ are not necessary), $$W_{\alpha'}\in\mathcal{F}$$. And $$\mathcal{F}$$ is non-principal, so, $$\bigcap\mathcal{F}\subsetneq W_{\alpha'}$$. Thus, it is possible to do the same as before: for some $$\gamma\in W_{\alpha'}-\bigcap\mathcal{F}$$, take some $$Z\in\mathcal{F}$$ such that $$\gamma\not\in Z$$ and then, define $$X_{\alpha'}=W_{\alpha'}\cap Z\in\mathcal{F}$$.

It is easy to see that, at each step, the ordinal $$\gamma$$ is an element of $$\alpha$$, because $$X_{\alpha'}\subset \alpha$$. Then, for some $$\varepsilon\leq\alpha$$, the set $$X_{\varepsilon}$$ will be empty: $$X_{\alpha'}\subset\alpha$$ and at each step at least one different element of $$\alpha$$ has been removed from the previous $$X_{\alpha'}$$'s. But $$|\alpha|<\kappa$$ and the filter is $$\kappa$$-complete, so, $$X_{\varepsilon}\in\mathcal{F}$$, then, $$\varnothing=X_{\varepsilon}\in\mathcal{F}$$, which is absurd. And this proves that every tail set belongs to $$\mathcal{F}$$.

Some basic properties:

(1). If $$C\subset \kappa$$ and if $$\forall A\in\mathcal F\,(C\cap A\ne\emptyset)$$ then $$C\in\mathcal F.$$ Proof: $$C\in \mathcal G=\{(C\cap A)\cup B: A\in \mathcal F\land B\subseteq \kappa\}$$ and $$\mathcal G$$ is a filter on $$\kappa$$ with $$\mathcal G\supseteq\mathcal F,$$ so $$\mathcal G=\mathcal F$$ (by maximality of $$\mathcal F) .$$

(2). If $$C\subset \kappa$$ and $$C\not\in\mathcal F$$ then $$C\cap A=\emptyset$$ for some $$A\in\mathcal F.$$ This is immediate from (1).

(3). If $$C\subset \kappa$$ then $$(C\in\mathcal F)\lor (\kappa\setminus C\in\mathcal F).$$ Proof: By(2), if $$C\not\in\mathcal F,$$ there exists $$A\in\mathcal F$$ with $$A\subset\kappa\setminus C,$$ and we have $$(\kappa\setminus C\supseteq A\in\mathcal F\implies \kappa\setminus C\in\mathcal F).$$

(4). If $$\beta\in\kappa$$ then $$\{\beta\}\not\in \mathcal F$$ because $$\mathcal F$$ is non-principal.

Suppose $$\emptyset\ne\alpha\in\kappa.$$ By (4), no member of $$D_{\alpha}=\{\{\beta\}:\beta\in\alpha\}$$ belongs to $$\mathcal F.$$ So by this, and (3), $$\mathcal F\supset E_{\alpha}=\{\kappa\setminus C: C\in D_{\alpha}\}\ne\emptyset.$$

Now $$\mathcal F$$ is $$\kappa$$-closed, and therefore $$C_{\alpha}=\cap E_{\alpha}\in\mathcal F.$$

And for the case $$\alpha=\emptyset,$$ we have $$C_{\alpha}=\kappa\in\mathcal F$$ because $$\mathcal F$$ is a filter on $$\kappa.$$