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

Question

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!

Answer 1

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} $.

Answer 2

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 $\kappa.$