understanding the definition of measurable cardinal I have two silly (as usual) questions about measurable cardinals:
1) By definition an uncountable cardinal k is measurable if there exists a k-complete nonprincipal ultrafilter U on k. Now take exactly k distinct elements of this ultrafilter. Consider the intersection of all these k-many sets. I understand it  doesn't belong to the ultrafilter. But what happens to this intersection, it is always empty, it is non-empty, what is the cardinality of it?
2) If  k is the smallest measurable cardinal then 2^k is non-measurable by inaccessibility? If m is another measurable cardinal, and hence m>k, then m>2^k?
Thanks in advance.
Please let me know because these questions can be considered silly.
 A: *

*There is no general answer to this. Consider these examples


*

*Let $x_\xi := \kappa \setminus \xi$ for all $\xi < \kappa$. Since $U$ is nonprincipal and $\kappa$-complete, we have $x_\xi \in U$ for all $\xi < \kappa$. And clearly
$$
\bigcap_{\xi < \kappa} x_\xi = \kappa \setminus \big(\bigcup_{\xi < \kappa} x_\xi\big) = \kappa \setminus \kappa = \emptyset
$$

*Let $x \subseteq \kappa$ be bounded and let $y_\xi = x_\xi \cup x$, where $x_\xi$ is as about. Then
$$
\bigcap_{\xi < \kappa} y_\xi = x
$$
is bounded in $\kappa$ and thus not in $U$

*Let $x_\xi = \kappa \setminus \{ (2 \odot \eta) \oplus 1 \mid \eta < \xi \}$ for all $\xi < \kappa$, where $\odot$/$\oplus$ denote ordinal multiplication/addition respectively. ($\dagger$) Then
$$
 \{ \lambda < \kappa \mid \lambda \text{ is a limit ordinal} \} \subseteq\bigcap_{\xi < \kappa} x_{\xi}
$$
and since $\{ \lambda < \kappa \mid \lambda \text{ is a limit ordinal} \} \in U$ we also have $\bigcap_{\xi < \kappa} x_\xi \in U$.


*You are correct: Let $\kappa < \mu$ both be measurable. Then, as $\mu$ is inaccessible, $2^\kappa < \mu$ and, by the same reasoning $2^{2^\kappa}< \mu$ and even
$$
2^{2^{\ldots^{2^\kappa}}}
$$
If "$\ldots$" has length $< \mu$. (More precisely, if we define a sequence $\rho_i$ by $\rho_0 := \kappa$, $\rho_{i+1} := 2^{\rho_i}$ and $\rho_{\lambda} := \sup_{i < \lambda} \rho_i$ then $\rho_i < \mu$ for all $i < \mu$.)
($\dagger$) It's almost as if I can hear Asaf typing his complaints about this notation even while I'm still writing this answer.
A: (I).If $F$ is a $k$-complete  free ultra-filter on the uncountable cardinal $k$ then every member of F has cardinal $k$. Let $k=A\cup B$ where $A\cap B=\phi$ and $|A|=|B|=k.$ Exactly one of $A, B,$ say $A,$ belongs to $F.$ Then $G=\{A\cup \{b\}: b\in B\}$ is a family of members of $F$ with $|G|=k$ and $\cap G=A\in F.$  In fact $H=\{A\cup C:C\subset B\}$ is a family of members of $F$ with $|H|=2^{|B|}=2^k,$ and $\cap H=A\in F.$
On the other hand let $A\in F$ and $B'\subset F$ with $|B'|=|A\setminus B'|$. Let $I=\{A \setminus \{b\}: b\in B'\}.$ Then $I\subset F$ and $\cap I=A\setminus B'$ may or may not belong to $F$. Exactly one of $B',A\setminus B'$ belongs to $F.$
(II). A measurable cardinal $k$ is a strong limit cardinal (I do not think this is obvious): If $m<k$ then $2^m<k.$ So a power cardinal cannot be measurable. In particular $2^k$ is not measurable.
(III). BTW. A measurable cardinal is regular. And if $k$ is the least uncountable cardinal that has a countably-closed free ultrafilter $F$ then $F$ is  $k$-closed and $k$ is the least measurable cardinal. 
