TITLE: understanding the definition of measurable cardinal
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: (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.