Non-principal ultrafilter containing a finite set

Question

Let $I$ be an infinite set. Is there a non-principal ultrafilter $F$ over $I$ which contains a finite subset of $I$?

Answer 1

No, suppose $F$ contains some finite subsets of $I$, and let $I_0$ be $\subseteq$-minimum finite subset of $I$ which is in $F$. (Note that $I_0$ is indeed $\bigcap F$.)

Claim (1): $F=\{A\subseteq I : I_0\subseteq A\}$.

Proof of claim (1): Since $F$ is a filter and so closed under taking superset, $\{A\subseteq I : I_0\subseteq A\}\subseteq F$. Let $A\in F$, then $(A\cap I_0) \in F$ and $A\cap I_0\subseteq I_0$. Since $I_0$ is $\subseteq$-minimum, $A\cap I_0=I_0$. So $I_0\subseteq A$. Therefore $F=\{A\subseteq I : I_0\subseteq A\}$.

Claim (2): $I_0=\{i\}$ for some $i\in I$.

Proof of claim (2): If $I_0$ is not a singleton, then $I$ can be written as the disjoint union of $B$ and $C$ such that $B\neq\emptyset\neq C$. Since $I_0$ is $\subseteq$-minimum member of $F$, $B\notin F$ and $C\notin F$. Since $F$ is an ultrafilter $(I\setminus B)\in F$ and so $I_0\cap (I\setminus B)=C\in F$ which is a contradiction.

Hence, if $F$ has a finite subset of $I$, it must be a principal ultrafilter.