Ultrafilter with finite set While I was working in some exercise about filters, a question came to my mind: let $X$ a set and $F\subseteq X$ a non empty finite set. How many ultrafilters $U$ there are such that $F\in U$? I think that there exist a unique ultrafilter that contains $F$ but I can't see why or how to prove but my intuition says that it is true. Am I wrong?
My work: take $F$ a non empty and finite subset of $X$. Suposse that there exist two different ultrafilters $U$ and $V$ such that $F\in U$ and $F\in V$. Since $U\neq V$, w.l.g., we can take $A\in U\setminus V$. Then $A\notin V$ but $V$ is an ultrafilter and therefore $X\setminus A\in V$. Moreover, $F\cap (X\setminus A)\neq\emptyset$ and $F\cap A\neq\emptyset$. But then, from here, what can I do? If my intuition is wrong, then, is there a bound over the number of ultrafilters that contains a fixed finite set? Thanks.
 A: It is well known that any ultrafilter containing a finite set is principal (cf. this question). So any ultrafilter containing $F$ must be a principal ultrafilter on $x_0 \in F$. Since there are $|F|$ many choices of $x_0$, we get $|F|$ many ultrafilters containing $F$.

I hope this helps ^_^
A: Suppose that $F$ is a non-empty finite subset of $X$, and let $\mathscr{U}$ be an ultrafilter on $X$ such that $F\in\mathscr{U}$. Let $F=\{x_1,\ldots,x_n\}$, and for $k=1,\ldots,n$ let $A_k=X\setminus\{x_k\}$. Suppose that $\{x_k\}\notin\mathscr{U}$ for each $k\in\{1,\ldots,n\}$; then for each $k\in\{1,\ldots,n\}$ we must have $A_k\in\mathscr{U}$, and therefore $\bigcap_{k=1}^nA_k\in\mathscr{U}$. But $\bigcap_{k=1}^nA_k=X\setminus F$, which is certainly not in $\mathscr{U}$, since $F\in\mathscr{U}$. This contradiction shows that there must be some $k\in\{1,\ldots,n\}$ such that $\{x_k\}\in\mathscr{U}$, and in that case $\mathscr{U}$ is the fixed (or principal) ultrafilter over $x_k$:
$$\mathscr{U}=\{U\subseteq X:x_k\in U\}\;.$$
Conversely, if $x\in F$, then $\mathscr{U}_x=\{U\subseteq X:x\in U\}$ is an ultrafilter on $X$, and clearly $F\in\mathscr{U}$. Thus, the ultrafilters on $X$ that contain $F$ are precisely the fixed ultrafilters over the elements of $F$, and there are therefore $|F|$ of them.
