Is an ultrafilter free if and only if it contains the cofinite filter? A filter $\mathcal F$ is called free if $\bigcap \mathcal F=\emptyset$. Filter, which is not free is called principal. any principal ultrafilter has the form $$\mathcal F_a=\{A\subseteq X; a\in A\}$$ for some $a\in X$.
Fréchet filter (or. cofinite filter) on a set $X$ is the filter consisting of all cofinite set, i.e., it is equal to $$\mathcal F_{F}=\{A\subseteq X; X\setminus A\text{ is finite}\}.$$

Is an ultrafilter  free if and only if it contains the cofinite  filter ? Why?

 A: Remember that a filter $\mathcal{F}$ is an ultrafilter if and only if it has the property
$$\bigl(\forall A \subset X\bigr)\bigl( A \in \mathcal{F} \lor (X\setminus A)\in\mathcal{F}\bigr).$$
An ultrafilter is principal if and only if it contains a finite set. So by the above characterisation, an ultrafilter is not principal if and only if it contains the complements of all finite sets, i.e. if and only if it contains the Fréchet filter.
A: Yes. Let $\mathscr{U}$ be an ultrafilter on $X$, and let $a\in X$ be arbitrary. Since $\mathscr{U}$ is an ultrafilter, either $\{a\}\in\mathscr{U}$, or $X\setminus\{a\}\in\mathscr{U}$. If $\{a\}\in\mathscr{U}$, then $\mathscr{F}_a\subseteq\mathscr{U}$, and therefore $\mathscr{U}=\mathscr{F}_a$. If $X\setminus\{a\}\in\mathscr{U}$ for each $a\in X$, then $X\setminus F\in\mathscr{U}$ for each finite $F\subseteq X$; this is easily proved by induction on $|F|$.
Alternatively, you can argue as follows. Let $\mathscr{F}$ be the cofinite filter on $X$. If $\mathscr{U}\supseteq\mathscr{F}$, then $$\bigcap\mathscr{U}\subseteq\bigcap\mathscr{F}=\varnothing\;,$$ so $\mathscr{U}$ cannot be a principal filter: $\bigcap\mathscr{F}_a=\{a\}$ for each $a\in X$. If, on the other hand, $\mathscr{U}\nsupseteq\mathscr{F}$, then there is a finite $F\subseteq X$ such that $X\setminus F\notin\mathscr{U}$. Let $F=\{x_1,\ldots,x_n\}$; clearly $$F=\{x_1\}\cup\{x_2\}\cup\ldots\cup\{x_n\}\;,$$ a finite union, so there is exactly one $k\in\{1,\dots,n\}$ such that $\{x_k\}\in\mathscr{U}$, and it follows immediately that $\mathscr{U}=\mathscr{F}_{x_k}$, the principal ultrafilter over $x_k$.
