# every non- principal ultrafilter contains a cofinite filter.

I have several questions about filter and ultrafilter:

(a)Dose every non- principal ultrafilter contains a cofinite filter.? Why?

(b) What is the difference between Frechet filter and cofinite filter?

(c)Can we say " A ultrafilter is free iff it contains a Frechet filter?Why?

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The Frechet filter and the cofinite filter, in the usual definition, are the same. –  André Nicolas Aug 26 '13 at 5:13

Yes, every non-principal ultrafilter on an infinite set $X$ contains the cofinite filter on $X$. Let $\mathscr{U}$ be a non-principal ultrafilter on $X$, and let $x\in X$ be arbitrary. Since $\mathscr{U}$ is an ultrafilter, exactly one of the sets $\{x\}$ and $X\setminus\{x\}$ belongs to $\mathscr{U}$, and since $\mathscr{U}$ is non-principal, $\{x\}\notin\mathscr{U}$. Thus, $X\setminus\{x\}\in\mathscr{U}$ for each $x\in X$. Now let $F$ be any finite subset of $X$; then $$X\setminus F=\bigcap_{x\in F}\left(X\setminus\{x\}\right)$$ is the intersection of finitely many members of $\mathscr{U}$. Every filter is closed under finite intersections, so $X\setminus F\in\mathscr{U}$. The cofinite filter on $X$ is precisely $\{X\setminus F:F\subseteq X\text{ is finite}\}$, so we’ve just shown that $\mathscr{U}$ contains the cofinite filter on $X$.
Yes, an ultrafilter on $X$ is free iff it contains the Fréchet filter on $X$. In the first paragraph I proved that every free ultrafilter on $X$ contains the Fréchet filter on $X$. For the converse, suppose that $\mathscr{U}$ is a fixed ultrafilter on $X$; then there is an $x\in X$ such that $\{x\}\in\mathscr{U}$. But then $X\setminus\{x\}$ is an element of the Fréchet filter that is not in $\mathscr{U}$, so $\mathscr{U}$ does not contain the Fréchet filter.