every non-principal ultrafilter contains a cofinite filter.

I have several questions regarding filters and ultrafilters:

(a) Does every non-principal ultrafilter contain a cofinite filter? Why?

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

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

Solution 1:

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 $X$.

Fréchet filter and cofinite filter are two names for the same thing; there is no difference.

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.