I've just started learning about filters and non-principal ultrafilters. I'm getting confused on the requirement: $U$ contains no finite subsets of $J$; where $U$ is the ultrafilter and $J$ is a set.
I believe what this means is that $U$ can only contain sets that are infinite? (if it is non-principal).
Furthermore I'm getting somewhat confused by the proof that non-principal ultrafilters can exist. By the following " Take the filter of all cofinite sets and extend to an ultrafilter". I can't seem to find a definition of a cofinte set?
Sorry if this is incoherent, getting frustrated by this.
Thanks in advance.
Best Answer
Yes, an ultrafilter which contain a finite set is principal. To see why, note that if $A$ is this finite set, then either some $a\in A$ satisfies $\{a\}$ is in the ultrafilter, in which case it is principal; or else $X\setminus\{a\}$ is in the ultrafilter for all $a\in A$, so the finite intersection $A\cap\bigcap_{a\in A}(X\setminus\{a\})$ is also in the ultrafilter.
So a non-principal ultrafilter must contain only infinite sets. In particular, if $X$ is finite, then every ultrafilter on $X$ is principal.
The definition of a cofinite set is relative to $X$, and it simply means that $X\setminus A$ is finite. Co being short for "complement [of]". And if $X$ is an infinite set, then the collection of cofinite subsets of $X$ makes a filter. Moreover if you extend this filter in any way, you will never add a finite set to it. Therefore an ultrafilter extending the cofinite filter is necessarily non-principal.