An ultrafilter $\mathcal{F}$ is said to be free if $\cap \mathcal{F} = \emptyset$.

An ultrafilter $\mathcal{F}$ is an uniform ultrafilter in $X$ if $|F| = |X|$ for every $F \in \mathcal{F}$.

Which relationship is there between free and uniform ultrafilter?Are they equivalent on infinite and countable set? Why?


The intersection of all subsets of an ultrafilter, if it is nonempty, must be a singleton; therefore a free ultrafilter is the same as a non-principal ultrafilter. It follows that a uniform ultrafilter on an infinite set, because it cannot contain a singleton, must be free. (The only finite set that has a uniform ultrafilter is the one-element set; we won't count that.)

Equivalently, a free ultrafilter is one which contains the filter of cofinite sets. This means a free ultrafilter cannot contain a finite set, so if the given set is countable, then a free ultrafilter on that set is uniform.

However, a free ultrafilter on an uncountable set $X$ need not be uniform (if we accept that every proper filter can be extended to an ultrafilter, which follows from the axiom of choice). For if $C \subseteq X$ is cofinite and $A \subseteq X$ is some given countable subset, then we cannot have $A \cap C = \emptyset$ (for that would imply $A$ is contained in the complement of $C$, hence contained in a finite set). It follows that sets of the form $A \cap C$, where $C$ ranges over cofinite sets, generate a proper filter, which is contained in some ultrafilter. This ultrafilter would contain $A$ and $X$, hence two sets of different cardinalities.


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