Cardinality of ultrafilters

Question

I have read that If $\mathscr F$ is an ultrafilter on an infinite set $X$, then it can be shown that $|\mathscr F|=2^{|X|}$.

Question 1: Can we also determine cardinality of filters on a given set?

Question 2: Why does the ultrafilter have the same cardinality as power set of $X$?

Yes, for $X$, the ultrafilter contains all supersets of its elements, but it also contains all finite intersections.

Doesn´t that make the cardinality bigger than of the power set, which contains all subsets, while the ultrafilters contains supersets and intersections?

Answer 1

In general filters cannot be larger than the powerset $\mathcal P(X)$, since any filter is a subset of $\mathcal P(X)$: each element of the filter is a subset of $X$. To be precise, filters are defined as exactly those nonempty subsets $F\subseteq \mathcal P(X)$ that are closed under supersets and finite intersections. This makes $\mathcal P(X)$ a filter as well, but we generally don't consider it as interesting, and thus usually work with proper filters (a filter is proper if it's not equal to all of $\mathcal P(X)$).

A proper filter $\mathscr F$ is an ultrafilter if it is maximal, that is, if $\mathscr F\subseteq \mathscr G$ and $\mathscr G$ is a proper filter, then $\mathscr F=\mathscr G$. An equivalent definition is that $\mathscr F$ is an ultrafilter if and only if for all $A\subseteq X$ either $A\in \mathscr F$ or $X\setminus A\in \mathscr F$ (but not both).

For Question 1:

Arbitrary filters can have many sizes. For example:

• $\{X\}$ is a filter with only one element,
• the Fréchet filter is the filter of all cofinite subsets of $X$ and has cardinality $|X|^{<\aleph_0}$, which is equal to $|X|$ assuming $X$ is infinite.
• Given $x\in X$, the principal ultrafilter $\{A\subseteq X\mid x\in A\}$ has an obvious bijection with $\mathcal P(X\setminus \{x\})$, and thus has cardinality $2^{|X|}$

For Question 2:

For any $A\subseteq X$, by maximality of ultrafilters $A\in\mathscr F$ if and only if $X\setminus A\notin \mathscr F$, so the map $A\mapsto X\setminus A$ is a bijection between $\mathscr F$ and $\mathcal P(X)\setminus \mathscr F$. Since $\mathcal P(X)$ is infinite, and $\mathcal P(X)=\mathscr F\cup (\mathcal P(X)\setminus \mathscr F)$, we see that $|\mathscr F|=|\mathcal P(X)\setminus \mathscr F|=|\mathcal P(X)|=2^{|X|}$.