= Solution
Put $\kappa=|X|$, an infinite <cardinal number>. The original PDF has the double exponent $2^{2^{|X|}}$; the single-exponent expression in the TeX aid is a transcription error.
We explicitly construct an <independent family of subsets> with $2^\kappa$ members on a set of size $\kappa$. Let
$$
D=\{(s,H):s\in[\kappa]^{<\omega},\ H\subseteq\mathcal P(s)\}.
$$
For each finite $s$ there are finitely many possibilities for $H$. The collection of finite subsets of an infinite cardinal has size $\kappa$, so $|D|\le\kappa$; the points $(\{\xi\},\varnothing)$ give the reverse bound. Thus $|D|=\kappa$.
For every $B\subseteq\kappa$ define
$$
A_B=\{(s,H)\in D:B\cap s\in H\}.
$$
To verify independence, take distinct $B_1,\ldots,B_n$ and bits $\varepsilon_1,\ldots,\varepsilon_n$. For each pair $i<j$, choose a point of $B_i\mathbin{\triangle}B_j$ and let $s$ contain all these finitely many points. The traces $B_i\cap s$ are pairwise distinct. Set
$$
H=\{B_i\cap s:\varepsilon_i=1\}.
$$
Then $(s,H)\in A_{B_i}$ exactly for those $i$ with $\varepsilon_i=1$. Every finite Boolean pattern is therefore realized. The cases $n=0,1$ work by the same definition, allowing $s=\varnothing$. In particular the sets $A_B$ are all distinct, so the <finite-trace independent family construction> gives exactly $2^\kappa$ independent members.
For each <function> $\varepsilon:\mathcal P(\kappa)\to\{0,1\}$, prescribe $A_B$ when $\varepsilon(B)=1$ and $D\setminus A_B$ when $\varepsilon(B)=0$. Independence gives the <finite intersection property>. Their finite intersections generate a proper <filter on a set>: its members are the subsets of $D$ containing one of those intersections.
To extend this to an <ultrafilter>, order its proper filter extensions by inclusion. The union of a chain is again a proper filter, so <Zorn's lemma> supplies a maximal one, $U_\varepsilon$. If neither a subset $Y$ nor its complement belonged to this filter, adjoining $Y$ would still generate a proper filter: otherwise some existing member would be disjoint from $Y$ and force its complement into the filter already. Maximality therefore gives the ultrafilter dichotomy. Using the ambient <axiom of choice>, choose such an extension for every $\varepsilon$.
Distinct functions differ at some $B$. Their ultrafilters respectively contain $A_B$ and its complement, and cannot be equal. There are $2^{2^\kappa}$ such functions. Transporting the ultrafilters along a <bijection> $D\to X$ proves the lower bound. The upper bound is immediate because every ultrafilter is a subset of $\mathcal P(X)$:
$$
2^{2^\kappa}\le|\operatorname{Ult}(X)|
\le|\mathcal P(\mathcal P(X))|=2^{2^\kappa}.
$$
Thus the <number of ultrafilters on an infinite set> is
$$
\boxed{|\operatorname{Ult}(X)|=2^{2^{|X|}}.}
$$
Back to article page
