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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 $\kappa$. LetFor 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 $B\subseteq \kappa$ defineTo verify independence, take distinct $B_1,\dots ,B_n$ and bits ${\epsilon }_1,\dots ,{\epsilon }_n$. For each pair $i<j$, choose a point of $B_i△B_j$ and let $s$ contain all these finitely many points. The traces $B_i\cap s$ are pairwise distinct. SetThen $(s,H)\in A_{B_i}$ exactly for those $i$ with ${\epsilon }_i=1$. 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 $\epsilon :\mathcal{P}(\kappa )\to \{0,1\}$, prescribe $A_B$ when $\epsilon (B)=1$ and $D\setminus A_B$ when $\epsilon (B)=0$. 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, $U_{\epsilon }$. 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 $\epsilon$.

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 $\mathcal{P}(X)$:Thus the number of ultrafilters on an infinite set is