PlanetMath: number of ultrafilters

Proof. If

is finite then each ultrafilter on

is principal, and so there are exactly $\vert X\vert$ ultrafilters. In the rest of the proof we will assume that

is infinite.

Let be the set of all finite subsets of , and let $\Phi$ be the set of all finite subsets of .

For each $A\subseteq X$ define $B_A=\{(f,\phi)\in F\times\Phi\mid A\cap f\in\phi\}$ , and $B_A^\complement=(F\times\Phi)\setminus B_A$ . For each $% latex2html id marker 360 $ {\cal S}\subseteq\mathcal{P}(X)$$ define ${\cal B}_{\cal S}= \{B_A\mid A\in{\cal S}\}\cup\{B_A^\complement\mid A\notin{\cal S}\}$ .

Let $% latex2html id marker 364 $ {\cal S}\subseteq\mathcal{P}(X)$$ , and suppose $A_1,\dots,A_m\in\cal S$ and $% latex2html id marker 368 $ D_1,\dots,D_n\in\mathcal{P}(X)\setminus\cal S$$ , so that we have $B_{A_1},\dots,B_{A_m},B_{D_1}^\complement,\dots, B_{D_n}^\complement\!\!\!\in{\cal B}_{\cal S}$ . For $i\in\{1,\dots,m\}$ and $j\in\{1,\dots,n\}$ let $a_{i,j}$ be such that either $a_{i,j}\in A_i\setminus D_j$ or $a_{i,j}\in D_j\setminus A_i$ . This is always possible, since $A_i\ne D_j$ . Let $f=\{a_{i,j}\mid1\le i\le m,\,1\le j\le n\}$ , and put $\phi=\{A_i\cap f\mid1\le i\le m\}$ . Then $(f,\phi)\in B_{A_i}$ , for $i=1,\dots,m$ . If for some $j\in\{1,\dots,n\}$ we have $D_j\cap f\in\phi$ , then $D_j\cap f=A_i\cap f$ for some $i\in\{1,\dots,m\}$ , which is impossible, as $a_{i,j}$ is in one of these sets but not the other. So $D_j\cap f\notin\phi$ , and therefore $(f,\phi)\in B_{D_j}^\complement$ . So $(f,\phi)\in B_{A_1}\cap\cdots\cap B_{A_m}\cap B_{D_1}^\complement\cap\cdots\cap B_{D_n}^\complement$ . This shows that any finite subset of ${\cal B}_{\cal S}$ has nonempty intersection, and therefore ${\cal B}_{\cal S}$ can be extended to an ultrafilter ${\cal U}_{\cal S}$ .

Suppose $% latex2html id marker 414 $ {\cal R},{\cal S}\subseteq\mathcal{P}(X)$$ are distinct. Then, relabelling if necessary, ${\cal R}\setminus {\cal S}$ is nonempty. Let $A\in{\cal R}\setminus {\cal S}$ . Then $B_A\in{\cal U}_{\cal R}$ , but $B_A\notin{\cal U}_{\cal S}$ since $B_A^\complement\in{\cal U}_{\cal S}$ . So ${\cal U}_{\cal R}$ and ${\cal U}_{\cal S}$ are distinct for distinct $\cal R$ and $\cal S$ . Therefore $% latex2html id marker 434 $ \{{\cal U}_{\cal S}\mid{\cal S}\subseteq\mathcal{P}(X)\}$$ is a set of $2^{2^{\vert X\vert}}$ ultrafilters on $F\times\Phi$ . But $\vert F\times\Phi\vert=\vert X\vert$ , so $F\times\Phi$ has the same number of ultrafilters as . So there are at least $2^{2^{\vert X\vert}}$ ultrafilters on , and there cannot be more than $2^{2^{\vert X\vert}}$ as each ultrafilter is an element of $% latex2html id marker 452 $ \mathcal{P}(\mathcal{P}(X))$$ . $\qedsymbol$