# How many ultrafilters there are in an infinite space?

I'm stuck with the next exercise from the book Rings of Continuous Functions by Gillman.

If $$X$$ is infinite, there exist $$2^{2^{|X|}}$$ ultrafilters on $$X$$ all of whose members are of cardinal $$X$$.

The exercise have a hint based on the next proof (here $$\beta X$$ is the Stone–Čech compactification)

In the proof, the author constructs $$2^{2^{X}}$$ distinct ultrafilters on $$X$$. The hint of the exercise says

In the proof of Theorem 9.2, observe that every finite intersection of members of $$\mathfrak{B}_{\mathscr{S}}$$ is of cardinal $$|X|$$. Adjoin to each family $$\mathfrak{B}_{\mathscr{S}}$$ all subsets of $$\mathscr{F}\times\Phi$$ with complement of power less than $$|X|$$.

I'm stuck in the two parts of the hint. I don't know how can I prove that every finite intersection of elements of $$\mathfrak{B}_{\mathscr{S}}$$ is of cardinal $$|X|$$. I only know that because $$\mathfrak{b}_{S_{i}}\subseteq \mathscr{F}\times \Phi$$ and $$-\mathfrak{b}_{S_{j}}\subseteq \mathscr{F}\times \Phi$$ then $$|\mathfrak{b}_{S_{i}}|\leq|X|$$ and then $$|-\mathfrak{b}_{S_{j}}|\leq |X|$$. Therefore $$|\mathfrak{b}_{S_1}\cap\mathfrak{b}_{S_2}\cap\dots\cap,\mathfrak{b}_{S_k}\cap-\mathfrak{b}_{S_{k+1}}\cap\dots\cap-\mathfrak{b}_{S_n}|\leq|\mathfrak{b}_{S_1}|\leq|X|$$But, how can I conclude the another inequality?, i.e., $$|\mathfrak{b}_{S_1}\cap\mathfrak{b}_{S_2}\cap\dots\cap,\mathfrak{b}_{S_k}\cap-\mathfrak{b}_{S_{k+1}}\cap\dots\cap-\mathfrak{b}_{S_n}|\geq|X|$$And, how can it helps to consider the subsets of $$\mathscr{F}\times\Phi$$ with complement of power less than $$|X|$$? I think the approach I've taken is so hard or there are something that I can't see because the proof looks so hard for me.

To show that all finite intersections of sets in $$\mathfrak{B}_{\mathscr{S}}$$ have cardinality $$|X|$$ it suffices just to construct $$|X|$$-many elements in the intersection. (This is because, as you have noticed, there cannot be more than $$|X|$$-many elements in the intersection.)

In the proof given, we have one particular element of this intersection: $$( F = \{ x_{ij} : i \neq j \} , \varphi = \{ F \cap S_1 , \ldots , F \cap S_k \} ).$$ Suppose that $$( F , \psi ) \in \mathscr{F} \times \Phi$$ is such that $$\phi \supseteq \varphi$$ is finite. Given any $$i \leq k$$ note that we clearly have that $$S_i \cap F \in \psi$$, and so $$( F , \psi ) \in \mathfrak{B}_{S_i}$$. Given $$k < j \leq n$$ note that $$( F , \psi ) \in - \mathfrak{b}_{S_j}$$ as long as $$S_j \cap F \notin \psi$$. Therefore as long as $$\psi \supseteq \phi$$ is chosen so that $$F \cap S_{k+1} , \ldots , F \cap S_{n} \notin \psi$$, then $$( F , \psi )$$ will belong to the intersection. There are clearly $$|X|$$-many ways to choose appropriate $$\psi$$.

Let $$\mathfrak{B} = \{ \mathscr{A} \subseteq \mathscr{F} \times \Phi : | ( \mathscr{F} \times \Phi ) \setminus \mathscr{A} | < |X| \}$$ denote the family of all subsets of $$\mathscr{F} \times \Phi$$ with complement of power $$< |X|$$. Note that not only does $$\mathfrak{B}$$ have the finite intersection property, it is actually closed under finite intersections.

With this observation and the work above it becomes relatively easy to show that given $$\mathscr{S} \subseteq \mathcal{P} ( X )$$ the family $$\mathfrak{B}_{\mathscr{S}} \cup \mathfrak{B}$$ has the finite intersection property. To see this, suppose that $$\mathfrak{b}_{S_1} , \ldots , \mathfrak{b}_{S_k} , - \mathfrak{b}_{S_{k+1}} , \ldots , - \mathfrak{b}_{S_n} , \mathscr{A}_1 , \ldots , \mathscr{A}_m$$ are given. Then

• by the work above the set $$\mathfrak{b} = \mathfrak{b}_{S_1} \cap \cdots \cap \mathfrak{b}_{S_k} \cap - \mathfrak{b}_{S_{k+1}} \cap \cdots \cap - \mathfrak{b}_{S_n}$$ has power $$|X|$$, and
• by the observation above the complement of $$\mathscr{A} = \mathscr{A}_1 \cap \cdots \cap \mathscr{A}_m$$ has power $$< |X|$$.

Thus $$\mathfrak{b} \cap \mathscr{A} \neq \emptyset$$.

Therefore this family can be extended to an ultrafilter $$\mathfrak{U}_{\mathscr{S}}$$, and since $$\mathfrak{B} \subseteq \mathfrak{U}_{\mathscr{S}}$$, we know that $$\mathfrak{U}_{\mathscr{S}}$$ cannot include any sets of power $$< |X|$$.