Krdytkyu

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 $mathscr{F}timesPhi$ 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 $$|mathfrak{b}_{S_1}capmathfrak{b}_{S_2}capdotscap,mathfrak{b}_{S_k}cap-mathfrak{b}_{S_{k+1}}capdotscap-mathfrak{b}_{S_n}|leq|mathfrak{b}_{S_1}|leq|X|$$But, how can I conclude the another inequality?, i.e., $$|mathfrak{b}_{S_1}capmathfrak{b}_{S_2}capdotscap,mathfrak{b}_{S_k}cap-mathfrak{b}_{S_{k+1}}capdotscap-mathfrak{b}_{S_n}|geq|X|$$And, how can it helps to consider the subsets of $mathscr{F}timesPhi$ 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 $< |X|$.