Moore-Smith Convergence - Universal nets (J. Kelley)

Question

I'm reading General Topology by J.Kelley and working through exercises at the end of each chapter.
This is Ex. J (c) from Ch 2. I found solution to exact problem here, but I don't see how the set constructed in point $1$ contains the net frequently - it seems to be eventually in the set, moreover I already started my own proof.

I tried to construct a set $\mathcal C$ which satisfies conditions of Lemma below, but need some advice on how to construct a set which satisfies the first point.

A net in a set $X$ is said to be universal iff for each subset $A$ of $X$ the net is eventually in $A$ or eventually in $X\setminus A$.

$(c)$ Lemma; If $S$ is a net in $X$, then there is a family $\mathcal C$ of subsets of $X$ s.t:

• $S$ is frequently in each member of $\mathcal C$;
• $\forall B,C \in \mathcal C \quad \left(B\cap C \right) \in \mathcal C$; and
• for each subset $A\subseteq X$ either $A \in \mathcal C$ or $(X\setminus A) \in \mathcal C$.

For intersection, in point nr. 2, let $\{D, \geq\}$ be the ordering on domain of $S$,
and let $A$ be a set s.t. for any $n$ let $S_n$ be in some member of $A$. And there is $m$ s.t. for every element of $A$, let $S_m$ be included in it.
$A =\{ \mathcal a | \forall n\in D\;\; \exists a\in \mathcal a:\; S_n \in a\text{, and } \quad \exists m\in D:\; \forall a\in \mathcal a S_m \in a \}$.
Then $\forall a,b \in A \;\; (a\cap b) \in A$.

Now for point nr. 1, my hypothesis is to construct $C$ s.t.
remove from $A$ any one element, then for some period of ordering in $D$: $m \geq \ldots \geq n$, remove each $S_n$.

Answer 1

The idea of the proof is to consider the poset of all "good" families of subsets (when the net $S: D \to X$ is fixed throughout).

A family $\mathcal{C}$ of subsets is "good" when

1. $\forall C \in \mathcal{C}$: $S$ is frequently in $C$.
2. $\mathcal{C}$ is closed under binary (hence finite) intersections.

Well, a very simple "good" family is $\mathcal{C}=\{X\}$ which clearly obeys both properties. In fact you can always add $X$ to any "good" family and it will remain so. Then you need to give an argument that this poset (partially ordered by inclusion) also obeys the condition for Zorn's lemma. (This is actually quite easy: the union of a chain of "good" families is again "good" and an upperbound). So there is a maximal "good" family and then you give an argument (as @Brian M.Scott did in the linked post) that such a maximal "good" family obeys your desired property 3. So that then shows the existence of a family of subsets obeying 1-3 from which you then construct a universal net (again see linked post).

So you don't really "construct" such a family, it "magically" follows from Zorn (or AC).