This is a follow-up to my previous question, which has a negative answer. Here is the most general version that I'm interested:
Does there exist a Boolean algebra $A$, an infinite cardinal $\kappa$, and a multi-valued map $f$ from $A$ to $\mathcal{P}(\kappa)$ (i.e., a map $f: A \to \mathcal{P}(\mathcal{P}(\kappa))$), s.t.,
- $f(0) = \{\varnothing\}$, $f(1) = \{\kappa\}$;
- $f(a)$ is nonempty for any $a \in A$;
- $f$ is "weakly monotone", i.e., for any $a,b \in A$, any $S_a \in f(a)$ and $S_b \in f(b)$, there exists $S \in f(a \wedge b)$ s.t. $S \subset S_a \cap S_b$ (note that this is exactly saying $f$ is monotone if it happens to be single-valued);
- $f$ is "compact", i.e., for any $a_1, \cdots, a_n \in A$ with $\bigvee_{i=1}^n a_i = 1$, and for any $S_i \in f(a_i)$, we have $\bigcup_{i=1}^n S_i = \kappa$;
- $f$ is "free", i.e., for any ultrafilter $\mathcal{U}$ on $A$, we have $\bigcap_{a \in \mathcal{U}} \bigcap_{S \in f(a)} S = \varnothing$.
Note that condition $1$ is somewhat redundant: $f(1) = \{\kappa\}$ follows from condition $4$. While $f(0) = \{\varnothing\}$ does not seem to follow from other conditions, if one is given an $f$ satisfying all the other conditions, simply modifying it so that $f(0) = \{\varnothing\}$ yields a new map that satisfies all the given conditions.
Since the above is quite esoteric and likely looks unmotivated, let me mention the precise reason I want such a map $f$. Again, this is not relevant to the question itself. It only serves to provide a motivation for the conditions required on $A$, $\kappa$, and $f$.
As noted in my previous question, this relates to a topology question, which asks whether every weakly locally compact, semiregular space is Baire. It is known this is true if semiregular is upgraded to regular, and since locally compact Hausdorff space is regular, there can be no Hausdorff counterexample.
In the MSE post, I wrote an answer which provides a $T_0$ counterexample. The approach is to simply take a semiregular, non-Baire space, say $X = \mathbb{Q}$, and adjoin a single point whose only neighborhood is the entire space. As long as $X$ does not have a point whose only neighborhood (in $X$) is $X$, this yields a compact, semiregular space which is not Baire. As long as $X$ is $T_0$, the result space is $T_0$ as well. However, this approach cannot yield a $T_1$ counterexample. The newly adjoined point, $\infty$, needs to have a local basis of regular open sets, and, for the entire space to be compact, those regular open sets must have compact complements. If the new space is $T_1$, this is enough to imply $X$ is nearly weakly locally compact itself (to be precise, it has an open dense subset $Y$ s.t. every point in $Y$ has a compact neighborhood in $X$). So, in a sense, if $X \cup \{\infty\}$ is a $T_1$ counterexample, $X$ is likely very close to being a $T_1$ counterexample itself. At the very least, $X = \mathbb{Q}$, or any other $T_1$, semiregular, non-Baire space I can think of, does not satisfy the requirements.
The same issue arises if I try to adjoin finitely many points to compactify $X$. This is where I began to think of adjoining an entire compact Hausdorff space to compactify $X$. The specific idea, related to the current question, is as follows:
Suppose $A$, $\kappa$, and $f$, as required in the question, exist. Then take $\mathbb{Q} \times \kappa$ equipped with the product topology (where $\mathbb{Q}$ is equipped with its usual topology and $\kappa$ is equipped with the discrete topology). Let the Stone space of $A$ be $S(A)$. Now take $X = (\mathbb{Q} \times \kappa) \sqcup S(A)$, in which $\mathbb{Q} \times \kappa$ is open, and, for any $\mathcal{U} \in S(A)$, a local basis is given by $(\mathbb{Q} \times S) \sqcup V_a$, where $a \in \mathcal{U}$, $V_a = \{\mathcal{V} \in S(A): a \in \mathcal{V}\}$, and $S \in f(a)$. Condition $3$ ensures that this is a well-defined topology. Condition $4$ ensures $X$ is compact. And condition $5$ ensures $X$ is $T_1$.
