Finite intersections and countable unions

Let $$\mathscr{M}\subseteq 2^\Omega$$ be a family of subsets of $$\Omega$$. Let $$d$$ and $$\sigma$$ be the closures to finite intersections, respectively countable unions. Is it true that $$\mathscr{M}_{d\sigma}=\mathscr{M}_{\sigma d}$$? What if $$\mathscr{M}$$ is countable?

I saw this property used in a proof, and tried to prove or disprove it myself unsuccessfully.

• Define what the subscripts mean. – William Elliot Sep 5 at 2:22
• I would prove this by showing that they are both separately equal to the set generated by $\mathcal{M}$ which is closed under finite intersections and countable unions. Note that if $\emptyset , \Omega \in \mathcal{M}$ this is just the topology generated by $\mathcal{M}$. – Charles Hudgins Sep 5 at 2:43
• @WilliamElliot I'm guessing that $\mathscr{M}_{d\sigma} = (\mathscr{M}_d)_\sigma$, the closure of $\mathscr{M}_d$ under countable unions, where $\mathscr{M}_d$ is the closure of $\mathscr{M}$ under finite intersections. – Theo Bendit Sep 5 at 2:47
• That's what I assumed. Though, as I'm contemplating it more, it seems like it couldn't possibly be true. – Charles Hudgins Sep 5 at 2:48
• @CharlesHudgins No, it's not the topology per se, because we only consider countable unions. – Henno Brandsma Sep 5 at 4:17

I can show that $$\mathscr{M}_{\sigma d} \subseteq \mathscr{M}_{d\sigma}$$, but the sets need not be equal in general.

To prove $$\mathscr{M}_{\sigma d} \subseteq \mathscr{M}_{d\sigma}$$, let $$A \in \mathscr{M}_{\sigma d}$$. Then there exist $$A_1, \ldots, A_n \in \mathscr{M}_\sigma$$ such that $$A = A_1 \cap \ldots \cap A_n.$$ Further, for each $$1 \le i \le n$$, there exists a family $$(A_i^j)_{j=1}^\infty$$ of subsets in $$\mathscr{M}$$ such that $$A_i = \bigcup_{j=1}^\infty A_i^j.$$ (If a finite union is wanted, then simply make the sequence $$(A_i^j)_{j=1}^\infty$$ periodic.)

Let $$\Lambda = \Bbb{N}^n$$, which we will treat as an index set. For $$\lambda \in \Lambda$$, define $$B_\lambda = A^{\lambda_1}_1 \cap \ldots \cap A^{\lambda_n}_n \in \mathscr{M}_d$$. I claim that $$A = \bigcup_{\lambda \in \Lambda} B_\lambda.$$ Note that $$\Lambda$$ is countable, hence this would imply $$A \in \mathscr{M}_{d\sigma}$$ as required.

To prove the claim, suppose $$x \in A$$. Then, $$x \in A_i$$ for all $$1 \le i \le n$$. For each such $$i$$, we know there exists some $$j(i) \in \Bbb{N}$$ such that $$x \in A^{j(i)}_i$$. Simply let $$\lambda = (j(1), \ldots, j(n))$$, and then $$x \in B_\lambda \subseteq \bigcup_{\lambda \in \Lambda} B_\lambda$$.

Conversely, suppose $$x \in \bigcup_{\lambda \in \Lambda} B_\lambda$$. Then there exists some $$\lambda \in \Lambda$$ such that $$x \in A^{\lambda_1}_1 \cap \ldots \cap A^{\lambda_n}_n$$. That is, for each $$i$$, $$x \in A_i^{\lambda_i} \subseteq A_i$$. Thus, $$x \in A_1 \cap \ldots \cap A_n = A$$. This proves the claim, completing the proof that $$\mathscr{M}_{\sigma d} \subseteq \mathscr{M}_{d\sigma}$$.

Now we present a counterexample to show $$\mathscr{M}_{d\sigma} \subseteq \mathscr{M}_{\sigma d}$$ need not be true in general (even when $$\mathscr{M}$$ is countable).

Let $$\Omega = \Bbb{N} = \{1, 2, \ldots\}$$. For $$n \in \Bbb{N}$$, let $$\mathscr{M} = \{p^k \Bbb{N} : p, k \in \Bbb{N} \text{ and p is prime}\}.$$ Note that $$\mathscr{M}_d = \{n \Bbb{N} : n \in \Bbb{N} \setminus \{1\}\}.$$

Let $$C \subseteq \Bbb{N}$$ be the set of composite numbers. Then $$C \in \mathscr{M}_{d\sigma}$$, since we can write $$C = \bigcup_{n \in C} n\Bbb{N}$$. I claim that $$C \notin \mathscr{M}_{\sigma d}$$.

Suppose $$A \in \mathscr{M}_\sigma$$ contains $$C$$. For any distinct primes $$p, q$$, we have $$pq \in C \subseteq A$$. The only sets in $$\mathscr{M}$$ that contain $$pq$$ are $$p\Bbb{N}$$ and $$q\Bbb{N}$$, so one or the other must appear explicitly in the countable union forming $$A$$. Since this is true of every pair of distinct primes, we must have $$p\Bbb{N}$$ appear in the union for all primes $$p$$, except possibly barring one prime $$p_0$$.

So, if a finite intersection of such sets in $$\mathscr{M}_\sigma$$ contained $$C$$, then each set excludes at most one prime. Thus, their intersection excludes a finite number of primes, leaving an infinite number of primes in the intersection. Therefore, no finite intersection could leave $$C$$, as claimed.