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.

  • $\begingroup$ Define what the subscripts mean. $\endgroup$ – William Elliot Sep 5 at 2:22
  • $\begingroup$ 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 $\mathcal{M}$. $\endgroup$ – Charles Hudgins Sep 5 at 2:43
  • $\begingroup$ @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. $\endgroup$ – Theo Bendit Sep 5 at 2:47
  • $\begingroup$ That's what I assumed. Though, as I'm contemplating it more, it seems like it couldn't possibly be true. $\endgroup$ – Charles Hudgins Sep 5 at 2:48
  • 1
    $\begingroup$ @CharlesHudgins No, it's not the topology per se, because we only consider countable unions. $\endgroup$ – 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.


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