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can anybody come up with a concrete example for the following statement (i.e., an example where $\mathcal{M}_{\sigma\delta}\neq\mathcal{M}_{\delta\sigma}$):

"In general, $\mathcal{M}_{\delta\sigma}\neq \mathcal{M}_{\sigma\delta}$, where $\mathcal{M}\subset 2^{\Omega}$, for some set $\Omega$, and $\mathcal{M}_{\delta\sigma}:=\{\bigcap_{i\in \mathbb{N}}\bigcup_{j\in \mathbb{N}}M_{i,j}\mid M_{i,j}\in \mathcal{M}, \forall i,j\}$ and $\mathcal{M}_{\sigma\delta}:=\{\bigcup_{i\in \mathbb{N}}\bigcap_{j\in \mathbb{N}}M_{i,j}\mid M_{i,j}\in \mathcal{M}, \forall i,j\}$."

many thanks for your help!!!

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Are you looking for an example where $\cal M_{\delta\sigma}=M_{\sigma\delta}$ or where the equality does not hold? –  Asaf Karagila Jun 2 '12 at 12:26
    
where the equality does NOT hold. –  s_2 Jun 2 '12 at 12:52

1 Answer 1

I can't think of a simple example, so here's a more complicated (but "real-life") one.

Take $\Omega = \mathbb{R}$, and let $\mathcal{M}$ be the collection of all open subsets of $\mathbb{R}$. Since any union of open sets is open, $\mathcal{M}_{\delta \sigma}$ consists of all the countable intersections of open sets, better known as the $G_\delta$ sets. The set of rationals, $\mathbb{Q}$, is not $G_\delta$; this follows from the Baire category theorem. However, $\mathbb{Q}$ is $G_{\sigma \delta}$, i.e. a countable union of countable intersections of open sets. This is easy to see: enumerate the rationals as $q_1, q_2, \dots$, and let $M_{i,j}$ be the open interval $(q_i - \frac{1}{j}, q_i + \frac{1}{j})$. Then $\bigcap_{j \in \mathbb{N}} M_{i,j} = \{q_i\}$ so $\bigcup_{i \in \mathbb{N}} \bigcap_{j \in \mathbb{N}} M_{i,j} = \mathbb{Q}$. Thus we have $\mathbb{Q} \in \mathcal{M}_{\sigma \delta} \setminus \mathcal{M}_{\delta \sigma}$.

If there is an elementary example I'd like to see it.

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I too was really hoping to avoid the Baire category theorem, but no luck... –  Qiaochu Yuan Jun 2 '12 at 15:22
    
Thanks a lot, @NateEldredge, excellent answer. –  s_2 Jun 4 '12 at 7:27

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