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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!!!

  • 0
    Are you looking for an example where $\cal M_{\delta\sigma}=M_{\sigma\delta}$ or where the equality does not hold?2012-06-02
  • 0
    where the equality does NOT hold.2012-06-02

1 Answers 1