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I need to show that $\limsup_n \liminf_k A_n \cap A_k^c=\phi$. Thus

$\bigcap_n\bigcup_{r\geq n} \bigcup_k \bigcap_{m\geq k} A_r\cap A_m^c=\phi$?


I am trying to show that $\lim_n P(\liminf_k A_n \cap A_k^c)=0$. I need the above step and then $\limsup_n P\leq P(\limsup_n)$.

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  • $\begingroup$ @AsafKaragila I have added (elementary-set-theory) per recommendation from the limsup tag-wiki. $\endgroup$ – Martin Sleziak Feb 6 '15 at 7:44
  • $\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. $\endgroup$ – Martin Sleziak Feb 6 '15 at 7:56
  • $\begingroup$ I am trying to show that $\lim_n P(\liminf_k A_n \cap A_k^c)=0$. I need the above step and then $\limsup_n P\leq P(\limsup_n)$. $\endgroup$ – Meemo Feb 6 '15 at 10:17
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If I did not make a mistake somewhere, you could do it like this. (Basically just using distributivity and de Morgan laws.) $$\bigcap_n\bigcup_{r\geq n} \bigcup_k \bigcap_{m\geq k} (A_r\cap A_m^c) = \bigcap_n\bigcup_{r\geq n} \left(A_r \cap \left(\bigcup_k \bigcap_{m\geq k} A_m^c\right)\right) = \bigcap_n\bigcup_{r\geq n} \left(A_r \cap \left(\bigcap_k \bigcup_{m\geq k} A_m\right)^c\right) = \left(\bigcap_n\bigcup_{r\geq n} A_r \right) \cap \left(\bigcap_k \bigcup_{m\geq k} A_m\right)^c = \limsup A_n \cap (\limsup A_n)^c = \emptyset$$

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