# lim sup (liminf) (An\Ak)

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 $\limsup_n P\leq P(\limsup_n)$.

• @AsafKaragila I have added (elementary-set-theory) per recommendation from the limsup tag-wiki. – Martin Sleziak Feb 6 '15 at 7:44
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• 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 $\limsup_n P\leq P(\limsup_n)$. – Meemo Feb 6 '15 at 10:17

If I did not make a mistake somewhere, you could do it like this. (Basically just using distributivity and de Morgan laws.) $$\bigcap_n\bigcup_{r\geq n} \bigcup_k \bigcap_{m\geq k} (A_r\cap A_m^c) = \bigcap_n\bigcup_{r\geq n} \left(A_r \cap \left(\bigcup_k \bigcap_{m\geq k} A_m^c\right)\right) = \bigcap_n\bigcup_{r\geq n} \left(A_r \cap \left(\bigcap_k \bigcup_{m\geq k} A_m\right)^c\right) = \left(\bigcap_n\bigcup_{r\geq n} A_r \right) \cap \left(\bigcap_k \bigcup_{m\geq k} A_m\right)^c = \limsup A_n \cap (\limsup A_n)^c = \emptyset$$