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Why does $\bigcap_{m = 1}^\infty ( \bigcup_{n = m}^\infty A_n)$ mean the limit superior of sequence of set?

I'm not getting it. ${A_n}$ is a sequence of set in $S$. I do know what limsup means for a bounded sequence; however I am totally baffled at the sequence of set. Can anyone help?

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    This is called the lim sup of sets, and there's an explanation of it here: http://math.stackexchange.com/questions/107931/lim-sup-and-lim-inf-of-sequence-of-sets – Zach Effman May 24 '15 at 02:12

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Consider the statement $x \in \bigcap_{m=1}^\infty \bigcup_{n=m}^\infty A_n$. This means exactly that $x \in \bigcup_{n=m}^\infty A_n$ for every $m$. (Here I have expanded the definition of the intersection.) This means exactly that for every $m$, $x \in A_n$ for some $n \geq m$. (Here I have expanded the definition of the union.) This means exactly that $x$ is in infinitely many of the $A_n$.

Ian
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  • Thanks to the poor connectivity, I couldn't contact you. Sir, what I've asked is that why this means limsup of sequence of set. –  May 25 '15 at 11:58
  • @user36790 One view is that the indicator function of your set is the pointwise limsup (in the sense of real numbers) of the indicator functions of the $A_n$. There is a corresponding view based on the subset partial ordering. – Ian May 25 '15 at 13:40