lim sup and lim inf of sequence of sets.

I was wondering if someone would be so kind to provide a very simple explanation of lim sup and lim inf of s sequence of sets. For a sequence of subsets A_n of a set X, the \limsup A_n= \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N} A_n \right) and \liminf A_n = \bigcup_{N=1}^\infty \left(\bigcap_{n \ge N} A_n\right). But I am having a hard time imagining what that really means unions of intersections and intersections of unions I think maybe causing the trouble. I read the version on Wikipedia but that didn’t resolve this either.
Any help would be much appreciated.

Answer

A member of

\bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n

is a member of at least one of the sets

\bigcap_{n\ge N} A_n,

meaning it’s a member of either A_1\cap A_2 \cap A_3 \cap \cdots or A_2\cap A_3 \cap A_4 \cap \cdots or A_3\cap A_4 \cap A_5 \cap \cdots or A_4\cap A_5 \cap A_6 \cap \cdots or \ldots etc. That means it’s a member of all except finitely many of the A.

A member of

\bigcap_{N=1}^\infty \bigcup_{n\ge N} A_n

is a member of all of the sets

\bigcup_{n\ge N} A_n,

so it’s a member of A_1\cup A_2 \cup A_3 \cup \cdots and of A_2\cup A_3 \cup A_4 \cup \cdots and of A_3\cup A_4 \cup A_5 \cup \cdots and of A_4\cup A_5 \cup A_6 \cup \cdots and of \ldots etc. That means no matter how far down the sequence you go, it’s a member of at least one of the sets that come later. That means it’s a member of infinitely many of them, but there might also be infinitely many that it does not belong to.

Attribution
Source : Link , Question Author : Comic Book Guy , Answer Author : Michael Hardy

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