Limit superior of union and intersection of sets.

limitslimsup-and-liminf

How can be proved these statments?

  1. $\lim sup (A_n \cap B_n) = \lim sup (A_n) \cap \lim sup (B_n) $

  2. $\lim sup (A_n \cup B_n) = \lim sup (A_n) \cup \lim sup (B_n) $

My attempt:

  1. Let $x\in\lim sup (A_n \cup B_n)$, then $x\in \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N} A_n\cup B_n \right)$. Hence, $x\in \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N}A_n\right)$ or $x\in \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N}B_n\right)$. So $x\in\lim sup (A_n) \cap \lim sup (B_n)$. $\lim sup (A_n \cup B_n)\subseteq\lim sup (A_n) \cup \lim sup (B_n) $

Let $x\in \lim sup (A_n) \cup \lim sup (B_n)$, then $x\in \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N}A_n\right)$ or $x\in \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N}B_n\right)$. Hence, $x\in \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N} A_n\cup B_n \right)$. So $\lim sup (A_n) \cup \lim sup (B_n)\subseteq\\lim sup (A_n \cup B_n)$

It is correct?

  1. Can be proved the second statement? Is it false? Why?

Best Answer

Actually 1) is wrong.

Let $A$ and $B$ be disjoint non-empty sets, $A_n=A$ for $n$ odd , $A_n=B$ for $n$ even and $B_n=A$ for $n$ even , $B_n=B$ for $n$ odd. Then $A_n\cap B_n$ is empty for each $n$ so LHS is empty. But $\lim \sup A_n=\lim \sup B_n=A\cup B$. So RHS is $A \cup B$.

  1. is true. Clearly RHS is contained in LHS. If $x \notin $ RHS the there exists $N$ such that $x $ is neither in $A$ nor in $B$ for $n>N$. Hence $x \notin $LHS.
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