limit superior of union and intersection of sets.

Question

How can be proved these statments?

$lim sup (A_n cap B_n) = lim sup (A_n) cap lim sup (B_n) $

$lim sup (A_n cup B_n) = lim sup (A_n) cup lim sup (B_n) $

My attempt:

Let $xinlim sup (A_n cup B_n)$, then $xin bigcap_{N=1}^infty left( bigcup_{nge N} A_ncup B_n right)$. Hence, $xin bigcap_{N=1}^infty left( bigcup_{nge N}A_nright)$ or $xin bigcap_{N=1}^infty left( bigcup_{nge N}B_nright)$. So $xinlim sup (A_n) cap lim sup (B_n)$. $lim sup (A_n cup B_n)subseteqlim sup (A_n) cup lim sup (B_n) $

Let $xin lim sup (A_n) cup lim sup (B_n)$, then  $xin bigcap_{N=1}^infty left( bigcup_{nge N}A_nright)$ or $xin bigcap_{N=1}^infty left( bigcup_{nge N}B_nright)$. Hence, $xin bigcap_{N=1}^infty left( bigcup_{nge N} A_ncup B_n right)$. So $lim sup (A_n) cup lim sup (B_n)subseteq\lim sup (A_n cup B_n)$
It is correct?

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

Kavi Rama Murthy · 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  $A_ncap B_n$ is empty for each $n$ so LHS is empty. But $lim sup A_n=lim sup B_n=Acup B$.  So RHS is $A cup B$.

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.

limit superior of union and intersection of sets.

One Answer

Add your own answers!

Ask a Question