Lim sup of sequence of sets and theirs unions

Question

I have to prove the following equality:

$$\left(\limsup_{n \to \infty} A_n\right)\cup\left(\limsup_{n \to \infty} B_n\right) = \limsup_{n \to \infty} \left(A_n\cup B_n\right)$$

Can somebody help me to prove this?

Answer 1

If $(C_n,n\geqslant 1)$ is a collection of sets, then $x\in\limsup_n C_n$ means that $I(x):=\{n\geqslant 1,x\in C_n\}$ is infinite.

Since $A_n\subset A_n\cup B_n$ for each $n$, we certainly have $\limsup_nA_n\subset\limsup_n (A_n\cup B_n)$.

For the reverse inclusion, let $x\in \limsup_n (A_n\cup B_n)$. Then $x\in A_n\cup B_n$ for infinitely many $n$. Defining $I:=\{n\geqslant 1,x\in A_n\}$ and $J:=\{n\geqslant 1,x\in B_n\}$, we have that $I\cup J$ is infinite, hence $I$ or $J$ is infinite.

Answer 2

I just answered the question Understanding the supremum limit of a set which has at least some to do with this. It is proven there that $x$ belongs to $\limsup_n A_n$ is equivalent to that $x$ belongs to infinitely many $A_n$'s.

Using this description and defining $C_n=A_n\cup B_n$ one easily checks that if $x$ belongs to $$ \left(\limsup_n A_n\right)\cup\left(\limsup_n B_n\right) $$ then $x$ belongs to either infinitely many $A_n$ or infinitely many $B_n$ and thus to infinitely many $C_n$. This gives the inclusion ''$\subseteq$''.

For the other inclusion consider some $x$ in $\limsup_n C_n$ so that $x$ belongs to infinitely many $C_n$. Then assume for contradiction that $x$ only belongs to finitely many $A_n,B_n$. Then there exist a $k$ so that $x$ does not belong to $A_n$ nor $B_n$ for $n>k$. But this implies that $x$ does not belong to $C_n$ for $n>k$ either which is a contradiction. The inclusion ''$\supseteq$'' follows.