# Inclusion for pairwise disjoint sets and $\limsup A_n = \emptyset$

Spin-off from here.

1 Please give an example of how the following does not hold for a collection that is not pairwise disjoint.

$$\bigcup_{k \ge n+1} A_k = A\setminus (A_1 \cup\cdots \cup A_n)$$

2 Does anyone have any references for $\limsup A_n = \emptyset$ if $(A_n)_{n=1}^{\infty}$ is a pairwise disjoint collection?

• What have you tried? Did you try to use the definition of $\limsup A_n$ for $2$...?
– Pedro
Jul 6, 2014 at 23:07
• To anyone with the same question: You can also try any ascending or descending sequence of sets. That might do the trick
– BCLC
Jul 11, 2014 at 3:29

a.) take $A_{i}=[\frac{-1}{i},\frac{1}{i}]$. We see that $$\bigcup_{i=1}^{\infty}A_{i}=[-1,1]=A$$ From this we have that $$\bigcup_{k\geq n+1}^{\infty}A_{k}=[\frac{-1}{n+1},\frac{1}{n+1}]$$ but [ $$A\backslash(A_{1}\cup...\cup A_{n})=A\backslash A=\emptyset$$ which is do to fact $A_{1}=[-1,1]$
Or consider any ascending or descending set. That is, $A_n \subseteq A_{n+1} or A_n \supseteq A_{n+1} \forall n \in \mathbb{N}$. I like @user159813's answer better though.