Nested countably infinite union of countably infinite intersection of sets

Question

I came across an interesting question on intersections and unions which I am having difficulty solving. I am having some trouble with it as I have not worked a lot with infinite intersections/unions of sets.

Here goes :-

"There is a countably infinite family of indexed sets , $A_i$ , (where $i$ is a positive integer). A set $B$ is given such that :- $$ B = \bigcup_{n=1}^\infty \bigcap_{i=n}^\infty A_i $$ "

We are asked as to whether a) elements of $B$ are the ones belonging to a single $A_i$ or b) to a finite number of $A_i$ or c) to all but a finite number of $A_i$ (there were other possibilities given in it too but I don't recall the complete question)

My Attempt :-

Consider $\bigcap_{i=1}^\infty A_i$ , lets say it is $b_1$. The set $b_1$ contains elements which are common to all the $A_i$s. The set $b_2 = \bigcap_{i=2}^\infty A_i$ contains elements common to all sets except $A_1$. The union $b_1 \bigcup b_2$ would be the same as $b_2$ , as $b_1$ is a subset of $b_2$. Thus this union consists of elements common to all sets except $A_1$.

Similarly,the union $b_1 \bigcup b_2 \bigcup b_3$ would consist of elements common to all sets except $A_1$ and $A_2$.

On generalisation $\bigcup_{n=1}^M \bigcap_{i=n}^\infty A_i$ would consist of elements common to all sets except $A_1 ,A_2, A_3, ..... A_{M-1}$ .

From here onwards I am facing problem. I am able to understand the outer union as long it ranges from $ n=1$ to $n=M$ . But how to handle the case when it ranges from $n=1$ to $n=\infty$ .

(If the union and intersection were done over a finite collection of sets $B = \bigcup_{n=1}^N \bigcap_{i=n}^N A_i$, $B$ would equal the set $A_N$ . But I am totally confused as the range of the union and intersection is a countably infinite collection of sets )

Thanks.

Answer 1

Using your notation, set $b_n=\cap_{i=n}^\infty A_i$. Then, if $x\in B=\cup_{n=1}^\infty b_n$, we must have $x\in b_k$ for some $k\in\mathbb{N}$. Thus $x\in\cap_{i=k}^\infty A_i$, and hence $x\notin A_i$ for at most finitely many $i$. $x$ could be in some of the $A_m$ for $1\leq m\leq k$, however!

Answer 2

A subset $I$ of the natural integers has a finite complement if and only if there exists $n$ such that $k\in I$ for every $k\geqslant n$ (both implications seem relatively clear). Thus, the correct answer is option (c). (And $B$ is called the liminf of the sequence $(A_n)_{n\geqslant1}$, usually denoted $B=\liminf A_n$.)