limsup of sets - proof

Question

Suppose I have a sequence of sets $A_n$. How to prove that elements of the set $\limsup A_n$ defined as $$\limsup A_n = \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N} A_n \right)$$

appear in infinitely many $A_n$s?

Proving that they are in all $\bigcup_{n\ge N} A_n$ is easy, it follows from the definition of set intersection directly. How to conclude that they appear in infinitely many $A_n$?

I'd do it by contradiction - suppose the elements of above set appear in finitely many sets $A_n$, let's say in $k$ sets. We order those sets so that they become the first $k$ sets in the sequence of $A_n$s (is it always possible to order a finite number of sets? I guess so). Knowing they appear in all sets $\bigcup_{n\ge N} A_n$ (it means for all $N$) we conclude they belong to $\bigcup_{n\ge k+1} A_n$ as well, so there's at least one set containing the elements that's not among $k$ sets - a contradiction.

Could anyone verify my proof? I'd like to make it as formal as possible.

Answer 1

I think you've got the right idea, but I don't think the proof is adequately justified: as mentioned by Henning Malkhom, you need to show that changing the order of the $A_n$'s does not change the set $\limsup A_n$.

Luckily, you do not need to change the order of the sets. If you suppose, by contradiction, that there exists $x \in \limsup A_n$ such that the set $K(x) := \{n : x \in A_n\}$ is finite, we can take $k = \max K(x)$ (which exists because $K(x)$ is a finite set of natural numbers) and conclude that $$x \notin \bigcup_{n \geq k+1} A_n \subset \limsup A_n,$$ which contradicts the fact that $x \in \limsup A_n$.

Answer 2

For brevity, let $B_k=\cup_{n\geq k}A_n.$ Let $S=\{x :\sup \{n\in N : x\in A_n\}=\infty\}.$ We show that $\forall x\in S\;(x\in \lim \sup A_n)$ and we show that $\forall x\not \in S\; (x\not \in \lim \sup A_n).$ For the first part, by the def'n of $S$ we have $$x\in S\implies (\quad\forall k\in N\;\exists j\geq k\;(x\in A_j)\quad )\implies$$ $$\implies \forall k\in N\;(x\in B_k)\implies$$ $$\implies x\in \cap_{n\in N}B_k=\lim \sup A_n.$$ For the second part, by the def'n of $S$ we have $$x\not \in S\implies \exists k\in N\;\forall j\geq k\;(x\not \in A_j)\implies$$ $$\implies \exists k\in N\;(x\not \in B_k)\implies x\not \in \cap_{k\in N}B_k=\lim \sup A_n.$$ We could shorten this because every "$\implies$" in the first part can be replaced by "$\iff$", which makes the second part redundant, but also makes the proof harder to follow. It is often easier to discover a proof of $p=q$ in two parts : Show $p\subset q$, then show $q\subset p .$