Let $(A_n)_{n\in\mathbb{N}}$ be a series of sets. Define $$ A^+:=\limsup\limits_{n\to\infty}A_n:=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k,~~~~~A^-:=\liminf\limits_{n\to\infty}A_n:=\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_k $$ Show that $A^+$ and $A^-$ are not changing, when one changes finite many sets $A_i$.
Hello, I tell you my idea concerning $A^+$ and would be very thankful to get a feedback from you if my idea is right or wrong, thanks!
Consider the changed series $(\tilde{A}_n)_{n\in\mathbb{N}}$ with $\tilde{A}_n\neq A_n$ for a finite set $I=\left\{i_1,i_2,\ldots,i_m\right\}, m\in\mathbb{N}$ and $\tilde{A}_n=A_n$ when $n\notin I$. What is to show is $$ A^+=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}\tilde{A}_k=:\tilde{A}^+. $$
To show the conclusion "$\subseteq$", suppose that $x\in A^+$. $$ \Rightarrow\forall~n\in\mathbb{N}: x\in\bigcup_{k=n}^{\infty}A_k\Rightarrow\forall~n\in\mathbb{N}~\exists~j\in\left\{n,n+1,n+2,\ldots\right\}: x\in A_j\\\Rightarrow\forall~n\in\mathbb{N}~\exists~s>\max\limits_{i_l\in I, 1\leq l\leq m}\left\{i_l\right\}: x\in \tilde{A}_s (=A_s)\\\Rightarrow\forall~n\in\mathbb{N}: x\in\bigcup_{k=n}^{\infty}\tilde{A}_k\Rightarrow x\in\tilde{A}^+ $$ To show the other conclusion, assume that $x\in\tilde{A}^+$. $$ \Rightarrow\forall~n\in\mathbb{N}: x\in\bigcup_{k=n}^{\infty}\tilde{A}_k\Rightarrow\forall~n\in\mathbb{N}~\exists~j\in\left\{n,n+1,n+2,\ldots\right\}: x\in\tilde{A}_j\\\Rightarrow\forall~n\in\mathbb{N}~\exists s>\max\limits_{i_l\in I, 1\leq l\leq m}\left\{i_l\right\}: x\in\tilde{A}_s=A_s\\\Rightarrow~\forall~n\in\mathbb{N}: x\in\bigcup_{k=n}^{\infty}A_k\Rightarrow x\in A^+ $$
With kind regards,
math12
An easier thing to do is to note that $x \in \limsup A_n$ if and only if $x$ is in infinitely many of the $A_n$, and $x \in \liminf A_n$ if and only if $x$ is in all but finitely many of the $A_n$. From these descriptions, it is trivial to see that changing finitely many of the $A_n$ does not affect either the limsup or liminf.