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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

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    $\begingroup$ Looks good. But you need not go through both directions, one direction is enough, because it shows that $A^+\subseteq\tilde A^+$. Since $A$ is just $\tilde A$ with finitely many $\tilde A$'s changed, this directions already tells you that $\tilde A^+\subseteq A^+$. $\endgroup$ – Stefan Hamcke Oct 23 '13 at 14:52
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    $\begingroup$ That's because you can just consider $\tilde A$ the original family and $A$ the new family. If $\tilde A$ is $A$ with finitely many sets replaced, then $A$ is $\tilde A$ with finitely many sets replaced. $\endgroup$ – Stefan Hamcke Oct 23 '13 at 15:09
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    $\begingroup$ It is the same argument with the roles reversed :-) $\endgroup$ – Stefan Hamcke Oct 23 '13 at 15:11
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    $\begingroup$ I want to add the proof for $A^-=\tilde{A}^-$. Assume $x\in A^-. \Rightarrow\exists~n\in\mathbb{N}: x\in\bigcap_{k=n}^{\infty}A_k\Rightarrow\exists~n\in\mathbb{N}: x\in A_k~\forall~k\geq n\Rightarrow\exists j>\max\limits_{i_l\in I, 1\leq l\leq m}\left\{i_l\right\}: x\in\tilde{A}_k~\forall~k\geq j\Rightarrow x\in\bigcap_{k=j}^{\infty}\tilde{A}_k\Rightarrow x\in \tilde{A}^-$ - Right? $\endgroup$ – math12 Oct 23 '13 at 15:42
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    $\begingroup$ That proof is correct. $\endgroup$ – Stefan Hamcke Oct 23 '13 at 15:53
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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.

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