For any sets $A_n,n\in\mathbb{N}$ consider $$ A^+:=\limsup_{n\to\infty}A_n:=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k,~~~~~E_m:=\bigcup_{n\geq m}A_n. $$ Show that the sets $E_m, m\geq 1$ can be written as a disjoint union $$ E_m=A^+\uplus\biguplus_{n\geq m}(E_n\setminus E_{n+1}). $$

I do not have a working idea. I started with writing $E_m$ as a disjoint union, i.e. $$ E_m=A_m\uplus\biguplus_{i=m+1}^{\infty}A_i\setminus\bigcup_{j=m}^{i-1}A_j=A_m\uplus\biguplus_{i=m+1}^{\infty}\bigcap_{j=m}^{i-1}A_i\setminus A_j $$

and additionally I see that $$ A^+=\bigcap_{n=1}^{\infty}E_n. $$ But I do not know if this is helpful...

Would be great to get a help resp. answer.

With kind regards,



Hint: For $x \in E_m$ consider $$\sup_{A_k \ni x} k.$$ Show that if this supremum is infinite, then $x \in A^{+}$. Otherwise, denote this supremum by $n$ and show that $x \in E_n \setminus E_{n+1}$.

  • $\begingroup$ Thank you! I write an answer to my question (see below), trying to use your hint, because I have not enough place in a comment to do it in a comment. Please have a look if my usage is correct. $\endgroup$ – math12 Oct 25 '13 at 8:55

I try to use the hint of njguliyev to create an own answer to my question. It would be really great if you had a look on this and gave me a feedback whether I am right or not.

I have to show two inclusions.

(1) "$\subseteq$":

Let $x\in E_m=\bigcup_{n\geq m}A_n$. Consider $$ q:=\sup_{n\geq m}\left\{n|x\in A_n\right\}. $$ First case: $q<\infty$

Then $\exists~k>q: x\notin A_l, l\geq k$, so $x\notin\bigcup_{j\geq k}A_j$ and therefore $x\notin\bigcap_{n\geq 1}\bigcup_{s\geq n}A_s=A^+$. But it is $$ x\in A_q\subset\bigcup_{j\geq q}A_j\wedge x\notin\bigcup_{i\geq q+1}A_i\\\Rightarrow x\in\bigcup_{j\geq q}A_j\setminus\bigcup_{i\geq q+1}A_i=E_q\setminus E_{q+1}\subset\biguplus_{n\geq m}(E_n\setminus E_{n+1}). $$ So, to sum this up, $x\notin A^+$ and $x\in\biguplus_{n\geq m}(E_n\setminus E_{n+1})$, which means $$x\in A^+\uplus\biguplus_{n\geq m}(E_n\setminus E_{n+1}). $$ Second case: $q=\infty$

Then $x\in A_n~\forall~n\geq m$, so $x\in\bigcup_{k=n}^{\infty}A_k~\forall~n\in\mathbb{N}$ and therefore $x\in\bigcap_{n\geq 1}\bigcup_{k\geq n}A_k=A^+$. Furthermore it is $$ x\in\bigcup_{j\geq n}A_j\wedge x\in\bigcup_{i\geq n+1}A_i~\forall~n\in\mathbb{N}\\\Rightarrow\not\exists~n\geq m: x\notin E_n\setminus E_{n+1}\\\Rightarrow x\notin\biguplus_{n\geq m}(E_n\setminus E_{n+1}) $$

(2) "$\supseteq$":

Consider $x\in A^+\uplus\biguplus_{n\geq m}(E_n\setminus E_{n+1})$.

First case: $x\in A^+$ $$ x\in A^+\Rightarrow x\in\bigcap_{n\geq 1}\bigcup_{k\geq n}A_k\Rightarrow x\in\bigcup_{k\geq n}A_k~\forall~n\geq 1\Rightarrow x\in\bigcup_{k\geq m}A_k=E_m $$

Second case: $x\in\biguplus_{n\geq m}(E_n\setminus E_{n+1})$ $$ x\in\biguplus_{n\geq m}(E_n\setminus E_{n+1})\Rightarrow\exists~j\geq m: x\in E_j\setminus E_{j+1}=\bigcup_{s\geq j}A_s\setminus\bigcup_{t\geq j+1}A_t\\\Rightarrow x\in\bigcup_{s\geq j}A_s\\\Rightarrow\exists~w\geq j\geq m: x\in A_w\\\Rightarrow x\in\bigcup_{n\geq m}A_n=E_m $$

To my opinion, all is shown and so $\Box$.


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