Suppose $(A_n)_{n\in\Bbb N}$ is a sequence of sets. Determine the relationship between the sets $\lim\limits_{n\to\infty}\inf(A_0\triangle\ldots\triangle A_n)$ and $\bigcup\limits_{n\in\Bbb N}\left((A_0\triangle\ldots\triangle A_n)\setminus\bigcup\limits_{k\ge n+1} A_k\right).$
My attempt:
Let $L:=\lim\limits_{n\to\infty}\inf(A_0\triangle\ldots\triangle A_n)\overset{\text{by definition }}{=}\bigcup\limits_{n\in\Bbb N}\bigcap\limits_{k\ge n}(A_0\triangle\ldots\triangle A_k)$ and $R:=\bigcup\limits_{n\in\Bbb N}\left((A_0\triangle\ldots\triangle A_n)\setminus\bigcup\limits_{k\ge n+1} A_k\right).$
First, a few observations:
- $x\in L\implies\exists n_0\in\Bbb N\setminus\{0\}, x\in A_0\triangle\ldots\triangle A_k,\forall k\ge n_0$
- $x\in R\implies\exists n_0\in\Bbb N\setminus\{0\}, x\in A_0,\triangle\ldots\triangle A_{n_0}\land x\notin\bigcup\limits_{k>n}A_k$
- $A_0\triangle\ldots\triangle A_{n+k}=(A_0\triangle\ldots\triangle A_n)\setminus(A_{n+1}\triangle\ldots\triangle A_{n+k})\cup(A_{n+1}\triangle\ldots\triangle A_{n+k})\setminus(A_0\triangle\ldots\triangle A_n)$ for $k,n\in\Bbb N\setminus\{0\}.$
- $A_1\triangle\ldots\triangle A_n\subseteq\bigcup_{k\in\Bbb N,\\k\le n}A_k,\forall n\in\Bbb N\\\implies S\setminus\bigcup\limits_{k\in\Bbb N,k\le n}A_k\subseteq S\setminus(A_0\triangle\ldots\triangle A_n),\forall n\in\Bbb N\text{ and any set }S.$
- $(A_0\triangle\ldots\triangle A_n)\setminus\bigcup_{n>k}A_k\subseteq (A_0\triangle\ldots\triangle A_n)\setminus A_{n+1}\subseteq A_0\triangle\ldots\triangle A_n\triangle A_{n+1}.$
I tried to use some of the above conclusions to prove $R\subseteq L.$
Suppose $x\in R.$ Then, there is some $n_0\in\Bbb N$ such that $x\in A_0\triangle\ldots\triangle A_{n_0}$ and $x\not\in\bigcup_{k>n_0}A_k,$ but $A_{n+1}\triangle\ldots\triangle A_{n+k}\triangle\ldots\subseteq\bigcup_{k>n}A_k,$ so $(A_0\triangle\ldots\triangle A_{n_0})\setminus\bigcup_{k>n}A_k\subseteq (A_0\triangle\ldots\triangle A_{n_0})\setminus(A_{n_0+1}\triangle A_{n_0+2}\triangle\ldots)\subseteq (A_0\triangle\ldots A_{n_0})\triangle(A_{n_0+1}\triangle A_{n_0+2}\triangle\ldots),$ hence $x\in L.$ However, I'm not sure about my assertion and I couldn't prove the converse or find a counterexample to $L\subseteq R,$ either.
Any help would be appreciated.
First note that a symmetric difference of $A_0, ..., A_n$ is the set of those $x$ which belong to an odd number of sets $A_0, ..., A_n$. This can be proven by induction. For two sets it's clear. Now, if $x\in A_0\Delta ...\Delta A_n$, then $x$ belongs to precisely one set $A_0\Delta ...\Delta A_{n-1}$ or $A_n$. If $x$ belongs to $A_n$ and not $A_0 \Delta ... A_{n-1}$, then $x$ belongs to even amount of sets from $A_0, ..., A_{n-1}$ and to $A_n$, so it belongs to odd amount of sets among $A_0, ..., A_n$. Otherwise, if $x\in A_0\Delta ...\Delta A_{n-1}$ but $x\notin A_n$, then $x$ belongs to an odd amount of sets among $A_0, ..., A_n$. Conversely, if $x$ belongs to odd amount of sets from $A_0, ..., A_n$ then either of the two cases must hold. So $x\in A_0\Delta ...\Delta A_n$.
If $x\in \liminf_n (A_0\Delta ...\Delta A_n)$, then there is $N$ such that $x\in A_0\Delta ...\Delta A_n$ for $n\geq N$. Then $x$ belongs to odd amount of $A_0, ..., A_N$ as well as odd amount of $A_0, ..., A_{N+1}$. By induction, $x\notin A_n$ for all $n\geq N+1$. This proves $x\in A_0\Delta ...\Delta A_N\setminus \left(\bigcup_{n\geq N+1} A_n\right)$.
Conversely, if $x\in A_0\Delta ...\Delta A_N\setminus \left(\bigcup_{n\geq N+1} A_n\right)$, then $x$ belongs to an odd number of $A_0, ..., A_N$ but $x\notin A_n$ for all $n\geq N+1$. In particular, $x$ belongs to an odd number of $A_0, ..., A_n$ for all $n\geq N$. That is, $x\in\liminf_n (A_0\Delta ...\Delta A_n)$. This proves $$\liminf_n (A_0\Delta ...\Delta A_n) = \bigcup_{n\in\mathbb{N}} \left( A_0\Delta ...\Delta A_n \setminus \left(\bigcup_{k\geq n+1} A_k\right)\right).$$
P.S. I'm sorry for writing a wrong answer before, it should be correct now.