Proof that $A_1=\bigcap_{i=1}^{\infty}A_i \cup \bigcup_{i=1}^{\infty}(A_i\backslash A_{i+1})$ Let $A_{i+1} \subseteq A_i$, $F_{i}=A_i\backslash A_{i+1}$ which implies $A_i=A_{i+1}\cup F_i$ with $A_{i+1} \cap F_i = \varnothing$.
Proof that $$A_1=\bigcap_{i=1}^{\infty}A_i \cup \bigcup_{i=1}^{\infty}(A_i\backslash A_{i+1}) \cdots (*)$$.
I have tried some ways.
First, I tried this way
\begin{align*}
\bigcap_{i=1}^{\infty}A_i \cup \bigcup_{i=1}^{\infty}(A_i\backslash A_{i+1})
&=(A_1 \cap A_2 \cap A_3 \cap \cdots) \cup ((A_1\backslash A_2)\cup (A_2\backslash A_3)\cup (A_3\backslash A_4) \cup \cdots)\\
&=(A_1 \cap A_2 \cap A_3 \cap \cdots) \cup ((A_1\cap {A_2}^c)\cup (A_2\cap {A_3}^c)\cup (A_3\cap {A_4}^c) \cup \cdots)\\
&=(A_1 \cap A_1 \cap {A_2}^c) \cup (A_2 \cap A_2 \cap {A_3}^c) \cup (A_3 \cap A_3 \cap {A_4}^c) \cup \cdots
\end{align*}
then I didn't know what must i do next and I tried the second way, I tried to expand each terms like this
\begin{align*}
\bigcap_{i=1}^{\infty}A_i&=A_1 \cap A_2 \cap A_3 \cap \cdots\\
\bigcup_{i=1}^{\infty}(A_{i}\cap (A_{i+1})^c)&=(A_1 \cap {A_2}^c) \cup(A_2 \cap {A_3}^c) \cup (A_3 \cap {A_4}^c) \cup \cdots\\
&=A_1 \cap ({A_2}^c \cup A_2) \cap ({A_3}^c \cup A_3) \cap ({A_4}^c \cup A_4) \cap \cdots\\
&=A_1 \cap \cdots \text{(i didn't know what's the next step)}
\end{align*}
Is it right if I think ${A_i}^c \cup A_i=A_1$ for $i=2,3,4,\cdots$? If it is right, I get this result from my second way:
\begin{align*}
\bigcup_{i=1}^{\infty}(A_{i}\cap (A_{i+1})^c)&=(A_1 \cap {A_2}^c) \cup(A_2 \cap {A_3}^c) \cup (A_3 \cap {A_4}^c) \cup \cdots\\
&=A_1 \cap ({A_2}^c \cup A_2) \cap ({A_3}^c \cup A_3) \cap ({A_4}^c \cup A_4) \cap \cdots\\
&=A_1 \cap A_1 \cap A_1 \cap \cdots\\
&=A_1
\end{align*}
then
\begin{align*}
\bigcap_{i=1}^{\infty}A_i \cup \bigcup_{i=1}^{\infty}(A_{i}\cap (A_{i+1})^c) &= (A_1 \cap A_2 \cap A_3 \cap \cdots) \cup A_1\\
&= (A_1 \cap A_1) \cup (A_2 \cap A_1) \cup (A_3 \cap A_1) \cup \cdots\\
&= A_1 \cup \varnothing \cup \varnothing \cup \cdots\\
&= A_1
\end{align*}
But, i was not sure about that way. So, how to proof equation  (*)? Thanks for any help.
 A: Let $B=\bigcap_{i=1}^{\infty}A_i $ and $C= \bigcup_{i=1}^{\infty}(A_i\backslash A_{i+1})$.
Intuitivelly, $B$ is the "limit" of the decreasing sequence $A_i$, that is the elements that are in all $A_i$. And $C$ contains the elements of $A_1$ that are removed at some step, between $A_i$ and $A_{i+1}$.
Let $x\in A_1$. Then either $x\in A_i$ for all $i$, and then $x\in B$. Or for some $i$, $x\in A_i$ and $x\notin A_{i+1}$, and then $x\in C$. Hence $A_1\subseteq B\cup C$. The converse is obvious, as $A_i\subseteq A_1$ for all $i$.
A: ("$\subset$"): Let $x\in A_1$. There are two cases:


*

*$x\in\bigcap_{i=1}^\infty A_1$. In this case the conclusion holds.

*$x\notin\bigcap_{i=1}^\infty A_1$. In the second case, there exists some $k$ s.t. $x\notin A_k$ (clearly $k>1$). We assume $k$ is the smallest such that $x\notin A_k$. Then $x\in A_{k-1}\backslash A_k\subset\bigcup_{i=1}^\infty(A_i\backslash A_{i+1})$.
Either way, $x\in\bigcap_{i=1}^\infty A_1\cup\bigcup_{i=1}^\infty(A_i\backslash A_{i+1})$.
("$\supset$"): Let $x\in\bigcap_{i=1}^\infty A_1\cup\bigcup_{i=1}^\infty(A_i\backslash A_{i+1})$. Then there are two cases:


*

*$x\in\bigcap_{i=1}^\infty A_i$. Clearly, $x\in A_1$.

*$x\in\bigcup_{i=1}^\infty(A_i\backslash A_{i+1})$. Then $x\in A_k\backslash A_{k+1}$ for some $k\geq1$. Since $A_{i+1}\subset A_i$, we have $x\in A_k\subset A_{k-1}\subset\cdots\subset A_1$.
Either way, $x\in A_1$. Hence, $\bigcap_{i=1}^\infty A_1\cup\bigcup_{i=1}^\infty(A_i\backslash A_{i+1})\subset A_1$
