TITLE: Sequences of sets property
QUESTION [8 upvotes]: I'm having trouble to prove the following question:
Supose $\{A_n\}_{n\in\mathbb{N}}$ is a family of sets such that $A_1\subset A_2\subset A_3\subset\ldots$ (it's possible to have $A_n=A_{n+1}$). I need to prove that $$\lim_{n\to\infty}A_n = A_1\cup\bigcup_{n=2}^\infty(A_{n}\backslash A_{n-1}).$$
I'm not sure if this is useful, but I proved that $A_{n}\backslash A_{n-1}$ and $A_{n'}\backslash A_{n'-1}$ are disjoint if $n\neq n'$.
Thank you.

REPLY [4 votes]: Define $A_0=\varnothing$, keep your $A_1,A_2,\ldots$ and put $B_n=A_n-A_{n-1}$ for $n=1,2,\ldots$ You want to show that $$\tag 1 \bigcup_{n\geqslant 1}A_n=\bigcup_{n\geqslant 1}B_n$$
Since $B_n\subseteq A_n$ for each $n$; one inclusion is clear, namely that  $$\bigcup_{n\geqslant 1}A_n\supseteq \bigcup_{n\geqslant 1}B_n$$
Now, suppose that $x\in \displaystyle\bigcup_{n\geqslant 1}A_n$. Then for some $n=1,2,\ldots$ we have that $x\in A_n$. If $x\in A_1$, then $x\in B_1=A_1$. If not, $x\in A_n$ for some $n=2,3,\ldots$. If $x\in A_2$; then since $x\notin A_1$; $x\in A_2-A_1=B_2$. If $x\notin A_2$, then $x\in A_n$ for some $n=3,4,\ldots$. Continuing, I claim that the process must end for some $n$. If not, we would obtain that $x\notin A_n$ for every $n=1,2,\ldots$, which contradicts that $x\in \displaystyle\bigcup_{n\geqslant 1}A_n$. Thus   $$\bigcup_{n\geqslant 1}A_n\subseteq \bigcup_{n\geqslant 1}B_n$$ and $(1)$ is proven.
ADD The above can be modified even when the $A_n$ are not increasing, namely, let $B_n=A_n\smallsetminus (A_1\cup\cdots \cup A_{n-1})$. Then $B_n\cap B_k=\varnothing$ if $n\neq k$ and $\bigcup A=\bigcup B$ exactly by the reasoning above.