Sequences of sets property 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.
 A: When asked what meaning one should give to $\lim$, the OP mentioned one could assume $\lim \limits_{n\to\infty}A_n=\bigcup \limits_{n=1}^\infty A_n$. So the equality to prove is $\bigcup \limits_{n=1}^\infty A_n=A_1\cup\bigcup \limits_{n=2}^\infty(A_{n}\backslash A_{n-1})$.

In this section I provide a possible definition of limit of a sequence of sets and prove its limit is what the OP suggested. Skip it if you just wish to see the equality above proved.
Definition: Given a set $\mathcal A$ and a sequence $(A_n)_{n\in \mathbb N}$ of subsets of $\mathcal A$ and $A\subseteq X$, it is said $(A_n)_{n\in \mathbb N}$ converges to $A$ and it is denoted by $\lim \limits_{n\to +\infty}\left(A_n\right)=A$, if, and only if,


*

*$\forall x\in A\exists p\in \mathbb N\forall n\in \mathbb N\left(n\ge p\implies x\in A_n\right)$

*$\forall x\not \in A\exists p\in \mathbb N\forall n\in \mathbb N\left(n\ge p\implies x\not \in A_n\right)$


Proposition: The limit of a sequence of sets $(A_n)_{n\in \mathbb N}$, when it exists, is unique. That is, if there exist sets $A$ and $B$ such that $$\lim \limits_{n\to +\infty}(A_n)=A\land \lim \limits_{n\to +\infty}(A_n)=B,$$ then $A=B.$
Proof: Suppose the antecedent and assume $A\setminus B\neq \varnothing$. Take $x\in A\setminus B$. 
Since $x\in A$, there exists $p\in \mathbb N$ such that $\forall n\in \mathbb N\left(n\ge p\implies x\in A_n\right)$.
Since $x\not \in B$, there exists $q\in \mathbb N$ such that $\forall n\in \mathbb N\left(n\ge q\implies x\not \in A_n\right)$.
Thus $\forall n\in \mathbb N\left(n\ge \max\left(p,q\right)\implies (x\in A_n\land x\not \in A_n)\right)$, which is a contradiciton.
Therefore $A\setminus B=\varnothing$ and similarly $B\setminus A=\varnothing$, hence $A=B$.$\,\square$
Proposition: If $(A_n)_{n\in \mathbb N}$ is an increasing sequence of sets in a set $\mathcal A$, then $\lim \limits_{n\to +\infty}(A_n)=\bigcup \limits_{n\in \mathbb N}(A_n)$.
Proof: Set $A:=\bigcup \limits_{n\in \mathbb N}(A_n)$.


*

*Let $a\in A$. By definition of $A$ there exists $p\in \mathbb N$ such that $a\in A_p\color{grey}{\subseteq A_{p+1}\subseteq A_{p+2}\ldots}$. Since $(A_n)_{n\in \mathbb N}$ is increasing it follows that $\forall n\in \mathbb N(n\ge p\implies a\in A_n)$.

*Let $a\not \in A$. By definition of $\forall n\in \mathbb N(a\not \in A_n)$, thus taking $p=1$ yields $\forall n\in \mathbb N(n\ge p\implies a\not \in A_n)$.


Therefore $\lim \limits_{n\to +\infty}(A_n)=\bigcup \limits_{n\in \mathbb N}(A_n)$.$\, \square$

The inclusion $\supseteq$ is just following the definition.
For the other one, let $x\in \bigcup \limits_{n=1}^\infty A_n$. There exists a minimal $m\in \mathbb N$ such that $\forall n\in \mathbb N(n\ge m\implies x\in A_n)$.
Assume $m\ge 2$, (so $m-1\ge 1$. The case $m=1$ is trivial).
Suppose, hoping to find a contradiction, that $x\not \in A_1\cup\bigcup \limits_{n=2}^\infty(A_{n}\backslash A_{n-1})$.
Then $x\not \in A_1$ and $x\not \in \bigcup \limits_{n=2}^\infty(A_{n}\backslash A_{n-1})$, in particular $x\not \in A_m\setminus A_{m-1}$. But since $x\in A_{m}$, this means $x\in A_{m-1}$, contradicting the minimality of $m$.
A: 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.
