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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.

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1  
How are you defining $\lim$ here? –  Git Gud Jan 11 at 20:27
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In this case $\lim_{n\to\infty}A_n=\bigcup_{n=1}^\infty A_n$. –  Integral Jan 11 at 22:05

3 Answers 3

up vote 4 down vote accepted

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,

  1. $\forall a\in A\exists p\in \mathbb N\forall n\in \mathbb N\left(n\ge p\implies x\in A_n\right)$
  2. $\forall a\not \in A\exists p\in \mathbb N\forall n\in \mathbb N\left(n\ge p\implies x\not \in A_n\right)$

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 $A:=\bigcup \limits_{n\in \mathbb N}(A_n)$.

  1. 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 $\forall n\in \mathbb N(n\ge p\implies a\in A_n)$.
  2. 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 $m$.

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I always find it easier to communicate this definition of a limit by saying that $A$ is the limit of the $A_n$ if the characteristic functions of $A_n$ converge pointwise to that of $A$. Of course, it is not really easier, but it reduces it to terms that many know already. –  Carsten Schultz Jan 12 at 22:21
    
@CarstenSchultz Thank you for your comment. It being there should be useful enough to whoever might prefer that approach. –  Git Gud Jan 12 at 22:24

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 no 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.

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This looks to me like you should simply use 'Mathematical Induction':

i) $$A_2=A_1 \cup (A_2\setminus A_1)$$ $$A_3=A_1 \cup (A_3\setminus A_2) \cup (A_2\setminus A_1)$$

Both are true! Since for every element in $A_2$ that excluded because it was in $A_1$ it will be added again when we add $A_1$, same for $A_3$

From this logic, we will now construct our induction:

ii) $n\rightarrow n+1$

$$A_{n+1} = A_1\cup\bigcup_{n=2}^{n+1}(A_{n+1}\backslash A_{n})$$

$$A_{n+1} = A_1\cup(A_{n+1}\backslash A_{n})\cup\bigcup_{n=2}^{n}(A_{n}\backslash A_{n-1}) = A_1\cup(A_{n+1}\backslash A_{n})\cup A_n$$

Now with the same reasoning we used in i), we can say: for every element that we remove from $A_{n+1}$ when we substract $A_n$, we will add the same element when we merge the set again with $A_n$!

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I don't think you can conclude some limiting result from induction, which give you results only for finite $n$. At least, not in this case. –  Integral Jan 11 at 22:06
    
Well, the normal induction may not be supposed for this, but I still think this is solid. But you are right, it's an edge case that deserves some further thinking. –  Nils Ziehn Jan 12 at 0:47
    
It's easy to find properties that are true for all $n$, which fail when limits are involved. A trivial example is $\dfrac 1{n+1}<\dfrac{1}{n}$ which at $+\infty$ becomes an equality. –  Git Gud Jan 12 at 4:16
    
Well, this is the nature of in equalities, that there may be an edge case in the limit! –  Nils Ziehn Jan 13 at 3:32

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