TITLE: Proof Verification, family of sets
QUESTION [1 upvotes]: Given $\left\{ A_i \right\}_{i\in\mathbb{N}}$ a family of non empty sets, $A$ a non empty set such that $\forall i\, \in \mathbb{N}, A_i\cap A\neq\emptyset$. The family $\left\{B_j \right\}_{j\in\mathbb{N}}$ is defined by: $$B_1= A \cap A_1 \quad \wedge \quad\forall k\geq 2,\; B_k = A\; \cap \;\left(A_k-\bigcup_\limits{i=1}^{k-1}A_i\right). $$ Prove that $\bigcup_\limits{j\in\mathbb{N}}{B_j}=A\cap\bigcup_\limits{i\in\mathbb{N}}{A_i}$.

I'm trying to complete the proof but i don't know exactly how to proceed: 

First we want to prove that $\bigcup_\limits{j\in\mathbb{N}}{B_j} \subseteq A\cap\bigcup_\limits{i\in\mathbb{N}}{A_i}, \; \;$so let $x\in\bigcup_\limits{j\in\mathbb{N}}{B_j} $, then $x\in A\cap A_1 $ or $x\in A\; \cap \;\left(A_k-\bigcup_\limits{i=1}^{k-1}A_i\right)$ for some $j\geq2$. There are two cases, if $x\in A \cap A_1,\;$  then $x\in A \cap \bigcup_\limits{i\in\mathbb{N}}{A_i}$. 
In the other case $x\in A\; \cap \;\left(A_k-\bigcup_\limits{i=1}^{k-1}A_i\right)$, thereupon we can say that $x\in A \,\cap A_k$, so $x\in A \cap \bigcup_\limits{i\in\mathbb{N}}{A_i}$. Then we proved the first inclusion.
Now we want to prove that $A\cap\bigcup_\limits{i\in\mathbb{N}}{A_i}\subseteq \bigcup_\limits{j\in\mathbb{N}}{B_j}, \; \;$ so let $x \in A\cap\bigcup_\limits{i\in\mathbb{N}}{A_i}$, then $x\in A\,\cap A_1$ or $x\in A\;\cap\;\bigcup_\limits{i\in\mathbb{N}}A_i$ or $\exists j\in \mathbb{N}$ such that $x\in A$...

I'm stuck on this part, but i want to know if there's some shorter way to prove it.
thanks in advance.

REPLY [1 votes]: To prove 
$$A\cap\bigcup_\limits{i\in\mathbb{N}}{A_i}\subseteq \bigcup_\limits{j\in\mathbb{N}}{B_j}$$
Take $x \in A\cap\bigcup_\limits{i\in\mathbb{N}}{A_i}$. By hypothesis, $x \in A$ and it exists $j \in \mathbb N$ such that $x \in A_j$. Denote by $j_0$ the smallest such $j$. Then $x \in B_{j_0}$ and you’re done.
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