help with the proof - regular points of a union of sets I have such a proof from a leture which I don't understand.
$M$ - complex manifold.
Let $\left\{ A_j\right\}_{j\in J}$ be a nonempty, locally finite family of analytic subset of $M$, such that for $i\neq j$ we have that $A_i\cap A_j$ is nowheredense both in $A_i$ and $A_j$. 
Then $$\text{Reg}\left(\bigcup_{j\in J}A_j\right)=\left(\bigcup_{j\in J}\text{Reg}A_j\right)\setminus\left( \bigcup_{j\neq i}A_j\cup A_i \right). $$
Proof.
Fix $j_0 \in J, \ a\in A_{j_0}$
I've done the case when $\forall j\neq j_0$ we have that $a\notin A_j$.
And now I'm trying to do the second case, namely:
$\exists j_1\neq j_0$ such that $a\in A_i\cap A_j$. Then by assumptions germs $\left(A_{j_1}\right)\neq\left(A_{j_0}\right)$. Due to local finiteness the germ (set germ) $\left(\bigcup_{j\in J} A_j\right)_a$ decomposes into finitely many germs:
$$\left(\bigcup_{j\in J}A_j\right)_a=\left(A_{j_1}\right)_a\cup\ldots\cup\left(A_{j_r}\right)_a .$$
Why is this the end of the proof? How it exatcly works?
 A: it's my first time when I answer my own question. I don't know if I'm doing it right, but I have managed above problem, so since there are not any answers I'll put one.
Generally, problem is parted into two cases:
Fix $j_0$ and let $a\in A_{j_0}$
In the first case we concider a situation when $a\in A_{j_0} $ and we have not any sets $A_j$ such that element $a$ belogns to $A_j$. Then, as I've already written, we can separate such a set $A_{j_0}$ from the rest, namely from $\bigcup_j A_j$. From the definition, if $a\in A_{j_0}$, then $a\in \bigcup_j A_j$. On the other hand if $a\in\bigcup_j A_j$ and we know that it belongs only into $A_{j_0}$, then, clearly, $a\in A_{j_0}$ and it ends the first case.
More explanations are needed in the second case.
Let us suppose that there exists such a $j_1$ that $a\in A_{j_1}\cap A_{j_0}$ (of course we are intuitively concidering now a case when $a$ belongs to some $A_j$'s).
Then again we can separate those sets from the rest. Since an intersection of any pair $A_j\cap A_i$ is nowheredense both in $A_i$ and $A_j$, we get from the definition that germs in $a$ of those sets are different, namely  $(A_{j_1})_a\neq (A_{j_0})_a$. The family is locally finite so if we take those $A_j$'s which $a$ belongs into, we can write a germ of a sum as a sum of germs"
$$(\bigcup_j A_j)_a =(A_{j_1})_a\cup\ldots\cup (A_{j_r})_a,$$
but we know, that on the right hand side we have at least two different germs, which implies that the germ of a sum is not irreducible. This fact gives us that such a germ is not smooth, which means that $a$ doesn't belong to the regular part of a set.
Finally, we've showed that the sum is smooth in $a$ if and only if $a$ is the regular point of some $A_{j_0}$ and there are not any $j$'s such that $a$ belongs to the intersection $A_j\cap A_{j_0}$.
