Question about proof for intersection of a set family with union of a set I am studying this proof and there are a few things I need help understanding.
Let $A$ be a set and $B_i$, for $i \in I$ be a family of sets
Prove
$ A \cup (\cap_{i \in I}B_i)$=$\cap_{i \in I}(A \cup B_i)$
proof: suppose $x \in A$ then $x \in (A \cup B_i)$ for all $i \in I$
Also if $x \in \cap_{i \in I}B_i$, $x \in (A \cup B_i)$ for all $i \in I$
thus $x \in \cap_{i \in I}(A \cup B_i)$.
Can someone please explain to me what's in the set  $ \cap_{i \in I}(A \cup B_i)$.
Is it the intersection of $A$ included with the intersection of all the $B_i's$ 
or is it the intersection of all the $B_i's$ with the union of A
In other words is it the intersection of B included with the intersection of the family
Or is it the intersection of the family added with the set A??
I don't get most why the set A can be incorporated in the parenthesis with the intersection of the family!
It just doesn't seem right to me, I need to know if the notation $\cap_{i \in I}B_i$, the intersection of all these sets  is restricted to the family.
 A: 
I don't get most why the set A can be incorporated in the parenthesis with the intersection of the family!

How you state the question: "Let $A$ be a set and $B_i$, for $i\in I$ be a family of sets", it is not clear, why you can make 'sense' out of $A\cup B_i$ or other set operations. I suppose, that $A$ and $B_i$ are subsets of the same set $X$. Else you can not really compare these sets in general, and the results might be trivial.
So you might check that first. 
$\bigcap_{i\in I} (A\cup B_i)$ is the intersection of $(A\cup B_i)$ for every $i\in I$.
For example: Take $I=\{1,2,3\}$ And let $A, B_1,B_2, B_3\subseteq X$ with $X=\{1,2,3\}$. And $A=\{1,2,3\}, B_1=\{1\}, B_2=\{1,2\}, B_3=\{1,2,3\}$, then:
$\bigcap_{i=1}^3 (A\cup B_i)=(A\cup B_1)\cap (A\cup B_2)\cap (A\cup B_3)=A$.
The example given is not really insightful, but I just wanted to show, how this intersection works.
A: $x\in\bigcap_{i\in I}(A\cup B_i)$ exactly when $\forall i\in I~[x\in(A\cup B_i)]$. That is $\forall i\in I~[x\in A\lor x\in B_i]$.
In words: For all $i$ in the indices $I$, $x$ is in $A$ or $x$ is in $B_i$.
